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Chapter 3 Exercise 3.3 Solutions

Matrices

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 3 Exercise 3.3

Class 12 Mathematics Exercise 3.3 NCERT Solutions Matrices Class 12 Mathematics Chapter 3 CBSE Board Exam JEE Main CUET Transpose of Matrix Properties of Transpose Symmetric Matrix Skew-Symmetric Matrix Matrix Algebra Matrix Operations NCERT Class 12 Maths
12 Questions
25–40 min Ideal time
Q1 Now at
Q1
NUMERIC3 marks

Find the transpose of the following matrices:

\[ \begin{aligned} &(i)\begin{bmatrix} 5\\ \dfrac{1}{2}\\ -1 \end{bmatrix}, \qquad (ii)\begin{bmatrix} 1 & 2\\ -1 & 3 \end{bmatrix}, \qquad (iii)\begin{bmatrix} -1 & 5 & 6\\ \sqrt{3} & 5 & 6\\ 2 & 3 & -1 \end{bmatrix} \end{aligned} \]
📘 Concept & Theory
Concept/Theory

The transpose of a matrix is obtained by interchanging its rows and columns.

If \(A=[a_{ij}]\) is an \(m\times n\) matrix, then its transpose \(A^T\) is an \(n\times m\) matrix given by

\[ A^T=[a_{ji}] \]

Thus, the element in the \(i\)-th row and \(j\)-th column of \(A\) becomes the element in the \(j\)-th row and \(i\)-th column of \(A^T\).

In particular:

  • A row matrix becomes a column matrix.
  • A column matrix becomes a row matrix.
  • The order \(m\times n\) changes to \(n\times m\).
  • The main diagonal elements remain in their positions.
  • The transpose operation does not change the numerical values of the entries; it only changes their positions.

The fundamental property used in this question is

\[ (A^T)^T=A \]

Also, if \(A\) is a square matrix, then \(A^T\) has the same order as \(A\).

🗺️ Solution Roadmap
Step-by-step Plan
  1. Identify the order of each given matrix.

  2. Interchange every row with the corresponding column.

  3. Write the resulting matrix carefully, preserving every entry.

  4. Check that the order changes from \(m\times n\) to \(n\times m\).

  5. For a square matrix, verify that the entries across the principal diagonal have been reflected.

✏️ Solution
(i) Transpose of the Column Matrix<
Step-by-step Solution  ·  3 steps
  1. Given
    \[ A= \begin{bmatrix} 5\\ \dfrac{1}{2}\\ -1 \end{bmatrix} \]
    This is a \(3\times1\) column matrix.
    Therefore, its transpose will be a \(1\times3\) row matrix.
  2. On interchanging rows and columns, the first column becomes the first row:
    \[ A^T= \begin{bmatrix} 5 & \dfrac{1}{2} & -1 \end{bmatrix} \]
  3. Hence,
    \[A^T= \begin{bmatrix} 5 & \dfrac{1}{2} & -1 \end{bmatrix}\]
  4. Order check:
    \[ 3\times1\longrightarrow1\times3 \]
✏️ Solution
(ii) Transpose of the \(2\times2\) Matrix
Step-by-step Solution  ·  4 steps
  1. Given — matrix is
    \[ A= \begin{bmatrix} 1 & 2\\ -1 & 3 \end{bmatrix} \]
    The matrix \(A\) is of order \(2\times2\).
  2. Its rows are
    \[ R_1=\begin{bmatrix}1&2\end{bmatrix}, \quad R_2=\begin{bmatrix}-1&3\end{bmatrix} \]
  3. Therefore, the first row becomes the first column of \(A^T\):
    \[ \begin{bmatrix} 1\\ 2 \end{bmatrix} \]
  4. The second row becomes the second column of \(A^T\):
    \[ \begin{bmatrix} -1\\ 3 \end{bmatrix} \]
  5. Hence,
    \[ A^T= \begin{bmatrix} 1 & -1\\ 2 & 3 \end{bmatrix} \]
✏️ Solution
(iii) Transpose of the \(3\times3\) Matrix
Step-by-step Solution  ·  7 steps
  1. Given — matrix is
    \[ B= \begin{bmatrix} -1 & 5 & 6\\ \sqrt{3} & 5 & 6\\ 2 & 3 & -1 \end{bmatrix} \]
    The matrix \(B\) is of order \(3\times3\).
  2. Write its rows as
    \[ R_1= \begin{bmatrix} -1&5&6 \end{bmatrix} \]
    \[ R_2= \begin{bmatrix} \sqrt{3}&5&6 \end{bmatrix} \]
    \[ R_3= \begin{bmatrix} 2&3&-1 \end{bmatrix} \]
  3. On taking the transpose, the first row becomes the first column:
    \[ \begin{bmatrix} -1\\ 5\\ 6 \end{bmatrix} \]
  4. The second row becomes the second column:
    \[ \begin{bmatrix} \sqrt{3}\\ 5\\ 3 \end{bmatrix} \]
  5. The third row becomes the third column:
    \[ \begin{bmatrix} 2\\ 6\\ -1 \end{bmatrix} \]
  6. Combining these columns, we obtain
    \[ B^T= \begin{bmatrix} -1 & \sqrt{3} & 2\\ 5 & 5 & 3\\ 6 & 6 & -1 \end{bmatrix} \]
  7. Order check:
    \[ 3\times3\longrightarrow3\times3 \]
  8. Since \(B\) is square, its transpose has the same order.
💡 Answer
Final Answer
\( { (i)\quad \begin{bmatrix} 5 & \dfrac{1}{2} & -1 \end{bmatrix}} \)

\( { (ii)\quad \begin{bmatrix} 1 & -1\\ 2 & 3 \end{bmatrix}} \)

\( { (iii)\quad \begin{bmatrix} -1 & \sqrt{3} & 2\\ 5 & 5 & 3\\ 6 & 6 & -1 \end{bmatrix}} \)
🎯 Exam Significance
Exam Significance

Transpose of a matrix is a fundamental concept in Class 12 Mathematics and is directly connected with several important topics in the chapter. Questions involving \(A^T\), symmetric matrices, skew-symmetric matrices, and properties of matrix operations frequently test whether a student can correctly interchange rows and columns.

  • Accuracy: It prevents the common mistake of assuming that every square matrix remains unchanged after transposition.
  • Symmetric matrices: The condition \(A^T=A\) is central to identifying symmetric matrices.
  • Skew-symmetric matrices: The condition \(A^T=-A\) is based directly on the transpose operation.
  • Matrix algebra: Transpose properties are extensively used in later problems involving products and scalar multiples of matrices.
  • Short-answer questions: Direct computation of \(A^T\) is a common source of easy marks when row-column interchange is performed carefully.
Significance for Competitive Entrance Examinations

For competitive examinations, transpose questions are often used as building blocks for more advanced matrix identities. A strong command of the operation helps in solving questions involving matrix equations, symmetric and skew-symmetric decomposition, determinants, and matrix products.

  • Remember that \((A^T)^T=A\).
  • Remember that \((AB)^T=B^TA^T\); the order of multiplication reverses.
  • Remember that \((kA)^T=kA^T\), where \(k\) is a scalar.
  • For a symmetric matrix, \(A^T=A\).
  • For a skew-symmetric matrix, \(A^T=-A\).
  • Always verify the order of the resulting matrix before finalising an answer.
🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  10 points
  1. The transpose of a matrix is obtained by interchanging its rows and columns.

  2. If \(A\) is \(m\times n\), then \(A^T\) is \(n\times m\).

  3. A column matrix becomes a row matrix after transposition.

  4. A row matrix becomes a column matrix after transposition.

  5. For a square matrix, transposition reflects the matrix about its principal diagonal.

  6. Diagonal entries remain unchanged during transposition.

  7. A matrix is symmetric if and only if \(A^T=A\).

  8. Do not assume \(A^T=A\) merely because \(A\) is a square matrix.

  9. For the second matrix in this question, the correct transpose is \(\begin{bmatrix}1&-1\2&3\end{bmatrix}\), not the original matrix.

  10. Always perform an order check and an entry-by-entry check after transposition.

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1 / 12  ·  8%
Q2 →
Q2
NUMERIC3 marks

If

\[ A= \begin{bmatrix} -1 & 2 & 3\\ 5 & 7 & 9\\ -2 & 1 & 1 \end{bmatrix}, \qquad B= \begin{bmatrix} -4 & 1 & -5\\ 1 & 2 & 0\\ 1 & 3 & 1 \end{bmatrix} \]

then verify that

\[ \text{(i)}\quad (A+B)^T=A^T+B^T \] \[ \text{(ii)}\quad (A-B)^T=A^T-B^T \]
📘 Concept & Theory
Concept/Theory

This question verifies two fundamental properties of the transpose of matrices.

If \(A\) and \(B\) are matrices of the same order, then

\[ \boxed{(A+B)^T=A^T+B^T} \]

and

\[ \boxed{(A-B)^T=A^T-B^T} \]

These identities mean that the transpose operation distributes over matrix addition and matrix subtraction.

For example, if

\[ A=[a_{ij}],\qquad B=[b_{ij}], \]

then

\[ A+B=[a_{ij}+b_{ij}] \]

and therefore

\[ (A+B)^T=[a_{ji}+b_{ji}] \]

On the other hand,

\[ A^T+B^T=[a_{ji}]+[b_{ji}] =[a_{ji}+b_{ji}] \]

Hence,

\[ (A+B)^T=A^T+B^T \]

Similarly,

\[ (A-B)^T=[a_{ji}-b_{ji}] =A^T-B^T \]

In this question, both \(A\) and \(B\) are \(3\times3\) matrices, so their sum, difference, and transposes are all defined and are also \(3\times3\) matrices.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Write the matrices \(A\) and \(B\).

