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Chapter 3 Miscellaneous Exercise Solutions

Matrices

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

Class 12 Mathematics Miscellaneous Exercise NCERT Solutions Matrices Class 12 Mathematics Chapter 3 CBSE Board Exam JEE Main CUET Matrix Algebra Types of Matrices Symmetric Matrices Skew-Symmetric Matrices Transpose of a Matrix Matrix Multiplication Identity Matrix Matrix Equations Idempotent Matrix
11 Questions
25–35 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks
If A and B are symmetric matrices, prove that AB − BA is a skew-symmetric matrix.
📘 Concept & Theory
Concept/Theory
›

This problem uses the fundamental properties of transpose of a matrix and the definitions of symmetric and skew-symmetric matrices.

1. Symmetric Matrix

A square matrix A is called symmetric if its transpose is equal to the matrix itself:

\[A^T=A\]

Similarly, if B is symmetric, then:

\[B^T=B\]

2. Transpose of a Product

For any two matrices A and B for which the product is defined, the transpose of their product is given by:/p>

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

Notice carefully that the order of multiplication is reversed when taking the transpose.

3. Transpose of a Difference

For matrices of the same order,

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

4. Skew-Symmetric Matrix

A square matrix M is called skew-symmetric if:

\[M^T=-M\]

Therefore, to prove that

\[AB-BA\]
is skew-symmetric, it is sufficient to prove that its transpose is equal to its negative.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the matrix to be tested be \(M=AB-BA\).

  2. Take the transpose of \(M\).

  3. Apply the transpose-of-a-difference property.

  4. Apply the transpose-of-a-product property.

  5. Use the fact that \(A\) and \(B\) are symmetric, so \(A^T=A\) and \(B^T=B\).

  6. Simplify the resulting expression.

  7. Show that \(M^T=-M\).

  8. Conclude that \(M=AB-BA\) is skew-symmetric.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  26 steps
  1. Let
    \[A \text{ and } B\]
    be symmetric matrices.
  2. Since \(A\) and \(B\) are symmetric matrices, by definition:
    \[A^T=A\]
    and
    \[B^T=B\]
  3. We have to prove that:
    \[AB-BA\]
    is a skew-symmetric matrix.
  4. Let the given matrix be \(M\),
  5. Put
    \[M=AB-BA\]
  6. Our objective is to prove:
    \[M^T=-M\]
  7. If this relation is established, then \(M\) will be skew-symmetric by definition.
  8. Take the transpose of \(M\)
  9. Since
    \[M=AB-BA,\]
    taking transpose on both sides gives:
    \[M^T=(AB-BA)^T\]
  10. Apply the transpose-of-a-difference property
  11. Using
    \[(X-Y)^T=X^T-Y^T,\]
    we obtain:
    \[M^T=(AB)^T-(BA)^T\]
  12. Apply the transpose-of-a-product property
  13. We know that:
    \[(AB)^T=B^TA^T\]
    and
    \[(BA)^T=A^TB^T\]
  14. Therefore:
    \[M^T=B^TA^T-A^TB^T\]
  15. Use the symmetry of \(A\) and \(B\)
  16. Since \(A\) and \(B\) are symmetric:
    \[A^T=A\]
    and
    \[B^T=B\]
  17. Substituting these relations into the expression for \(M^T\), we get:
    \[M^T=BA-AB\]
  18. Factor out the negative sign
  19. We can rewrite \(BA-AB\) as:
    \[BA-AB=-(AB-BA)\]
  20. Hence:
    \[M^T=-(AB-BA)\]
  21. Substitute \(M=AB-BA\)
  22. Since
    \[M=AB-BA,\]
    we have:
    \[M^T=-M\]
  23. Apply the definition of a skew-symmetric matrix
  24. A square matrix \(M\) is skew-symmetric if:
    \[M^T=-M\]
  25. We have proved exactly this condition for
    \[M=AB-BA\]
  26. Therefore,
    \[\boxed{AB-BA\text{ is a skew-symmetric matrix.}}\]
🎯 Exam Significance
Exam Significance
›

This result is important because it combines several high-frequency concepts from the chapter Matrices in a single proof: transpose, symmetric matrices, matrix multiplication, and skew-symmetric matrices.