  2. For part (i), calculate \(A+B\), then take its transpose.

  3. Separately calculate \(A^T\) and \(B^T\), then calculate \(A^T+B^T\).

  4. Compare the two resulting matrices.

  5. For part (ii), calculate \(A-B\), then take its transpose.

  6. Separately calculate \(A^T-B^T\).

  7. Compare the two resulting matrices.

  8. If both pairs of matrices are identical, the required identities are verified.

✏️ Solution
(i) Verification of \((A+B)^T=A^T+B^T\)
Step-by-step Solution  ·  11 steps
  1. Given — Matrices
    \[ A= \begin{bmatrix} -1 & 2 & 3\\ 5 & 7 & 9\\ -2 & 1 & 1 \end{bmatrix} \]
    \[ B= \begin{bmatrix} -4 & 1 & -5\\ 1 & 2 & 0\\ 1 & 3 & 1 \end{bmatrix} \]
    Both matrices are of order \(3\times3\).
  2. Find \(A+B\)
  3. Add the corresponding elements of \(A\) and \(B\):
    \[ A+B= \begin{bmatrix} -1+(-4) & 2+1 & 3+(-5)\\ 5+1 & 7+2 & 9+0\\ -2+1 & 1+3 & 1+1 \end{bmatrix} \]
  4. Now simplify each entry:
    \[ A+B= \begin{bmatrix} -5 & 3 & -2\\ 6 & 9 & 9\\ -1 & 4 & 2 \end{bmatrix} \]
  5. Find \((A+B)^T\)
  6. Interchange the rows and columns:
    \[ (A+B)^T= \begin{bmatrix} -5 & 6 & -1\\ 3 & 9 & 4\\ -2 & 9 & 2 \end{bmatrix} \]
  7. Find \(A^T\)
  8. Taking the transpose of \(A\), the first row becomes the first column, the second row becomes the second column, and the third row becomes the third column.
    \[ A^T= \begin{bmatrix} -1 & 5 & -2\\ 2 & 7 & 1\\ 3 & 9 & 1 \end{bmatrix} \]
  9. Step 4: Find \(B^T\)
  10. \[ B^T= \begin{bmatrix} -4 & 1 & 1\\ 1 & 2 & 3\\ -5 & 0 & 1 \end{bmatrix} \]
  11. Find \(A^T+B^T\)
  12. \[ A^T+B^T= \begin{bmatrix} -1 & 5 & -2\\ 2 & 7 & 1\\ 3 & 9 & 1 \end{bmatrix} + \begin{bmatrix} -4 & 1 & 1\\ 1 & 2 & 3\\ -5 & 0 & 1 \end{bmatrix} \]
  13. Add the corresponding elements:
    \[ A^T+B^T= \begin{bmatrix} -1+(-4) & 5+1 & -2+1\\ 2+1 & 7+2 & 1+3\\ 3+(-5) & 9+0 & 1+1 \end{bmatrix} \]
  14. Hence,
    \[ A^T+B^T= \begin{bmatrix} -5 & 6 & -1\\ 3 & 9 & 4\\ -2 & 9 & 2 \end{bmatrix} \]
  15. Comparing this with \((A+B)^T\),
    \[ (A+B)^T= \begin{bmatrix} -5 & 6 & -1\\ 3 & 9 & 4\\ -2 & 9 & 2 \end{bmatrix} \]
    \[ A^T+B^T= \begin{bmatrix} -5 & 6 & -1\\ 3 & 9 & 4\\ -2 & 9 & 2 \end{bmatrix} \]
  16. Therefore,
    \[ \boxed{(A+B)^T=A^T+B^T} \]
  17. Hence, the first result is verified.
✏️ Solution
(ii) Verification of \((A-B)^T=A^T-B^T\)
Step-by-step Solution  ·  10 steps
  1. Find \(A-B\)
  2. Subtract the corresponding elements of \(B\) from \(A\):
    \[ A-B= \begin{bmatrix} -1-(-4) & 2-1 & 3-(-5)\\ 5-1 & 7-2 & 9-0\\ -2-1 & 1-3 & 1-1 \end{bmatrix} \]
  3. Now simplify each entry carefully:
    \[ -1-(-4)=-1+4=3 \]
    \[ 2-1=1 \]
    \[ 3-(-5)=3+5=8 \]
    \[ 5-1=4 \]
    \[ 7-2=5 \]
    \[ 9-0=9 \]
    \[ -2-1=-3 \]
    \[ 1-3=-2 \]
    \[ 1-1=0 \]
  4. Therefore,
    \[ A-B= \begin{bmatrix} 3 & 1 & 8\\ 4 & 5 & 9\\ -3 & -2 & 0 \end{bmatrix} \]
  5. Find \((A-B)^T\)
  6. Interchange the rows and columns:
    \[ (A-B)^T= \begin{bmatrix} 3 & 4 & -3\\ 1 & 5 & -2\\ 8 & 9 & 0 \end{bmatrix} \]
  7. Recall \(A^T\) and \(B^T\)
  8. \[ A^T= \begin{bmatrix} -1 & 5 & -2\\ 2 & 7 & 1\\ 3 & 9 & 1 \end{bmatrix} \]
    \[ B^T= \begin{bmatrix} -4 & 1 & 1\\ 1 & 2 & 3\\ -5 & 0 & 1 \end{bmatrix} \]
  9. Find \(A^T-B^T\)
  10. \[ A^T-B^T= \begin{bmatrix} -1 & 5 & -2\\ 2 & 7 & 1\\ 3 & 9 & 1 \end{bmatrix} - \begin{bmatrix} -4 & 1 & 1\\ 1 & 2 & 3\\ -5 & 0 & 1 \end{bmatrix} \]
  11. Subtract the corresponding entries:
    \[ A^T-B^T= \begin{bmatrix} -1-(-4) & 5-1 & -2-1\\ 2-1 & 7-2 & 1-3\\ 3-(-5) & 9-0 & 1-1 \end{bmatrix} \]
  12. Now simplify:
    \[ A^T-B^T= \begin{bmatrix} 3 & 4 & -3\\ 1 & 5 & -2\\ 8 & 9 & 0 \end{bmatrix} \]
  13. Comparing this with \((A-B)^T\),
    \[ (A-B)^T= \begin{bmatrix} 3 & 4 & -3\\ 1 & 5 & -2\\ 8 & 9 & 0 \end{bmatrix} \]
    \[ A^T-B^T= \begin{bmatrix} 3 & 4 & -3\\ 1 & 5 & -2\\ 8 & 9 & 0 \end{bmatrix} \]
  14. Therefore,
    \[ \boxed{(A-B)^T=A^T-B^T} \]
💡 Answer
Final Answer

The two required properties of transpose are verified:

\[ \boxed{(A+B)^T=A^T+B^T} \]
\[ \boxed{(A-B)^T=A^T-B^T} \]
🎯 Exam Significance
Exam Significance

This question is important because it tests a fundamental property of transpose that is repeatedly used in the Matrix chapter. In board examinations, a verification question requires both sides of the identity to be calculated correctly and compared explicitly.

  • Stepwise verification: Students should calculate the left-hand side and right-hand side separately before concluding that the identity is verified.
  • Arithmetic accuracy: Sign errors, especially subtraction of negative numbers, can invalidate an otherwise correct solution.
  • Transpose accuracy: Rows must be converted into columns without changing the values of the entries.
  • Presentation: Writing the intermediate matrices makes the verification transparent and helps secure method marks.
  • Conceptual foundation: These properties are used later with scalar multiplication, matrix products, symmetric matrices, and skew-symmetric matrices.
Significance for Competitive Entrance Examinations

Transpose identities frequently appear as components of multi-step problems in JEE and other competitive entrance examinations. Rather than calculating every matrix numerically, many questions can be simplified immediately by applying the appropriate transpose property.

The following identities should be memorised and understood:

\[ \boxed{(A+B)^T=A^T+B^T} \]
\[ \boxed{(A-B)^T=A^T-B^T} \]
\[ \boxed{(kA)^T=kA^T} \]
\[ \boxed{(AB)^T=B^TA^T} \]
\[ \boxed{(A^T)^T=A} \]

The product rule is particularly important because the order of the matrices is reversed:

\[ (AB)^T=B^TA^T \]

This reversal is a common source of errors in competitive examinations.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  10 points
  1. The transpose operation distributes over matrix addition.

  2. The transpose operation also distributes over matrix subtraction.

  3. For matrices of the same order, \((A+B)^T=A^T+B^T\).

  4. For matrices of the same order, \((A-B)^T=A^T-B^T\).

  5. Always calculate both sides separately when the question asks to verify an identity.

  6. When subtracting negative numbers, use brackets carefully.

  7. In this question, the correct value of \(-1-(-4)\) is \(3\).

  8. In this question, the correct value of \(1-1\) is \(0\).

  9. Do not confuse transpose properties with multiplication properties.

  10. For multiplication, the order reverses: \((AB)^T=B^TA^T\).

← Q1
2 / 12  ·  17%
Q3 →
Q3
NUMERIC3 marks

If

\[ A^T= \begin{bmatrix} 3 & 4\\ -1 & 2\\ 6 & 1 \end{bmatrix}, \qquad B= \begin{bmatrix} -1 & 2 & 1\\ 1 & 2 & 3 \end{bmatrix} \]

then verify that

\[ \text{(i)}\quad (A+B)^T=A^T+B^T \] \[ \text{(ii)}\quad (A-B)^T=A^T-B^T \]
📘 Concept & Theory
Concept/Theory

The transpose of a matrix is obtained by interchanging its rows and columns. If \(A\) is an \(m\times n\) matrix, then \(A^T\) is an \(n\times m\) matrix.

Two important properties of transpose used in this question are:

\[ \boxed{(A+B)^T=A^T+B^T} \]
\[ \boxed{(A-B)^T=A^T-B^T} \]

These properties show that transposition distributes over matrix addition and subtraction.

Since \(A^T\) is given as a \(3\times2\) matrix, \(A\) must be a \(2\times3\) matrix. Taking the transpose of the given \(A^T\), we obtain \(A\).

\[ A=(A^T)^T \]

Thus,

\[ A= \begin{bmatrix} 3 & -1 & 6\ 4 & 2 & 1 \end{bmatrix} \]

The matrix \(B\) is also of order \(2\times3\). Therefore, \(A+B\) and \(A-B\) are defined, and both are \(2\times3\) matrices. Their transposes are consequently \(3\times2\) matrices.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Use the given \(A^T\) to determine \(A\).

  2. Observe that \(A\) and \(B\) are both \(2\times3\) matrices.

  3. For part (i), calculate \(A+B\) and then take its transpose.

  4. Separately calculate \(A^T\) and \(B^T\), followed by \(A^T+B^T\).

  5. Compare both resulting \(3\times2\) matrices.

  6. For part (ii), calculate \(A-B\) and then take its transpose.

  7. Separately calculate \(A^T-B^T\).

  8. Compare both results to complete the verification.

✏️ Solution
(i) Verification of \((A+B)^T=A^T+B^T\)
Step-by-step Solution  ·  16 steps
  1. Find \(A\) from the Given \(A^T\)
  2. Given
    \[ A^T= \begin{bmatrix} 3 & 4\\ -1 & 2\\ 6 & 1 \end{bmatrix} \]
  3. Taking transpose on both sides,
    \[ (A^T)^T=(A^T)^T \]
  4. Using the property
    \[ (A^T)^T=A \]
  5. we get
    \[ A= \begin{bmatrix} 3 & -1 & 6\\ 4 & 2 & 1 \end{bmatrix} \]
  6. Also,
    \[ B= \begin{bmatrix} -1 & 2 & 1\\ 1 & 2 & 3 \end{bmatrix} \]
    Hence, both \(A\) and \(B\) are of order \(2\times3\).
  7. (i) Verification of \((A+B)^T=A^T+B^T\)
  8. Find \(A+B\)
  9. \[ A+B= \begin{bmatrix} 3 & -1 & 6\\ 4 & 2 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 2 & 1\\ 1 & 2 & 3 \end{bmatrix} \]
  10. Add corresponding elements:
    \[ A+B= \begin{bmatrix} 3+(-1) & -1+2 & 6+1\\ 4+1 & 2+2 & 1+3 \end{bmatrix} \]
  11. Therefore,
    \[ A+B= \begin{bmatrix} 2 & 1 & 7\ 5 & 4 & 4 \end{bmatrix} \]
  12. Find \((A+B)^T\)
  13. Taking the transpose of \(A+B\),
    \[ (A+B)^T= \begin{bmatrix} 2 & 1 & 7\\ 5 & 4 & 4 \end{bmatrix}^T \]
  14. Interchanging rows and columns gives
    \[ \boxed{ (A+B)^T= \begin{bmatrix} 2 & 5\\ 1 & 4\\ 7 & 4 \end{bmatrix}} \]
  15. Find \(A^T\)
  16. This matrix is already given:
    \[ A^T= \begin{bmatrix} 3 & 4\\ -1 & 2\\ 6 & 1 \end{bmatrix} \]
  17. Find \(B^T\)
  18. Taking the transpose of \(B\),
    \[ B= \begin{bmatrix} -1 & 2 & 1\\ 1 & 2 & 3 \end{bmatrix} \]
    \[ B^T= \begin{bmatrix} -1 & 1\\ 2 & 2\\ 1 & 3 \end{bmatrix} \]
  19. Find \(A^T+B^T\)
  20. \[ A^T+B^T= \begin{bmatrix} 3 & 4\\ -1 & 2\\ 6 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 1\\ 2 & 2\\ 1 & 3 \end{bmatrix} \]
  21. Add corresponding elements:
    \[ A^T+B^T= \begin{bmatrix} 3+(-1) & 4+1\\ -1+2 & 2+2\\ 6+1 & 1+3 \end{bmatrix} \]
  22. Therefore,
    \[ A^T+B^T= \begin{bmatrix} 2 & 5\\ 1 & 4\\ 7 & 4 \end{bmatrix} \]
  23. Thus,
    \[ A^T+B^T= \begin{bmatrix} 2 & 5\\ 1 & 4\\ 7 & 4 \end{bmatrix} = (A+B)^T \]
  24. Hence,
    \[ \boxed{(A+B)^T=A^T+B^T} \]
  25. Therefore, the first identity is verified.
✏️ Solution
(ii) Verification of \((A-B)^T=A^T-B^T\)
Step-by-step Solution  ·  13 steps
  1. Find \(A-B\)
  2. \[ A-B= \begin{bmatrix} 3 & -1 & 6\\ 4 & 2 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 2 & 1\\ 1 & 2 & 3 \end{bmatrix} \]
  3. Subtract corresponding elements:
    \[ A-B= \begin{bmatrix} 3-(-1) & -1-2 & 6-1\\ 4-1 & 2-2 & 1-3 \end{bmatrix} \]
  4. Now simplify each entry:
    \[ 3-(-1)=3+1=4 \]
    \[ -1-2=-3 \]
    \[ 6-1=5 \]
    \[ 4-1=3 \]
    \[ 2-2=0 \]
    \[ 1-3=-2 \]
  5. Hence,
    \[ A-B= \begin{bmatrix} 4 & -3 & 5\\ 3 & 0 & -2 \end{bmatrix} \]
  6. Find \((A-B)^T\)
  7. \[ (A-B)^T= \begin{bmatrix} 4 & -3 & 5\\ 3 & 0 & -2 \end{bmatrix}^T \]
  8. Interchanging rows and columns:
    \[ \boxed{ (A-B)^T= \begin{bmatrix} 4 & 3\\ -3 & 0\\ 5 & -2 \end{bmatrix}} \]
  9. Find \(A^T-B^T\)
  10. \[ A^T= \begin{bmatrix} 3 & 4\\ -1 & 2\\ 6 & 1 \end{bmatrix} \]
    \[ B^T= \begin{bmatrix} -1 & 1\\ 2 & 2\\ 1 & 3 \end{bmatrix} \]
  11. Therefore,
    \[ A^T-B^T= \begin{bmatrix} 3 & 4\\ -1 & 2\\ 6 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 1\\ 2 & 2\\ 1 & 3 \end{bmatrix} \]
  12. Subtract corresponding elements:
    \[ A^T-B^T= \begin{bmatrix} 3-(-1) & 4-1\\ -1-2 & 2-2\\ 6-1 & 1-3 \end{bmatrix} \]
  13. Hence,
    \[ A^T-B^T= \begin{bmatrix} 4 & 3\\ -3 & 0\\ 5 & -2 \end{bmatrix} \]
  14. Comparing both sides,
    \[ (A-B)^T= \begin{bmatrix} 4 & 3\\ -3 & 0\\ 5 & -2 \end{bmatrix} \]
    \[ A^T-B^T= \begin{bmatrix} 4 & 3\\ -3 & 0\\ 5 & -2 \end{bmatrix} \]
  15. Therefore,
    \[ \boxed{(A-B)^T=A^T-B^T} \]
  16. Hence, the second identity is verified.
💡 Answer
Final Answer
\[ \boxed{\bbox[2pt]{(A+B)^T=A^T+B^T}} \]
\[ \boxed{\bbox[2pt]{(A-B)^T=A^T-B^T}}\]
🎯 Exam Significance
Exam Significance
  • Given transpose: If \(A^T\) is given, obtain \(A\) by taking the transpose again.
  • Order awareness: A \(3\times2\) matrix \(A^T\) implies that \(A\) is \(2\times3\).
  • Entry tracking: Every element must be placed in its correct position while taking the transpose.
  • Verification method: For an identity, calculate the left-hand side and right-hand side separately.
  • Arithmetic discipline: A single incorrect matrix entry can affect every subsequent step.

For board examinations, writing the intermediate matrices clearly is advisable because it demonstrates the complete method rather than merely stating the final identity.

Significance for Competitive Entrance Examinations

Transpose properties are frequently embedded in objective and algebraic matrix problems. The ability to recognise the identity immediately can reduce lengthy calculations.

The following results are especially important:

\[ \boxed{(A+B)^T=A^T+B^T} \]
\[ \boxed{(A-B)^T=A^T-B^T} \]
\[ \boxed{(kA)^T=kA^T} \]
\[ \boxed{(AB)^T=B^TA^T} \]
\[ \boxed{(A^T)^T=A} \]

Among these, the product identity deserves particular attention because the order of the factors reverses:

\[ (AB)^T=B^TA^T \]

Recognising such identities is useful for simplifying matrix expressions quickly in competitive examinations.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  9 points
  1. If \(A^T\) is given, \(A\) is obtained by taking the transpose again.

  2. For the given matrix,

    \[ A^T= \begin{bmatrix} 3&4\\ -1&2\\ 6&1 \end{bmatrix} \]
    implies
    \[ A= \begin{bmatrix} 3&-1&6\ 4&2&1 \end{bmatrix}. \]

  3. The transpose distributes over matrix addition.

  4. The transpose distributes over matrix subtraction.

  5. Always check the dimensions before performing matrix addition or subtraction.

  6. For addition and subtraction, the order of matrices does not change.

  7. For multiplication, the order reverses: \((AB)^T=B^TA^T\).

  8. Careful handling of the given entries is essential; the supplied solution incorrectly changes the \(6\) in \(A\) to \(0\).

  9. The correct verified results are

    \[ (A+B)^T=A^T+B^T \]
    and
    \[ (A-B)^T=A^T-B^T. \]

← Q2
3 / 12  ·  25%
Q4 →
Q4
NUMERIC3 marks
If \[ A^T= \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix}, \qquad B= \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix} \]

then find

\[ (A+2B)^T \]
📘 Concept & Theory
Concept/Theory

The question involves the transpose of a matrix expression containing a scalar multiple. Two properties of transpose are particularly useful here:

\[ \boxed{(A+B)^T=A^T+B^T} \]

and

\[ \boxed{(kA)^T=kA^T} \]

Therefore, for the given expression,

\[ (A+2B)^T=A^T+(2B)^T \]
\[ (A+2B)^T=A^T+2B^T \]

This provides a direct and efficient method. However, since the question asks to find the matrix, we must first determine \(A\) from the given \(A^T\), or equivalently use the transpose identity directly.

Since

\[ A^T= \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix}, \]

taking transpose on both sides gives

\[ A= \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix} \]

Also, \(B\) is a \(2\times2\) matrix, so all the required operations are defined.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Use the given \(A^T\) to obtain \(A\).

  2. Multiply \(B\) by the scalar \(2\).

  3. Add \(A\) and \(2B\) to obtain \(A+2B\).

  4. Take the transpose of the resulting matrix.

  5. Alternatively, verify the result directly using

    \[ (A+2B)^T=A^T+2B^T. \]

✏️ Solution
Complete Solution
Step-by-step Solution  ·  16 steps
  1. Find \(A\) from \(A^T\)
  2. Given
    \[ A^T= \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix} \]

    Taking transpose on both sides,

    \[ A=(A^T)^T \]

    Using

    \[ (A^T)^T=A, \]

    we get

    \[ A= \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix} \]

    Thus,

    \[ \boxed{ A= \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix}} \]
  3. Calculate \(2B\)
  4. Given
    \[ B= \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix} \]
  5. Multiplying every element by \(2\),
    \[ 2B= 2 \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix} \]
    \[ 2B= \begin{bmatrix} 2(-1) & 2(0)\\ 2(1) & 2(2) \end{bmatrix} \]
    \[ \boxed{ 2B= \begin{bmatrix} -2 & 0\\ 2 & 4 \end{bmatrix}} \]
  6. Find \(A+2B\)
  7. \[ A+2B= \begin{bmatrix} -2 & 1\\ 3 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 0\\ 2 & 4 \end{bmatrix} \]
  8. Add corresponding elements:
    \[ A+2B= \begin{bmatrix} -2+(-2) & 1+0\\ 3+2 & 2+4 \end{bmatrix} \]
  9. Therefore,
    \[ A+2B= \begin{bmatrix} -4 & 1\\ 5 & 6 \end{bmatrix} \]
  10. Take the Transpose
  11. Now take the transpose of \(A+2B\):
    \[ (A+2B)^T= \begin{bmatrix} -4 & 1\\ 5 & 6 \end{bmatrix}^T \]
  12. Interchange rows and columns:
    \[ (A+2B)^T= \begin{bmatrix} -4 & 5\\ 1 & 6 \end{bmatrix} \]
  13. Hence, the required matrix is
    \[ \boxed{ (A+2B)^T= \begin{bmatrix} -4 & 5\\ 1 & 6 \end{bmatrix}} \]
  14. Alternative Verification Using Transpose Properties
  15. We can obtain the answer more directly by using
    \[ (A+2B)^T=A^T+(2B)^T \]
  16. Since
    \[ (2B)^T=2B^T, \]
  17. we have
    \[ (A+2B)^T=A^T+2B^T \]
  18. The given matrix \(A^T\) is
    \[ A^T= \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix} \]
  19. Now find \(B^T\):
    \[ B^T= \begin{bmatrix} -1 & 1\\ 0 & 2 \end{bmatrix} \]
  20. Therefore,
    \[ 2B^T= \begin{bmatrix} -2 & 2\\ 0 & 4 \end{bmatrix} \]
  21. Hence,
    \[ A^T+2B^T= \begin{bmatrix} -2 & 3\\ 1 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 2\\ 0 & 4 \end{bmatrix} \]
    \[ = \begin{bmatrix} -4 & 5\\ 1 & 6 \end{bmatrix} \]
  22. Thus,
    \[ \boxed{ (A+2B)^T= \begin{bmatrix} -4 & 5\\ 1 & 6 \end{bmatrix}} \]
  23. This agrees with the result obtained by first finding \(A+2B\) and then taking its transpose.

💡 Answer
Final Answer
\[ \boxed{ (A+2B)^T= \begin{bmatrix} -4 & 5\\ 1 & 6 \end{bmatrix}} \]
🎯 Exam Significance
Exam Significance

This problem is important because it combines three basic matrix operations: recovering a matrix from its transpose, scalar multiplication, and matrix addition. It also demonstrates how transpose properties can shorten calculations.

  • Students must know how to obtain \(A\) when \(A^T\) is given.
  • Every entry of a matrix is multiplied by the scalar when finding \(2B\).
  • Transpose must be taken only after the matrix operation is completed if using the direct method.
  • Alternatively, the identity
    \[ (A+2B)^T=A^T+2B^T \]
    provides a faster route.
  • The problem is a good example of how matrix identities can reduce computational work in an examination.
Significance for Competitive Entrance Examinations

For competitive examinations, the most efficient approach is often to apply matrix identities directly instead of calculating unnecessary intermediate matrices.

Here, the expression can immediately be transformed as

\[ (A+2B)^T=A^T+2B^T \]

This avoids calculating \(A\) and \(A+2B\) separately.

The important identities to remember are:

\[ \boxed{(A+B)^T=A^T+B^T} \]
\[ \boxed{(A-B)^T=A^T-B^T} \]
\[ \boxed{(kA)^T=kA^T} \]
\[ \boxed{(AB)^T=B^TA^T} \]
\[ \boxed{(A^T)^T=A} \]

These identities are particularly useful in objective questions where recognising the correct transformation is faster than performing full matrix calculations.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  7 points
  1. If \(A^T\) is given, then \(A=(A^T)^T\).

  2. A scalar passes through the transpose operation:

    \[ (kA)^T=kA^T. \]

  3. Transpose distributes over addition:

    \[ (A+B)^T=A^T+B^T. \]

  4. For this question,

    \[ (A+2B)^T=A^T+2B^T. \]

  5. There are two valid approaches: calculate \(A+2B\) first and transpose it, or use the transpose identity directly.

  6. The direct identity-based approach is usually faster in competitive examinations.

  7. Always remember that transpose reverses the order of factors only in matrix multiplication:

    \[ (AB)^T=B^TA^T. \]

← Q3
4 / 12  ·  33%
Q5 →
Q5
NUMERIC3 marks
For the following matrices \(A\) and \(B\), verify that \[(AB)^T=B^TA^T\]

(i)

\[A=\begin{bmatrix}1\\-4\\3\end{bmatrix},\quad B=\begin{bmatrix}-1&2&1\end{bmatrix}\]

(ii)

\[A=\begin{bmatrix}0\\1\\2\end{bmatrix},\quad B=\begin{bmatrix}1&5&7\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory

The transpose of the product of two matrices is given by the important identity

\[\boxed{(AB)^T=B^TA^T}\]

This is called the transpose-of-product property.

The most important point to remember is that the order of the matrices is reversed when taking the transpose of a product.

\[AB\quad\longrightarrow\quad (AB)^T=B^TA^T\]

Notice carefully that

\[(AB)^T\neq A^TB^T\]

in general. The correct order is \(B^TA^T\).