For CBSE Board Examinations
  • The question tests whether you can correctly use the identity
    \[ (AB)^T=B^TA^T. \]
  • It tests the definition of a symmetric matrix:
    \[ A^T=A. \]
  • It tests the definition of a skew-symmetric matrix:
    \[ A^T=-A. \]
  • The reversal of order in the transpose of a product is a particularly important scoring step.
  • A complete proof should explicitly establish
    \[ (AB-BA)^T=-(AB-BA). \]
For JEE and Other Competitive Entrance Examinations
  • The expression \(AB-BA\) is related to the commutator of two matrices and frequently appears in matrix-property questions.
  • The result is useful for quickly identifying whether a matrix expression is symmetric or skew-symmetric.
  • The key observation is that transposition reverses the order:
    \[ (AB)^T=B^TA^T. \]
  • If \(A\) and \(B\) are symmetric, then:
    \[ (AB)^T=BA. \]
    Therefore \(AB\) itself need not be symmetric unless \(AB=BA\).
  • The expression \(AB-BA\) automatically becomes skew-symmetric when both \(A\) and \(B\) are symmetric.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. For a symmetric matrix \(A\):

    \[ A^T=A. \]

  2. For a skew-symmetric matrix \(A\):

    \[ A^T=-A. \]

  3. The transpose of a product reverses the order:

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

  4. If \(A\) and \(B\) are symmetric, then:

    \[ (AB)^T=BA. \]

  5. Consequently:

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

  6. Since

    \[ BA-AB=-(AB-BA), \]
    the matrix \(AB-BA\) is skew-symmetric.

  7. The most important exam test is:

    \[ \boxed{M^T=-M\Rightarrow M\text{ is skew-symmetric}.} \]
↑ Top
1 / 11  ·  9%
Q2 →
Q2
NUMERIC3 marks

Show that the matrix \(B'AB\) is symmetric or skew-symmetric according as \(A\) is symmetric or skew-symmetric.

Here, \(B'\) denotes the transpose of matrix \(B\), that is,

\[B'=B^T\]

📘 Concept & Theory
Concept/Theory
›

This question is based on the relationship between transpose, symmetric matrices, and skew-symmetric matrices. The central expression is

\[B^TAB\]
Such an expression is commonly called a congruence-type matrix expression.

1. Symmetric Matrix

A square matrix \(A\) is symmetric if:

\[A^T=A\]

2. Skew-Symmetric Matrix

A square matrix \(A\) is skew-symmetric if:

\[A^T=-A\]

3. Transpose of a Product

For matrices for which the products are defined:

\[(XYZ)^T=Z^TY^TX^T\]

Therefore, when we take the transpose of

\[B^TAB\]
the order of the matrices must be reversed.

4. Main Idea of the Proof

Let

\[C=B^TAB\]
We calculate \(C^T\):

\[C^T=(B^TAB)^T\]

On taking the transpose and reversing the order:

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

Since

\[(B^T)^T=B\]
we obtain:

\[C^T=B^TA^TB\]

Thus, the nature of \(C\) depends directly on whether \(A^T=A\) or \(A^T=-A\)

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(C=B^TAB\).