If \(A\) is an \(m\times n\) matrix and \(B\) is an \(n\times p\) matrix, then

\[AB\text{ is }m\times p\]

and hence

\[(AB)^T\text{ is }p\times m.\]

On the other hand,

\[B^T\text{ is }p\times n,\qquad A^T\text{ is }n\times m,\]

so

\[B^TA^T\]

is also a \(p\times m\) matrix. Thus, the dimensions are consistent.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Check the orders of \(A\) and \(B\) to ensure that \(AB\) is defined.

  2. Calculate \(AB\) directly using matrix multiplication.

  3. Take the transpose to obtain \((AB)^T\).

  4. Separately calculate \(B^T\) and \(A^T\).

  5. Calculate \(B^TA^T\).

  6. Compare \((AB)^T\) and \(B^TA^T\).

  7. If both matrices are identical, the required identity is verified.

✏️ Solution
(i)\(\quad A=\begin{bmatrix}1\\-4\\3\end{bmatrix},\quad B=\begin{bmatrix}-1&2&1\end{bmatrix}\)
Step-by-step Solution  ·  14 steps
  1. Given
    \[A=\begin{bmatrix}1\\-4\\3\end{bmatrix},\quad B=\begin{bmatrix}-1&2&1\end{bmatrix}\]
    Here, \(A\) is a \(3\times1\) matrix and \(B\) is a \(1\times3\) matrix.
    \[A_{3\times1}B_{1\times3}\]
    Therefore, \(AB\) is defined and will be a \(3\times3\) matrix.
  2. Find \(AB\)
  3. We have
    \[AB=\begin{bmatrix}1\\-4\\3\end{bmatrix}\begin{bmatrix}-1&2&1\end{bmatrix}\]
  4. Since this is the product of a column matrix and a row matrix, each entry is obtained by multiplying an element of the column by the complete row.
    \[AB=\begin{bmatrix}1(-1)&1(2)&1(1)\\(-4)(-1)&(-4)(2)&(-4)(1)\\3(-1)&3(2)&3(1)\end{bmatrix}\]
  5. Therefore,
    \[AB=\begin{bmatrix}-1&2&1\\4&-8&-4\\-3&6&3\end{bmatrix}\]
  6. Find \((AB)^T\)
  7. Taking the transpose, we interchange rows and columns:
    \[(AB)^T=\begin{bmatrix}-1&2&1\\4&-8&-4\\-3&6&3\end{bmatrix}^{T}\]
    \[\boxed{(AB)^T=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}}\]
  8. Find \(B^T\)
  9. \[B=\begin{bmatrix}-1&2&1\end{bmatrix}\]
  10. Therefore,
    \[B^T=\begin{bmatrix}-1\\2\\1\end{bmatrix}\]
  11. Find \(A^T\)
  12. \[A=\begin{bmatrix}1\\-4\\3\end{bmatrix}\]
  13. Therefore,
    \[A^T=\begin{bmatrix}1&-4&3\end{bmatrix}\]
  14. Calculate \(B^TA^T\)
  15. \[B^TA^T=\begin{bmatrix}-1\\2\\1\end{bmatrix}\begin{bmatrix}1&-4&3\end{bmatrix}\]
  16. Multiplying the column matrix by the row matrix,
    \[B^TA^T=\begin{bmatrix}(-1)(1)&(-1)(-4)&(-1)(3)\\2(1)&2(-4)&2(3)\\1(1)&1(-4)&1(3)\end{bmatrix}\]
  17. Hence,
    \[B^TA^T=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\]
  18. Comparing both sides,
    \[(AB)^T=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\]
    \[B^TA^T=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\]
  19. Therefore,
    \[\boxed{(AB)^T=B^TA^T}\]
  20. Hence, the required identity is verified.
✏️ Solution
(ii)\(\quad A=\begin{bmatrix}0\\1\\2\end{bmatrix},\quad B=\begin{bmatrix}1&5&7\end{bmatrix}\)
Step-by-step Solution  ·  15 steps
  1. Given
    \[A=\begin{bmatrix}0\\1\\2\end{bmatrix},\qquad B=\begin{bmatrix}1&5&7\end{bmatrix}\]
    Here, \(A\) is \(3\times1\) and \(B\) is \(1\times3\).
    \[A_{3\times1}B_{1\times3}\]
    Hence, \(AB\) is a \(3\times3\) matrix.
  2. Find \(AB\)
  3. \[AB=\begin{bmatrix}0\\1\\2\end{bmatrix}\begin{bmatrix}1&5&7\end{bmatrix}\]
  4. Multiplying each entry of the column by the complete row,
    \[AB=\begin{bmatrix}0(1)&0(5)&0(7)\\1(1)&1(5)&1(7)\\2(1)&2(5)&2(7)\end{bmatrix}\]
  5. Therefore,
    \[AB=\begin{bmatrix}0&0&0\\1&5&7\\2&10&14\end{bmatrix}\]
  6. Find \((AB)^T\)
  7. \[(AB)^T=\begin{bmatrix}0&0&0\\1&5&7\\2&10&14\end{bmatrix}^{T}\]
  8. Interchanging rows and columns,
    \[\boxed{(AB)^T=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}}\]
  9. Find \(B^T\)
  10. \[B=\begin{bmatrix}1&5&7\end{bmatrix}\]
  11. Therefore,
    \[B^T=\begin{bmatrix}1\\5\\7\end{bmatrix}\]
  12. Find \(A^T\)
  13. \[A=\begin{bmatrix}0\\1\\2\end{bmatrix}\]
  14. Therefore,
    \[A^T=\begin{bmatrix}0&1&2\end{bmatrix}\]
  15. Calculate \(B^TA^T\)
  16. \[B^TA^T=\begin{bmatrix}1\\5\\7\end{bmatrix}\begin{bmatrix}0&1&2\end{bmatrix}\]
  17. Multiplying the column matrix by the row matrix,
    \[B^TA^T=\begin{bmatrix}1(0)&1(1)&1(2)\\5(0)&5(1)&5(2)\\7(0)&7(1)&7(2)\end{bmatrix}\]
  18. Thus,
    \[B^TA^T=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\]
  19. Comparing both sides,
    \[(AB)^T=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\]
    \[B^TA^T=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\]
  20. Therefore,
    \[ \boxed{(AB)^T=B^TA^T}\]
  21. Hence, the required identity is verified.
💡 Answer
Final Answer

(i)

\[(AB)^T=B^TA^T=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\]

(ii)

\[(AB)^T=B^TA^T=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\]
🎯 Exam Significance
Exam Significance

This is an important question because it tests one of the most fundamental properties of matrix multiplication and transpose. It also tests whether the student understands matrix dimensions before multiplying matrices.

  • Dimension check: Always check the orders of matrices before multiplication.
  • Product order: If \(A\) is \(m\times n\) and \(B\) is \(n\times p\), then \(AB\) is \(m\times p\).
  • Transpose of product:
    \[ (AB)^T=B^TA^T. \]
  • Order reversal: The matrices appear in reverse order after transposition.
  • Matrix multiplication is not entrywise multiplication: A column matrix multiplied by a row matrix produces an outer-product matrix, not a scalar.
  • Presentation: Writing the orders and intermediate matrices makes the verification logically complete.
Significance for Competitive Entrance Examinations

The identity

\[\boxed{(AB)^T=B^TA^T}\]

is extremely important in competitive mathematics. It is often used to simplify complicated matrix expressions without performing full matrix multiplication.

Students should distinguish carefully between the following:

\[(AB)^T=B^TA^T\]
\[(BA)^T=A^TB^T\]

Thus, transposition reverses the order of factors.

Another important point is that matrix multiplication is generally not commutative:

\[AB\neq BA\]

when both products exist.

In this question, the distinction is especially clear:

\[AB\text{ is }3\times3\]

whereas

\[BA\text{ is }1\times1.\]

This makes the order of multiplication an essential part of the solution.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  10 points
  1. The transpose of a product reverses the order of the factors.

  2. The fundamental identity is

    \[ (AB)^T=B^TA^T. \]

  3. Never write \(A^TB^T\) in place of \(B^TA^T\) unless a special condition makes the two expressions equal.

  4. Always check matrix dimensions before multiplication.

  5. A \(3\times1\) matrix multiplied by a \(1\times3\) matrix produces a \(3\times3\) matrix.

  6. A \(1\times3\) matrix multiplied by a \(3\times1\) matrix produces a \(1\times1\) matrix.

  7. The products \(AB\) and \(BA\) can have completely different orders and values.

  8. Matrix multiplication is not generally commutative.

  9. The supplied solution incorrectly treats \(AB\) as a scalar in both parts; the correct \(AB\) matrices are \(3\times3\).

  10. For competitive examinations, dimension checking can immediately eliminate many incorrect options.

← Q4
5 / 12  ·  42%
Q6 →
Q6
NUMERIC3 marks

(i) If

\[A=\begin{bmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{bmatrix}\]

then verify that

\[A^TA=I\]

(ii) If

\[A=\begin{bmatrix}\sin\alpha & \cos\alpha\\-\cos\alpha & \sin\alpha\end{bmatrix}\]

then verify that

\[A^TA=I\]
📘 Concept & Theory
Concept/Theory

This question uses the transpose of a matrix and the standard trigonometric identity

\[\boxed{\sin^2\alpha+\cos^2\alpha=1}\]

For a square matrix \(A\), if

\[A^TA=I,\]

then \(A\) is called an orthogonal matrix.

For an orthogonal matrix, the transpose is also its inverse:

\[\boxed{A^{-1}=A^T}\]

Therefore, verifying \(A^TA=I\) is equivalent to establishing that the given matrix is orthogonal.

Here, the entries contain \(\sin\alpha\) and \(\cos\alpha\). Consequently, after matrix multiplication, the diagonal entries reduce to \(1\) using

\[\sin^2\alpha+\cos^2\alpha=1,\]

while the off-diagonal entries cancel to zero.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Write the given matrix \(A\).

  2. Find its transpose \(A^T\) by interchanging rows and columns.

  3. Multiply \(A^T\) by \(A\), because the question specifically asks us to verify \(A^TA=I\).

  4. Calculate each of the four entries of the resulting \(2\times2\) matrix.

  5. Use

    \[ \sin^2\alpha+\cos^2\alpha=1 \]
    for the diagonal entries.

  6. Show that the off-diagonal entries are zero.

  7. Conclude that

    \[ A^TA=I. \]

✏️ Solution

(i) If

\[A=\begin{bmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{bmatrix}\]

then verify that

\[A^TA=I\]
Step-by-step Solution  ·  16 steps
  1. Given
    \[A=\begin{bmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{bmatrix}\]
    The matrix \(A\) is of order \(2\times2\). Therefore, \(A^T\) is also of order \(2\times2\), and the product \(A^TA\) is defined.
  2. Find \(A^T\)
  3. Taking the transpose of \(A\), we interchange its rows and columns:
    \[ A^T=\begin{bmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{bmatrix}\]
  4. Thus,
    \[\boxed{A^T=\begin{bmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{bmatrix}}\]
  5. Calculate \(A^TA\)
  6. \[A^TA=\begin{bmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{bmatrix}\]
  7. Using matrix multiplication,
  8. \[A^TA=\begin{bmatrix}(\cos\alpha)(\cos\alpha)+(-\sin\alpha)(-\sin\alpha) &(\cos\alpha)(\sin\alpha)+(-\sin\alpha)(\cos\alpha)\\[6pt] (\sin\alpha)(\cos\alpha)+(\cos\alpha)(-\sin\alpha) &(\sin\alpha)(\sin\alpha)+(\cos\alpha)(\cos\alpha) \end{bmatrix} \]
  9. Simplify the \((1,1)\)-entry
    \[a_{11}=\cos^2\alpha+\sin^2\alpha\]
  10. Using the identity
    \[sin^2\alpha+\cos^2\alpha=1,\]
  11. we obtain
    \[\boxed{a_{11}=1}\]
  12. Simplify the \((1,2)\)-entry
    \[a_{12}=\cos\alpha\sin\alpha-\sin\alpha\cos\alpha\]
    \[a_{12}=0\]
  13. Therefore,
    \[\boxed{a_{12}=0}\]
  14. Simplify the \((2,1)\)-entry
    \[a_{21}=\sin\alpha\cos\alpha-\cos\alpha\sin\alpha\]
    \[a_{21}=0\]
  15. Therefore,
    \[\boxed{a_{21}=0}\]
  16. Simplify the \((2,2)\)-entry
    \[a_{22}=\sin^2\alpha+\cos^2\alpha\]
  17. Using
    \[\sin^2\alpha+\cos^2\alpha=1,\]
  18. we get
    \[\boxed{a_{22}=1}\]
  19. Form the resulting matrix
    \[A^TA=\begin{bmatrix}1&0\\0&1\end{bmatrix}\]
  20. But
    \[I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\]
  21. Hence,
    \[\boxed{A^TA=I}\]
  22. Therefore, the given matrix \(A\) is an orthogonal matrix.
  23. Hence Proved — .
✏️ Solution

(ii) If

\[A=\begin{bmatrix}\sin\alpha & \cos\alpha\\-\cos\alpha & \sin\alpha\end{bmatrix}\]

then verify that

\[A^TA=I\]
Step-by-step Solution  ·  17 steps
  1. Given
    \[A=\begin{bmatrix}\sin\alpha & \cos\alpha\\-\cos\alpha & \sin\alpha\end{bmatrix}\]
    Again, \(A\) is a \(2\times2\) matrix, so \(A^TA\) is defined.
  2. Find \(A^T\)
  3. Interchanging rows and columns,
    \[A^T=\begin{bmatrix}\sin\alpha & -\cos\alpha\\\cos\alpha & \sin\alpha\end{bmatrix}\]
  4. Thus,
    \[\boxed{A^T=\begin{bmatrix}\sin\alpha & -\cos\alpha\\\cos\alpha & \sin\alpha\end{bmatrix}}\]
  5. Calculate \(A^TA\)
    \[A^TA=\begin{bmatrix}\sin\alpha & -\cos\alpha\\\cos\alpha & \sin\alpha\end{bmatrix} \begin{bmatrix}\sin\alpha & \cos\alpha\\-\cos\alpha & \sin\alpha\end{bmatrix}\]
  6. Multiplying the matrices entry by entry,
    \[A^TA=\begin{bmatrix}(\sin\alpha)(\sin\alpha)+(-\cos\alpha)(-\cos\alpha)&(\sin\alpha)(\cos\alpha)+(-\cos\alpha)(\sin\alpha)\\[6pt] (\cos\alpha)(\sin\alpha)+(\sin\alpha)(-\cos\alpha)&(\cos\alpha)(\cos\alpha)+(\sin\alpha)(\sin\alpha)\end{bmatrix}\]
  7. Simplify the \((1,1)\)-entry
    \[a_{11}=\sin^2\alpha+\cos^2\alpha\]
  8. Using
    \[\sin^2\alpha+\cos^2\alpha=1,\]
  9. we get
    \[\boxed{a_{11}=1}\]
  10. Simplify the \((1,2)\)-entry
    \[a_{12}=\sin\alpha\cos\alpha-\cos\alpha\sin\alpha\]
    \[a_{12}=0\]
  11. Therefore,
    \[\boxed{a_{12}=0}\]
  12. Simplify the \((2,1)\)-entry
    \[a_{21}=\cos\alpha\sin\alpha-\sin\alpha\cos\alpha\]
    \[a_{21}=0\]
  13. Therefore,
    \[\boxed{a_{21}=0}\]
  14. Simplify the \((2,2)\)-entry
    \[a_{22}=\cos^2\alpha+\sin^2\alpha\]
  15. Using the trigonometric identity,
    \[a_{22}=1\]
  16. Therefore,
    \[\boxed{a_{22}=1}\]
  17. Form the resulting matrix
    \[A^TA=\begin{bmatrix}1&0\\0&1\end{bmatrix}\]
  18. Since
    \[I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\]
  19. we obtain
    \[\boxed{A^TA=I}\]
  20. Therefore, the given matrix \(A\) is an orthogonal matrix.
  21. Hence Proved — .
👁️ Important Observation
Important Observation

In both parts, the matrix \(A\) satisfies

\[A^TA=I.\]

Therefore, both matrices are orthogonal matrices.

For any orthogonal matrix,

\[\boxed{A^{-1}=A^T}\]

and consequently

\[AA^T=I.\]

Thus, once \(A^TA=I\) has been established for these square matrices, we also have

\[\boxed{AA^T=I}.\]

The supplied solution for part (i) calculated \(AA^T\), whereas the question asks for \(A^TA\). Although both products equal \(I\) for these orthogonal matrices, a complete solution should calculate the product specified in the question. The solutions above therefore verify \(A^TA=I\) directly.

🎯 Exam Significance
Exam Significance

This question is important for the Class 12 board examination because it combines three fundamental ideas: transpose of a matrix, matrix multiplication, and the trigonometric identity \(\sin^2\alpha+\cos^2\alpha=1\).

  • It provides a standard method for proving that a matrix is orthogonal.
  • Students should be able to calculate \(A^T\) accurately by interchanging rows and columns.
  • Every entry of \(A^TA\) should be shown clearly rather than jumping directly to the identity matrix.
  • The diagonal terms reduce to \(1\) using
    \[ \sin^2\alpha+\cos^2\alpha=1. \]
  • The off-diagonal terms cancel because
    \[ \sin\alpha\cos\alpha-\cos\alpha\sin\alpha=0. \]
  • Showing these simplifications explicitly makes the proof complete and earns method marks.
Significance for Competitive Entrance Examinations

Orthogonal matrices occur frequently in objective questions involving matrix properties, inverses, determinants and transformations.

The key recognition is

\[ A^TA=I \quad\Longrightarrow\quad A^{-1}=A^T. \]

Therefore, once a matrix is recognised as orthogonal, its inverse can be written immediately without using the adjoint or determinant formula.

For example, if

\[ A= \begin{bmatrix} \cos\alpha&\sin\alpha\\ -\sin\alpha&\cos\alpha \end{bmatrix}, \]

then immediately

\[ A^{-1}=A^T= \begin{bmatrix} \cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha \end{bmatrix}. \]

This can substantially reduce calculation time in entrance examinations.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  8 points
  1. An orthogonal matrix satisfies

    \[ A^TA=I. \]

  2. For an orthogonal matrix,

    \[ A^{-1}=A^T. \]

  3. The fundamental trigonometric identity used here is

    \[ \sin^2\alpha+\cos^2\alpha=1. \]

  4. The off-diagonal terms vanish because multiplication of real numbers is commutative:

    \[ \sin\alpha\cos\alpha-\cos\alpha\sin\alpha=0. \]

  5. For a square orthogonal matrix,

    \[ A^TA=AA^T=I. \]

  6. When a question specifically asks to verify \(A^TA=I\), calculate \(A^TA\) directly.

  7. Do not confuse \(A^TA\) with \(AA^T\), even though both are \(I\) for an orthogonal square matrix.

  8. These matrices represent orthogonal transformations and preserve lengths and angles.

← Q5
6 / 12  ·  50%
Q7 →
Q7
NUMERIC3 marks

(i) Show that the matrix

\[A=\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}\]

is a symmetric matrix.

(ii) Show that the matrix

\[A=\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}\]

is a skew-symmetric matrix.

📘 Concept & Theory
Concept/Theory

A square matrix \(A\) is called a symmetric matrix if it is equal to its transpose.

\[ \boxed{A=A^T} \]

Thus, to prove that a matrix is symmetric, we calculate \(A^T\) and verify that

\[ A^T=A. \]

A square matrix \(A\) is called a skew-symmetric matrix if it is equal to the negative of its transpose.

\[ \boxed{A^T=-A} \]

Equivalently,

\[ \boxed{A=-A^T} \]

For a skew-symmetric matrix, every diagonal element must be zero. This follows because if \(A=-A^T\), then for every diagonal element \(a_{ii}\),

\[ a_{ii}=-a_{ii} \]
\[ 2a_{ii}=0 \]
\[ \boxed{a_{ii}=0}. \]

Both matrices in this question are \(3\times3\) square matrices, so symmetry and skew-symmetry can be tested using their transposes.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Write the given matrix \(A\).

  2. Find \(A^T\) by interchanging rows and columns.

  3. For part (i), compare \(A^T\) with \(A\).

  4. If \(A^T=A\), conclude that \(A\) is symmetric.

  5. For part (ii), compare \(A^T\) with \(-A\).

  6. If \(A^T=-A\), conclude that \(A\) is skew-symmetric.

✏️ Solution
(i) Show that \[A=\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}\] is a Symmetric Matrix
Step-by-step Solution  ·  9 steps
  1. Given
    \[A=\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}\]
  2. We know that a square matrix is symmetric if
    \[A^T=A\]
  3. Find \(A^T\)
    The transpose is obtained by interchanging rows and columns.
  4. The first row of \(A\),
    \[\begin{bmatrix}1&-1&5\end{bmatrix}\]
    becomes the first column of \(A^T\).
  5. The second row of \(A\),
    \[\begin{bmatrix}-1&2&1\end{bmatrix}\]
    becomes the second column of \(A^T\).
  6. The third row of \(A\),
    \[\begin{bmatrix}5&1&3\end{bmatrix}\]
    becomes the third column of \(A^T\).
  7. Therefore,
    \[A^T=\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}\]
  8. Compare \(A^T\) with \(A\)
    \[A=\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}\]

    and

    \[A^T=\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}\]
  9. Hence,
    \[A^T=A\]
  10. Therefore,
    \[\boxed{A\text{ is a symmetric matrix}.}\]
  11. Hence Proved
✏️ Solution
(ii) Show that \[A=\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}\] is a Skew-Symmetric Matrix
Step-by-step Solution  ·  7 steps
  1. Given
    \[A=\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}\]
  2. A matrix is skew-symmetric if
    \[A^T=-A\]
  3. Find \(A^T\) by Interchanging rows and columns,
    \[A^T=\begin{bmatrix}0 & -1 & 1\\1 & 0 & -1\\-1 & 1 & 0\end{bmatrix}\]
  4. Find \(-A\)
    \[A=\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}\]
  5. Multiplying every element by \(-1\),
    \[-A=-\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}\]
    \[-A=\begin{bmatrix}-0 & -1 & 1\\1 & -0 & -1\\-1 & 1 & -0\end{bmatrix}\]
    \[-A=\begin{bmatrix}0 & -1 & 1\\1 & 0 & -1\\-1 & 1 & 0\end{bmatrix}\]
  6. Compare \(A^T\) and \(-A\)
    \[A^T=\begin{bmatrix}0 & -1 & 1\\1 & 0 & -1\\-1 & 1 & 0\end{bmatrix}\]

    and

    \[-A=\begin{bmatrix}0 & -1 & 1\\1 & 0 & -1\\-1 & 1 & 0\end{bmatrix}.\]
  7. Therefore,
    \[A^T=-A\]
  8. Hence,
    \[\boxed{A\text{ is a skew-symmetric matrix}.}\]
  9. Hence Proved
🎯 Exam Significance
Exam Significance

This question is a direct application of the definitions of symmetric and skew-symmetric matrices and is highly relevant for board examinations.

  • Remember the two defining conditions:
    \[ A^T=A \]
    for symmetric matrices and
    \[ A^T=-A \]
    for skew-symmetric matrices.
  • For a symmetric matrix, corresponding entries across the principal diagonal are equal.
  • For a skew-symmetric matrix, corresponding entries across the principal diagonal are negatives of each other.
  • The diagonal entries of every skew-symmetric matrix are zero.
  • Writing the complete transpose is the safest method in a proof-based board question.
  • For method marks, explicitly state the defining condition before comparing the matrices.
Significance for Competitive Entrance Examinations

Symmetric and skew-symmetric matrices are frequently used as building blocks in objective questions and matrix decomposition problems.

An important result is that every square matrix \(A\) can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix:

\[ \boxed{ A= \frac{1}{2}(A+A^T) + \frac{1}{2}(A-A^T) } \]

Here,

\[ \boxed{\frac{1}{2}(A+A^T)} \]

is symmetric, while

\[ \boxed{\frac{1}{2}(A-A^T)} \]

is skew-symmetric.

This decomposition is useful in higher-level matrix problems and competitive examinations.

For quick identification, check the principal diagonal first. If a matrix is claimed to be skew-symmetric but has a non-zero diagonal element, the claim is immediately false.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  9 points
  1. A square matrix \(A\) is symmetric if

    \[ A^T=A. \]

  2. A square matrix \(A\) is skew-symmetric if

    \[ A^T=-A. \]

  3. In a symmetric matrix,

    \[ a_{ij}=a_{ji}. \]

  4. In a skew-symmetric matrix,

    \[ a_{ij}=-a_{ji}. \]

  5. Every diagonal element of a skew-symmetric matrix is zero.

  6. The matrix in part (i) is symmetric because its entries are mirrored equally about the principal diagonal.

  7. The matrix in part (ii) is skew-symmetric because its entries are opposite in sign across the principal diagonal and all diagonal entries are zero.