  2. Take the transpose of \(C\).

  3. Use the rule for the transpose of a product.

  4. Use the property \((B^T)^T=B\).

  5. Obtain \(C^T=B^TA^TB\).

  6. If \(A\) is symmetric, substitute \(A^T=A\).

  7. Show that \(C^T=C\), proving that \(C\) is symmetric.

  8. If \(A\) is skew-symmetric, substitute \(A^T=-A\).

  9. Show that \(C^T=-C\), proving that \(C\) is skew-symmetric.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  30 steps
  1. Let
    \[C=B^TAB\]
  2. We shall prove separately that \(C\) is symmetric when \(A\) is symmetric and \(C\) is skew-symmetric when \(A\) is skew-symmetric.
  3. Take the transpose of \(C\) From
    \[C=B^TAB\]
    taking transpose on both sides gives:
    \[C^T=(B^TAB)^T\]
  4. Apply the transpose-of-a-product rule
  5. The expression \(B^TAB\) is a product of three matrices:
    \[B^T,\quad A,\quad B\]
  6. Using
    \[(XYZ)^T=Z^TY^TX^T\]
    we get:
    \[C^T=B^TA^T(B^T)^T\]
  7. Simplify the double transpose
  8. We know that:
    \[(B^T)^T=B\]
  9. Therefore:
    \[C^T=B^TA^TB\]
  10. This is the key result from which both cases follow.
  11. Case I: \(A\) is symmetric
  12. Suppose \(A\) is symmetric.
  13. By definition of a symmetric matrix:
    \[A^T=A\]
  14. Substitute \(A^T=A\)
  15. We have already obtained:
    \[C^T=B^TA^TB\]
  16. Substituting
    \[A^T=A\]
    we get:
    \[C^T=B^TAB\]
  17. Compare with \(C\)
  18. Since
    \[C=B^TAB,\]
    we have:
    \[C^T=C\]
  19. By the definition of a symmetric matrix:
    \[C^T=C\quad\Rightarrow\quad C\text{ is symmetric}\]
  20. Therefore, if \(A\) is symmetric:
    \[\boxed{B^TAB\text{ is symmetric}.}\]
  21. Case II: \(A\) is skew-symmetric
  22. Now suppose \(A\) is skew-symmetric.
  23. By definition of a skew-symmetric matrix:
    \[A^T=-A\]
  24. Substitute \(A^T=-A\)
  25. We have:
    \[C^T=B^TA^TB\]
  26. Substituting
    \[A^T=-A,\]
    we get:
    \[C^T=B^T(-A)B\]
  27. Taking the scalar \(-1\) outside:
    \[C^T=-B^TAB\]
  28. Compare with \(C\)
  29. Since
    \[C=B^TAB,\]
    we obtain:
    \[C^T=-C\]
  30. By the definition of a skew-symmetric matrix:
    \[C^T=-C\quad\Rightarrow\quad C\text{ is skew-symmetric}\]
  31. Therefore, if \(A\) is skew-symmetric:
    \[\boxed{B^TAB\text{ is skew-symmetric}.}\]
  32. Final Conclusion
  33. Combining both cases:
    \[\boxed{B^TAB\text{ is symmetric if }A\text{ is symmetric, and skew-symmetric if }A\text{ is skew-symmetric.}}\]
🎯 Exam Significance
Exam Significance
›
  • This question tests the fundamental definitions of symmetric and skew-symmetric matrices.
  • The identity
    \[(ABC)^T=C^TB^TA^T\]
    is an important scoring step and should not be skipped.
  • The result provides a standard proof pattern: calculate the transpose of the given matrix expression and compare it with the original expression.
  • For a complete answer, clearly establish either
    \[C^T=C\]
    or
    \[C^T=-C.\]
For JEE and Other Competitive Entrance Examinations
  • The expression \(B^TAB\) is a standard matrix transformation that preserves the symmetric/skew-symmetric character of \(A\).
  • The problem is useful for questions involving matrix properties, transpose operations, and algebraic simplification.
  • The most important speed-building observation is:
    \[ (B^TAB)^T=B^TA^TB. \]
  • Therefore, one only needs to replace \(A^T\) by \(A\) or \(-A\), depending on the nature of \(A\).
  • This result can be applied directly in objective questions without expanding the matrices element by element.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Always remember:

    \[ (ABC)^T=C^TB^TA^T. \]

  2. The double-transpose property is:

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

  3. For

    \[ C=B^TAB, \]
    its transpose is:
    \[ C^T=B^TA^TB. \]

  4. If \(A\) is symmetric:

    \[ A^T=A \]
    and consequently:
    \[ C^T=C. \]
    Hence \(C\) is symmetric.