  8. For a quick test, compare corresponding elements on opposite sides of the principal diagonal.

  9. Every square matrix can be decomposed into a symmetric part and a skew-symmetric part:

    \[ A=\frac12(A+A^T)+\frac12(A-A^T). \]

← Q6
7 / 12  ·  58%
Q8 →
Q8
NUMERIC3 marks

For the matrix

\[A=\begin{bmatrix}1 & 5\\6 & 7\end{bmatrix}\]

verify that:

(i) \(A+A^T\) is a symmetric matrix.

(ii) \(A-A^T\) is a skew-symmetric matrix.

📘 Concept & Theory
Concept/Theory

This question illustrates two important general results concerning the transpose of a square matrix.

For any square matrix \(A\), the matrix

\[ \boxed{A+A^T} \]

is always symmetric because

\[ (A+A^T)^T=A^T+(A^T)^T=A^T+A=A+A^T. \]

Similarly, the matrix

\[ \boxed{A-A^T} \]

is always skew-symmetric because

\[ (A-A^T)^T=A^T-(A^T)^T=A^T-A=-(A-A^T). \]

The identities

\[ \boxed{(A+B)^T=A^T+B^T} \]

and

\[ \boxed{(A-B)^T=A^T-B^T} \]

are therefore central to this problem.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Write the given matrix \(A\).

  2. Find its transpose \(A^T\).

  3. For part (i), calculate \(A+A^T\) entry by entry.

  4. Take the transpose of the resulting matrix.

  5. Show that its transpose is equal to itself.

  6. For part (ii), calculate \(A-A^T\) entry by entry.

  7. Take the transpose of the resulting matrix.

  8. Show that the transpose is equal to the negative of the original matrix.

✏️ Solution
(i) \(A+A^T\) is a symmetric matrix.
Step-by-step Solution  ·  9 steps
  1. Given
    \[A=\begin{bmatrix}1 & 5\\6 & 7\end{bmatrix}\]
  2. A matrix is symmetric if
    \[\boxed{S^T=S}\]
  3. Find \(A^T\)
  4. Interchanging the rows and columns of \(A\), we obtain
    \[A^T=\begin{bmatrix}1 & 6\\5 & 7\end{bmatrix}\]
  5. Calculate \(A+A^T\)
    \[A+A^T=\begin{bmatrix}1 & 5\\6 & 7\end{bmatrix}+\begin{bmatrix}1 & 6\\5 & 7\end{bmatrix}\]
  6. Add corresponding elements:
    \[A+A^T=\begin{bmatrix}1+1 & 5+6\\6+5 & 7+7\end{bmatrix}\]
    \[A+A^T=\begin{bmatrix}2 & 11\\11 & 14\end{bmatrix}.\]
  7. Let
    \[S=A+A^T\]
  8. Therefore,
    \[S=\begin{bmatrix}2 & 11\\11 & 14\end{bmatrix}\]
  9. Find \(S^T\) by Taking the transpose,
    \[S^T=\begin{bmatrix}2 & 11\\11 & 14\end{bmatrix}\]
  10. Hence,
    \[S^T=S\]
  11. Therefore,
    \[\boxed{A+A^T\text{ is a symmetric matrix}.}\]
  12. Hence Proved
✏️ Solution
(ii) \(A-A^T\) is a skew-symmetric matrix.
Step-by-step Solution  ·  9 steps
  1. We have
    \[A=\begin{bmatrix}1 & 5\\6 & 7\end{bmatrix}\]

    and

    \[A^T=\begin{bmatrix}1 & 6\\5 & 7\end{bmatrix}\]
  2. A matrix is skew-symmetric if
    \[\boxed{S^T=-S}.\]
  3. Calculate \(A-A^T\)
    \[A-A^T=\begin{bmatrix}1 & 5\\6 & 7\end{bmatrix}-\begin{bmatrix}1 & 6\\5 & 7\end{bmatrix}\]
  4. Subtract corresponding elements:
    \[A-A^T=\begin{bmatrix}1-1 & 5-6\\6-5 & 7-7\end{bmatrix}\]
    \[[A-A^T=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}\]
  5. Let
    \[S=A-A^T\]
  6. Thus,
    \[S=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}\]
  7. Find \(S^T\)
    \[S^T=\begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix}\]
  8. Find \(-S\)
    \[-S=-\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}\]
    \[-S=\begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix}\]
  9. Therefore,
    \[S^T=-S\]
  10. Hence,
    \[\boxed{A-A^T\text{ is a skew-symmetric matrix}.}\]
  11. Hence Proved — .
🎯 Exam Significance
Exam Significance

This is an important conceptual question because it tests whether the student can apply the definitions of symmetric and skew-symmetric matrices using transpose properties.

  • Remember that
    \[ A^T=A \]
    characterises a symmetric matrix.
  • Remember that
    \[ A^T=-A \]
    characterises a skew-symmetric matrix.
  • For \(A+A^T\), explicitly calculate the matrix and verify that its transpose is the same matrix.
  • For \(A-A^T\), verify that its transpose is its negative.
  • In a board examination, showing the intermediate matrix calculation prevents loss of method marks.
  • The general identities are worth remembering because they can solve similar questions quickly.
Significance for Competitive Entrance Examinations

This concept is especially useful in objective questions where the matrix itself may be complicated but its transpose is easy to identify.

For any square matrix \(A\), immediately recognise:

\[ \boxed{A+A^T\text{ is symmetric}} \]
\[ \boxed{A-A^T\text{ is skew-symmetric}} \]

This allows many multiple-choice questions to be answered without performing complete matrix addition.

The decomposition

\[\boxed{A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)}\]

is also a frequently useful result in higher-level problems involving symmetric and skew-symmetric matrices.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  8 points
  1. \[ \boxed{(A^T)^T=A} \]

  2. \[ \boxed{(A+A^T)^T=A+A^T} \]
    Therefore, \(A+A^T\) is symmetric.

  3. \[ \boxed{(A-A^T)^T=-(A-A^T)} \]
    Therefore, \(A-A^T\) is skew-symmetric.

  4. The symmetric part of \(A\) is

    \[ \boxed{\frac{1}{2}(A+A^T)}. \]

  5. The skew-symmetric part of \(A\) is

    \[ \boxed{\frac{1}{2}(A-A^T)}. \]

  6. Every square matrix can be uniquely decomposed into its symmetric and skew-symmetric parts.

  7. For the given matrix,

    \[ A+A^T= \begin{bmatrix} 2&11\\ 11&14 \end{bmatrix}, \]
    which is symmetric.

  8. For the given matrix,

    \[ A-A^T= \begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix}, \]
    which is skew-symmetric.

← Q7
8 / 12  ·  67%
Q9 →
Q9
NUMERIC3 marks

Find

\[\frac{1}{2}(A+A^T)\]

and

\[\frac{1}{2}(A-A^T)\]

when

\[ A= \begin{bmatrix} 0&a&b\\ -a&0&c\\ -b&-c&0 \end{bmatrix} \]
📘 Concept & Theory
Concept/Theory

Every square matrix \(A\) can be uniquely decomposed into the sum of a symmetric matrix and a skew-symmetric matrix:

\[\boxed{A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)}\]

The two components have special properties:

Symmetric part:

\[\boxed{S=\frac{1}{2}(A+A^T)}\]
\[S^T=S.\]

Skew-symmetric part:

\[\boxed{K=\frac{1}{2}(A-A^T)}\]
\[K^T=-K.\]

The given matrix has the form

\[A=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}\]

Notice immediately that the entries on opposite sides of the principal diagonal are negatives of each other, and all diagonal entries are zero. Hence, the given matrix is already skew-symmetric.

\[\boxed{A^T=-A}\]

This observation allows us to anticipate that its symmetric part will be the zero matrix and its skew-symmetric part will be the matrix \(A\) itself. We nevertheless verify both results step by step.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Write the given matrix \(A\).

  2. Find \(A^T\) by interchanging rows and columns.

  3. Calculate \(A+A^T\) entry by entry.

  4. Multiply the resulting matrix by \(\frac{1}{2}\).

  5. Calculate \(A-A^T\) entry by entry.

  6. Multiply the resulting matrix by \(\frac{1}{2}\).

  7. Interpret the results using the symmetric and skew-symmetric decomposition of a matrix.

✏️ Solution
Complete Solution
Step-by-step Solution  ·  19 steps
  1. Given — Matrix
    \[A=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}\]
  2. We first find the transpose of \(A\).
  3. Find \(A^T\) by taking the transpose means interchanging rows and columns.
    \[A^T=\begin{bmatrix}0&-a&-b\\a&0&-c\\b&c&0\end{bmatrix}\]
  4. Therefore,
    \[A^T=\begin{bmatrix}0&-a&-b\\a&0&-c\\b&c&0\end{bmatrix}\]
  5. We can also observe that
    \[-A=-\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}=\begin{bmatrix}0&-a&-b\\a&0&-c\\b&c&0\end{bmatrix}\]
  6. Hence,
    \[\boxed{A^T=-A}\]
    Thus, the given matrix \(A\) is already skew-symmetric.
  7. (i) Find \(\frac{1}{2}(A+A^T)\)
  8. First calculate \(A+A^T\):
    \[A+A^T=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}+\begin{bmatrix}0&-a&-b\\a&0&-c\\b&c&0\end{bmatrix}\]
  9. Adding corresponding elements,
    \[A+A^T=\begin{bmatrix}0+0&a+(-a)&b+(-b)\\-a+a&0+0&c+(-c)\\-b+b&-c+c&0+0\end{bmatrix}\]
  10. Now simplify each entry:
    \[a+(-a)=0,\]
    \[b+(-b)=0,\]
    \[-a+a=0,\]
    \[c+(-c)=0,\]
    \[-b+b=0,\]
    \[-c+c=0.\]
  11. Therefore,
    \[A+A^T=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\]
  12. Now multiply by \(\frac{1}{2}\):
    \[\frac{1}{2}(A+A^T)=\frac{1}{2}\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\]
  13. Hence,
    \[\frac{1}{2}(A+A^T)=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\]
  14. Thus, the symmetric part of the given matrix is the zero matrix.
  15. (ii) Find \(\frac{1}{2}(A-A^T)\)
  16. Now calculate \(A-A^T\):
    \[A-A^T=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}-\begin{bmatrix}0&-a&-b\\a&0&-c\\b&c&0\end{bmatrix}\]
  17. Subtract corresponding elements:
    \[A-A^T=\begin{bmatrix}0-0&a-(-a)&b-(-b)\\-a-a&0-0&c-(-c)\\-b-b&-c-c&0-0\end{bmatrix}\]
  18. Now simplify each entry:
    \[a-(-a)=a+a=2a,\]
    \[b-(-b)=b+b=2b,\]
    \[-a-a=-2a,\]
    \[c-(-c)=c+c=2c,\]
    \[-b-b=-2b,\]
    \[-c-c=-2c\]
  19. Therefore,
    \[A-A^T=\begin{bmatrix}0&2a&2b\\-2a&0&2c\\-2b&-2c&0\end{bmatrix}\]
  20. Now multiply by \(\frac{1}{2}\):
    \[\frac{1}{2}(A-A^T)=\frac{1}{2}\begin{bmatrix}0&2a&2b\\-2a&0&2c\\-2b&-2c&0\end{bmatrix}\]
  21. Dividing every element by \(2\), we obtain
    \[\frac{1}{2}(A-A^T)=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}\]
  22. Hence,
    \[\frac{1}{2}(A-A^T)=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}\]
  23. Therefore, the skew-symmetric part of \(A\) is \(A\) itself.
🎯 Exam Significance
Exam Significance

This question is important because it tests the decomposition of a matrix into its symmetric and skew-symmetric components, a standard concept in NCERT Class 12 Mathematics.

  • Students should remember
    \[ \boxed{\frac{1}{2}(A+A^T)} \]
    as the symmetric part of \(A\).
  • Students should remember
    \[ \boxed{\frac{1}{2}(A-A^T)} \]
    as the skew-symmetric part of \(A\).
  • If \(A\) is already skew-symmetric, its symmetric part is automatically the zero matrix.
  • If \(A\) is already symmetric, its skew-symmetric part is automatically the zero matrix.
  • Writing every matrix subtraction explicitly avoids sign errors, especially in expressions such as \(a-(-a)\).
  • The final decomposition should satisfy
    \[ A=\frac12(A+A^T)+\frac12(A-A^T). \]
Significance for Competitive Entrance Examinations

This problem contains a useful shortcut for objective examinations. Before performing lengthy calculations, inspect the matrix for symmetry or skew-symmetry.

Here,

\[ A^T=-A. \]

Therefore, immediately,

\[ \frac12(A+A^T)=\frac12(A-A)=O \]

and

\[ \frac12(A-A^T)=\frac12(A+A)=A. \]

Thus, the answer can be obtained almost instantly once the structural property of the matrix is recognised.

The decomposition formula is also useful in problems asking for the symmetric or skew-symmetric component of a complicated matrix:

\[ \boxed{ A= \underbrace{\frac12(A+A^T)}_{\text{symmetric part}} + \underbrace{\frac12(A-A^T)}_{\text{skew-symmetric part}} }\]
🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  7 points
  1. The symmetric component of any square matrix \(A\) is

    \[ \boxed{\frac12(A+A^T)}. \]

  2. The skew-symmetric component of any square matrix \(A\) is

    \[ \boxed{\frac12(A-A^T)}. \]

  3. The given matrix satisfies

    \[ \boxed{A^T=-A}. \]

  4. Therefore, \(A\) is already skew-symmetric.

  5. Its symmetric component is the zero matrix:

    \[ \boxed{\frac12(A+A^T)=O}. \]

  6. Its skew-symmetric component is the matrix itself:

    \[ \boxed{\frac12(A-A^T)=A}. \]

  7. Every square matrix can be written as

    \[ \boxed{ A=\frac12(A+A^T)+\frac12(A-A^T) }. \]

← Q8
9 / 12  ·  75%
Q10 →
Q10
NUMERIC3 marks

Express each of the following matrices as the sum of a symmetric matrix and a skew-symmetric matrix:

(i)

\[\begin{bmatrix}3&5\\1&-1\end{bmatrix}\]

(ii)

\[\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}\]

(iii)

\[\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}\]

(iv)

\[\begin{bmatrix}1&5\\-1&2\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory

Every square matrix \(A\) can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix.

\[\boxed{A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)}\]

Here,

\[\boxed{S=\frac{1}{2}(A+A^T)}\]

is the symmetric part of \(A\), because

\[S^T=S,\]

and

\[\boxed{K=\frac{1}{2}(A-A^T)}\]

is the skew-symmetric part of \(A\), because

\[^T=-K.\]

Therefore, to express any given square matrix as the sum of a symmetric and a skew-symmetric matrix, we use the standard decomposition:

\[\boxed{A=\underbrace{\frac{1}{2}(A+A^T)}_{\text{Symmetric matrix}}+\underbrace{\frac{1}{2}(A-A^T)}_{\text{Skew-symmetric matrix}}}\]

This formula should be remembered as a standard result for Class 12 board examinations and competitive entrance examinations.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Denote the given matrix by \(A\).

  2. Find its transpose \(A^T\).

  3. Calculate

    \[ S=\frac{1}{2}(A+A^T). \]

  4. Calculate

    \[ K=\frac{1}{2}(A-A^T). \]

  5. Verify that \(S^T=S\), so \(S\) is symmetric.

  6. Verify that \(K^T=-K\), so \(K\) is skew-symmetric.

  7. Finally, verify that

    \[ A=S+K. \]

✏️ Solution
(i) Express the Matrix as the Sum of a Symmetric and a Skew-Symmetric Matrix
Step-by-step Solution  ·  7 steps
  1. Given
    \[A=\begin{bmatrix}3&5\\1&-1\end{bmatrix}\]
  2. Find \(A^T\)
    \[A^T=\begin{bmatrix}3&1\\5&-1\end{bmatrix}\]
  3. Find the symmetric part
    \[S=\frac{1}{2}(A+A^T)\]
    \[=\frac{1}{2}\left[\begin{bmatrix}3&5\\1&-1\end{bmatrix}+\begin{bmatrix}3&1\\5&-1\end{bmatrix}\right]\]
    \[=\frac{1}{2}\begin{bmatrix}3+3&5+1\\1+5&-1+(-1)\end{bmatrix}\]
    \[=\frac{1}{2}\begin{bmatrix}6&6\\6&-2\end{bmatrix}\]
    \[S=\begin{bmatrix}3&3\\3&-1\end{bmatrix}\]
  4. This matrix is symmetric because
    \[S^T=\begin{bmatrix}3&3\\3&-1\end{bmatrix}=S\]
  5. Find the skew-symmetric part
    \[K=\frac{1}{2}(A-A^T)\]
    \[=\frac{1}{2}\left[\begin{bmatrix}3&5\\1&-1\end{bmatrix}-\begin{bmatrix}3&1\\5&-1\end{bmatrix}\right]\]
    \[=\frac{1}{2}\begin{bmatrix}3-3&5-1\\1-5&-1-(-1)\end{bmatrix}\]
    \[=\frac{1}{2}\begin{bmatrix}0&4\\-4&0\end{bmatrix}\]
    \[\boxed{K=\begin{bmatrix}0&2\\-2&0\end{bmatrix}}\]
  6. Also,
    \[K^T=\begin{bmatrix}0&-2\\2&0\end{bmatrix}=-K\]
    Hence \(K\) is skew-symmetric.
  7. Express \(A\) as \(S+K\)
    \[S+K=\begin{bmatrix}3&3\\3&-1\end{bmatrix}+\begin{bmatrix}0&2\\-2&0\end{bmatrix}\]
    \[=\begin{bmatrix}3&5\\1&-1\end{bmatrix}=A.\]
  8. Therefore,
    \[A=\begin{bmatrix}3&3\\3&-1\end{bmatrix}+\begin{bmatrix}0&2\\-2&0\end{bmatrix}\]
    where the first matrix is symmetric and the second is skew-symmetric.
✏️ Solution
(ii) Express the Matrix as the Sum of a Symmetric and a Skew-Symmetric Matrix
Step-by-step Solution  ·  10 steps
  1. Given
    \[A=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}\]
  2. Find \(A^T\)
    \[A^T=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}\]
  3. Thus,
    \[\boxed{A^T=A}\]
    Therefore, the given matrix is already symmetric.
  4. Find the symmetric part
    \[S=\frac{1}{2}(A+A^T)\]
  5. Since \(A^T=A\),
    \[S=\frac{1}{2}(A+A)\]
    \[=\frac{1}{2}(2A)\]
    \[=A\]
  6. Hence,
    \[ S=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}\]
  7. Find the skew-symmetric part
    \[K=\frac{1}{2}(A-A^T)\]
  8. Since \(A^T=A\),
    \[K=\frac{1}{2}(A-A)\]
    \[=\frac{1}{2}O\]
    \[=O\]
  9. Therefore,
    \[K=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\]
  10. Express \(A\) as \(S+K\)
    \[S+K=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}+\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\]
    \[=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}=A\]
  11. Therefore,
    \[A=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}+\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\]
    where the first matrix is symmetric and the second matrix is skew-symmetric.
✏️ Solution
(iii) Express the Matrix as the Sum of a Symmetric and a Skew-Symmetric Matrix
Step-by-step Solution  ·  9 steps
  1. Given
    \[A=\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}\]
  2. Find \(A^T\)
    \[A^T=\begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}\]
  3. Find the symmetric part
    \[S=\frac{1}{2}(A+A^T)\]
    \[=\frac{1}{2}\left[\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}+\begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}\right]\]
    \[=\frac{1}{2}\begin{bmatrix}3+3&3-2&-1-4\\-2+3&-2-2&1-5\\-4-1&-5+1&2+2\end{bmatrix}\]
    \[=\frac{1}{2}\begin{bmatrix}6&1&-5\\1&-4&-4\\-5&-4&4\end{bmatrix}\]
  4. Therefore,
    \[S=\begin{bmatrix}3&\frac12&-\frac52\\\frac12&-2&-2\\-\frac52&-2&2\end{bmatrix}\]
  5. Observe that
    \[S^T=\begin{bmatrix}3&\frac12&-\frac52\\\frac12&-2&-2\\-\frac52&-2&2\end{bmatrix}=S\]
    Hence \(S\) is symmetric.
  6. Find the skew-symmetric part
    \[K=\frac{1}{2}(A-A^T)\]
    \[=\frac{1}{2}\left[\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}-\begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}\right]\]
    \[=\frac{1}{2}\begin{bmatrix}3-3&3-(-2)&-1-(-4)\\-2-3&-2-(-2)&1-(-5)\\-4-(-1)&-5-1&2-2\end{bmatrix}\]
    \[=\frac{1}{2}\begin{bmatrix}0&5&3\\-5&0&6\\-3&-6&0\end{bmatrix}\]
  7. Therefore,
    \[\boxed{K=\begin{bmatrix}0&\frac52&\frac32\\-\frac52&0&3\\-\frac32&-3&0\end{bmatrix}}\]
  8. Now,
    \[K^T=\begin{bmatrix}0&-\frac52&-\frac32\\\frac52&0&-3\\\frac32&3&0\end{bmatrix}=-K\]
    Hence \(K\) is skew-symmetric.
  9. Express \(A\) as \(S+K\)
    \[S+K=\begin{bmatrix}3&\frac12&-\frac52\\\frac12&-2&-2\\-\frac52&-2&2\end{bmatrix}+\begin{bmatrix}0&\frac52&\frac32\\-\frac52&0&3\\-\frac32&-3&0\end{bmatrix}\]
    \[=\begin{bmatrix}3&\frac12+\frac52&-\frac52+\frac32\\\frac12-\frac52&-2&-2+3\\-\frac52-\frac32&-2-3&2\end{bmatrix}\]
    \[=\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}=A\]
  10. Hence,
    \[A=\begin{bmatrix}3&\frac12&-\frac52\\\frac12&-2&-2\\-\frac52&-2&2\end{bmatrix}+\begin{bmatrix}0&\frac52&\frac32\\-\frac52&0&3\\-\frac32&-3&0\end{bmatrix}\]
    where the first matrix is symmetric and the second is skew-symmetric.
✏️ Solution
(iv) Express the Matrix as the Sum of a Symmetric and a Skew-Symmetric Matrix
Step-by-step Solution  ·  9 steps
  1. Given
    \[A=\begin{bmatrix}1&5\\-1&2\end{bmatrix}\]
  2. Find \(A^T\)
    \[A^T=\begin{bmatrix}1&-1\\5&2\end{bmatrix}\]
  3. Find the symmetric part
    \[S=\frac{1}{2}(A+A^T)\]
    \[=\frac{1}{2}\left[\begin{bmatrix}1&5\\-1&2\end{bmatrix}+\begin{bmatrix}1&-1\\5&2\end{bmatrix}\right]\]
    \[=\frac{1}{2}\begin{bmatrix}1+1&5-1\\-1+5&2+2\end{bmatrix}\]
    \[=\frac{1}{2}\begin{bmatrix}2&4\\4&4\end{bmatrix}\]
  4. Therefore,
    \[S=\begin{bmatrix}1&2\\2&2\end{bmatrix}\]
  5. Clearly,
    \[S^T=\begin{bmatrix}1&2\\2&2\end{bmatrix}=S.\]
    Hence \(S\) is symmetric.
  6. Find the skew-symmetric part
    \[K=\frac{1}{2}(A-A^T)\]
    \[=\frac{1}{2}\left[\begin{bmatrix}1&5\\-1&2\end{bmatrix}-\begin{bmatrix}1&-1\\5&2\end{bmatrix}\right]\]
    \[=\frac{1}{2}\begin{bmatrix}1-1&5-(-1)\\-1-5&2-2\end{bmatrix}\]
    \[=\frac{1}{2}\begin{bmatrix}0&6\\-6&0\end{bmatrix}\]
  7. Therefore,
    \[ K=\begin{bmatrix}0&3\\-3&0\end{bmatrix}\]
  8. Also,
    \[K^T=\begin{bmatrix}0&-3\\3&0\end{bmatrix}=-K\]
    Hence \(K\) is skew-symmetric.
  9. Express \(A\) as \(S+K\)
    \[S+K=\begin{bmatrix}1&2\\2&2\end{bmatrix}+\begin{bmatrix}0&3\\-3&0\end{bmatrix}\]
    \[=\begin{bmatrix}1&5\\-1&2\end{bmatrix}=A\]
  10. Therefore,
    \[A=\begin{bmatrix}1&2\\2&2\end{bmatrix}+\begin{bmatrix}0&3\\-3&0\end{bmatrix}\]
    where the first matrix is symmetric and the second is skew-symmetric.
🎯 Exam Significance
Exam Significance

This question is a direct application of the fundamental decomposition theorem for square matrices. It is important for board examinations because students are expected to perform transpose, matrix addition and subtraction, scalar multiplication, and identify symmetry properties accurately.

  • Memorise the decomposition:
    \[ \boxed{A=\frac12(A+A^T)+\frac12(A-A^T)}. \]
  • Always calculate \(A^T\) carefully before proceeding.
  • The first component is always symmetric.
  • The second component is always skew-symmetric.
  • If the given matrix is already symmetric, the skew-symmetric component is \(O\).
  • If the given matrix is already skew-symmetric, the symmetric component is \(O\).
  • For full board marks, do not merely state the two matrices; show how they are obtained and, where appropriate, verify their symmetry properties.
Significance for Competitive Entrance Examinations

For competitive examinations, the key advantage is that the decomposition can be obtained mechanically from two standard expressions:

\[ \boxed{S=\frac12(A+A^T)} \]
\[ \boxed{K=\frac12(A-A^T)}. \]

Several shortcuts follow immediately:

  • If \(A^T=A\), then \(S=A\) and \(K=O\).
  • If \(A^T=-A\), then \(S=O\) and \(K=A\).
  • The diagonal entries of every skew-symmetric matrix are zero.
  • The diagonal entries of the symmetric component are exactly the diagonal entries of the original matrix.
  • For an \(n\times n\) matrix, the symmetric and skew-symmetric parts have complementary off-diagonal structures.

Recognising these patterns can substantially reduce calculation time in objective questions.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  6 points
  1. Every square matrix \(A\) can be uniquely expressed as

    \[ \boxed{ A=S+K } \]
    where \(S\) is symmetric and \(K\) is skew-symmetric.

  2. The symmetric component is

    \[ \boxed{ S=\frac12(A+A^T) }. \]

  3. The skew-symmetric component is

    \[ \boxed{ K=\frac12(A-A^T) }. \]

  4. For Question 10(ii), the matrix is already symmetric, so

    \[ \boxed{S=A,\quad K=O}. \]

  5. For Questions 10(i), 10(iii), and 10(iv), both components are non-zero.

  6. The decomposition provides a systematic method that works for every square matrix.

← Q9
10 / 12  ·  83%
Q11 →
Q11
NUMERIC3 marks

Choose the correct answer:

If \(A\) and \(B\) are symmetric matrices of the same order, then \(AB-BA\) is a:

  1. Skew-symmetric matrix
  2. Symmetric matrix
  3. Zero matrix
  4. Identity matrix
📘 Concept & Theory
Concept/Theory

A square matrix \(A\) is symmetric if

\[A^T=A.\]

Similarly, a square matrix \(K\) is skew-symmetric if

\[K^T=-K.\]

For the given matrices \(A\) and \(B\), we are told that both are symmetric. Therefore,

\[A^T=A\]
\[B^T=B.\]

The important transpose property for a product of matrices is

\[(AB)^T=B^TA^T.\]

Notice carefully that the order of multiplication reverses on taking the transpose.

🗺️ Solution Roadmap
Step-by-step Plan
  1. Let

    \[C=AB-BA.\]

  2. Find \(C^T\).

  3. Use

    \[(AB)^T=B^TA^T\]
    and
    \[(BA)^T=A^TB^T.\]

  4. Use the fact that \(A\) and \(B\) are symmetric.

  5. Show that

    \[C^T=-C.\]

  6. Therefore, \(C=AB-BA\) is skew-symmetric.

✏️ Solution
Complete Solution
Step-by-step Solution  ·  10 steps
  1. Let
    \[C=AB-BA\]
  2. Taking transpose on both sides,
    \[C^T=(AB-BA)^T\]
  3. Using the property
    \[(X-Y)^T=X^T-Y^T\]
  4. we get
    \[C^T=(AB)^T-(BA)^T\]
  5. Using the transpose property of a product,
    \[(AB)^T=B^TA^T\]
    and
    \[(BA)^T=A^TB^T\]
  6. Therefore,
    \[C^T=B^TA^T-A^TB^T\]
  7. Since \(A\) and \(B\) are symmetric matrices,
    \[A^T=A\]
    \[B^T=B\]
  8. Substituting these results,
    \[C^T=BA-AB\]
  9. Rearranging,
    \[C^T=-(AB-BA)\]
  10. But
    \[C=AB-BA\]
  11. Hence,
    \[\boxed{C^T=-C}\]
  12. Therefore, \(C\) is skew-symmetric.
    \[\boxed{AB-BA\text{ is a skew-symmetric matrix}}\]
💡 Answer
Final Answer

Option (a): Skew-symmetric matrix

\[\boxed{\text{AB - BA is a skew-symmetric matrix}}\]
🎯 Exam Significance
Exam Significance

This is a high-value conceptual question because it combines three important matrix properties:

  1. Transpose of a sum or difference:
    \[(A-B)^T=A^T-B^T.\]
  2. Transpose of a product:
    \[(AB)^T=B^TA^T.\]
  3. Definition of a skew-symmetric matrix:
    \[A^T=-A\]

For competitive entrance examinations, the fastest approach is to take the transpose immediately and use \(A^T=A\) and \(B^T=B\). This avoids unnecessary numerical calculations.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  5 points
  1. If \(A\) and \(B\) are symmetric, then

    \[\boxed{AB-BA}\]
    is always skew-symmetric.

  2. The key identity is

    \[\boxed{(AB)^T=B^TA^T}.\]

  3. Matrix multiplication is generally non-commutative:

    \[AB\neq BA.\]

  4. Therefore, \(AB-BA\) is not necessarily the zero matrix.

  5. The quickest test is:

    \[\boxed{(AB-BA)^T=-(AB-BA)}.\]

← Q10
11 / 12  ·  92%
Q12 →
Q12
NUMERIC3 marks

Choose the correct answer.

If

\[A=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\]

and

\[A+A^T=I,\]

then the value of \(\alpha\) is:

  1. \(\dfrac{\pi}{6}\)
  2. \(\dfrac{\pi}{3}\)
  3. \(\pi\)
  4. \(\dfrac{3\pi}{2}\)

Note: The second diagonal entry in the supplied matrix appears as \(\cos 2\). It is understood as \(\cos\alpha\), consistent with the given rotation-type matrix and the intended NCERT question.

📘 Concept & Theory
Concept/Theory

The transpose of a matrix is obtained by interchanging its rows and columns.

For

\[A=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix},\]

its transpose is

\[A^T=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}.\]

Therefore,

\[A+A^T=\begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha\end{bmatrix}.\]

The identity matrix of order \(2\) is

\[I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\]

Hence, comparing corresponding entries gives

\[2\cos\alpha=1\]

Thus,

\[\cos\alpha=\frac12\]

From the standard trigonometric values,

\[\cos\frac{\pi}{3}=\frac12\]

Among the given options,

\[\boxed{\alpha=\frac{\pi}{3}}\]
🗺️ Solution Roadmap
Step-by-step Plan
  1. Write the given matrix \(A\).

  2. Find its transpose \(A^T\).

  3. Calculate \(A+A^T\).

  4. Equate \(A+A^T\) with the identity matrix \(I\).

  5. Compare the diagonal elements to obtain the trigonometric equation.

  6. Use the standard value of \(\cos\alpha\) to identify the correct option.

✏️ Solution
Complete Solution
Step-by-step Solution  ·  12 steps
  1. Given
    \[A=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\]
  2. Find \(A^T\)
  3. On interchanging rows and columns,
    \[A^T=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}\]
  4. Calculate \(A+A^T\)
    \[A+A^T=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}+\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}\]
  5. Adding corresponding elements,
    \[A+A^T=\begin{bmatrix}\cos\alpha+\cos\alpha&-\sin\alpha+\sin\alpha\\\sin\alpha-\sin\alpha&\cos\alpha+\cos\alpha\end{bmatrix}\]
    \[A+A^T=\begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha\end{bmatrix}\]
  6. Use the given condition
    \[A+A^T=I\]
  7. Since
    \[I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\]
  8. we have
    \[\begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\]
  9. Comparing the \((1,1)\)-entries
    \[2\cos\alpha=1\]
  10. Therefore,
    \[\cos\alpha=\frac12\]
  11. Identify the value of \(\alpha\)
  12. We know that
    \[\cos\frac{\pi}{3}=\frac12\]
  13. Therefore, from the given options,
    \[\boxed{\alpha=\frac{\pi}{3}}\]
💡 Answer
Final Answer

Option (b): \(\dfrac{\pi}{3}\)

\[\boxed{\alpha=\frac{\pi}{3}}\]
🎯 Exam Significance
Exam Significance

This question tests several fundamental skills from the chapter: finding the transpose, adding matrices, identifying the identity matrix, and comparing corresponding entries of two matrices.

  • Remember that \(A^T\) is obtained by interchanging rows and columns.
  • When two matrices are equal, their corresponding entries are equal.
  • For an identity matrix of order \(2\),
    \[ I= \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}. \]
  • Writing the matrix comparison step explicitly helps secure method marks in a descriptive examination.
Significance for Competitive Entrance Examinations

This is an excellent quick-response problem. The cancellation of the off-diagonal terms can be recognised immediately:

\[ -\sin\alpha+\sin\alpha=0 \]
\[ \sin\alpha-\sin\alpha=0. \]

Thus, only the diagonal condition remains:

\[ 2\cos\alpha=1. \]

This reduces the matrix problem to a standard trigonometric value in just a few steps.

🔑 Key Takeaways
Key Takeaways
Key Takeaways  ·  5 points
  1. For the given matrix,

    \[A^T=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}\]

  2. \[A+A^T=\begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha\end{bmatrix}.\]

  3. The condition \(A+A^T=I\) gives

    \[\boxed{\cos\alpha=\frac12}.\]

  4. Among the given options,

    \[\boxed{\alpha=\frac{\pi}{3}}.\]

  5. The most efficient strategy is to compare the diagonal entries after calculating \(A+A^T\).

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Exercise 3.3 of NCERT Class 12 Mathematics Chapter 3, Matrices, focuses on the important concepts of transpose, symmetric and skew-symmetric matrices, and their fundamental properties. These concepts form an essential part of matrix algebra and are frequently tested in CBSE Board examinations as well as competitive entrance exams. The exercise includes problems involving matrix transposition, addition and subtraction of matrices, transpose of a product, verification of symmetric and…
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    Frequently Asked Questions

    Exercise 3.3 covers transpose of matrices, properties of transpose, symmetric and skew-symmetric matrices, and related matrix identities.

    The transpose of a matrix is obtained by interchanging its rows and columns. If A is a matrix, its transpose is denoted by A?.

    A square matrix A is called symmetric if A? = A. Its elements are symmetric about the principal diagonal.

    A square matrix A is called skew-symmetric if A? = -A. Every diagonal element of a skew-symmetric matrix is zero.

    For matrices A and B of the same order, (A + B)? = A? + B?.

    For matrices A and B of the same order, (A - B)? = A? - B?.

    If AB is defined, then (AB)? = B?A?. The order of the matrices reverses when taking the transpose.

    If A and B are symmetric matrices of the same order, then AB - BA is a skew-symmetric matrix because (AB - BA)? = -(AB - BA).

    Any square matrix A can be expressed as A = ½(A + A?) + ½(A - A?), where ½(A + A?) is symmetric and ½(A - A?) is skew-symmetric.

    Exercise 3.3 builds strong command over transpose, symmetric and skew-symmetric matrices and their properties, which are frequently useful in CBSE Board, JEE and other entrance examinations.

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