  5. If \(A\) is skew-symmetric:

    \[ A^T=-A \]
    and consequently:
    \[ C^T=-C. \]
    Hence \(C\) is skew-symmetric.

  6. The transformation

    \[ A\longmapsto B^TAB \]
    preserves the symmetric or skew-symmetric nature of \(A\).

← Q1
2 / 11  ·  18%
Q3 →
Q3
NUMERIC3 marks

Find the values of \(x,\ y,\ z\), if the matrix

\[A=\begin{bmatrix}0 & 2y & z\\x & y & -z\\x & -y & z\end{bmatrix}\]

satisfies the equation

\[A'A=I\]

Here, \(A'\) denotes the transpose of \(A\), so \[A'=A^T\]

📘 Concept & Theory
Concept/Theory
›
1. Transpose of a Matrix

The transpose of a matrix is obtained by interchanging its rows and columns. If \(A\) is a matrix, its transpose is denoted by \(A^T\) or \(A'\).

For the given matrix,

\[A=\begin{bmatrix}0 & 2y & z\\x & y & -z\\x & -y & z\end{bmatrix},\]
its transpose is:

\[A^T=\begin{bmatrix}0 & x & x\\2y & y & -y\\z & -z & z\end{bmatrix}.\]

2. Orthogonal Matrix Condition

A square matrix \(A\) is called an orthogonal matrix if:

\[ A^TA=I. \]

Thus, the given condition

\[A'A=I\]
means that \(A\) is an orthogonal matrix.

Since \(A'A=A^TA\), we need to calculate \(A^TA\) and compare it with the identity matrix:

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

3. Important Observation

Instead of multiplying every entry blindly, we can interpret the entries of \(A^TA\) as dot products of the rows of \(A^T\), or equivalently, dot products of the columns of \(A\).

The condition

\[ A^TA=I \]
means that the columns of \(A\) are mutually orthonormal. Therefore:

  • Each column must have length \(1\).
  • The dot product of any two distinct columns must be \(0\).

In this problem, the off-diagonal entries automatically become zero, while the diagonal entries provide the three equations needed to determine \(x,\ y,\ z\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given matrix \(A\).

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

  3. Form the product \(A^TA\).

  4. Calculate the first row of \(A^TA\) completely.

  5. Calculate the second row of \(A^TA\) completely.

  6. Calculate the third row of \(A^TA\) completely.

  7. Compare the resulting matrix with the identity matrix \(I\).

  8. Equate corresponding diagonal entries to obtain equations in \(x,\ y,\ z\).

  9. Solve each equation and obtain all possible values, including both positive and negative roots.

  10. State the final values of \(x,\ y,\ z\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  29 steps
  1. Given — matrix
    \[A=\begin{bmatrix}0 & 2y & z\\x & y & -z\\x & -y & z\end{bmatrix}\]
  2. The condition given in the question is:
    \[A'A=I\]
  3. Since \(A'=A^T\), this can be written as:
    \[A^TA=I\]
  4. Find the transpose \(A^T\)
  5. Interchanging the rows and columns of \(A\), we obtain:
    \[A^T=\begin{bmatrix}0 & x & x\\2y & y & -y\\z & -z & z\end{bmatrix}\]
  6. Form \(A^TA\)
  7. Therefore:
    \[A^TA=\begin{bmatrix}0 & x & x\\2y & y & -y\\z & -z & z\end{bmatrix}\begin{bmatrix}0 & 2y & z\\x & y & -z\\x & -y & z\end{bmatrix}\]
  8. We now calculate the entries row by row without omitting any multiplication.
  9. Calculate the first row
  10. The first row of \(A^T\) is:
    \[\begin{bmatrix}0 & x & x\end{bmatrix}\]
  11. Therefore, the first entry is:
    \[\begin{aligned}a_{11}&=0(0)+x(x)+x(x)\\&=x^2+x^2\\&=2x^2\end{aligned}\]
  12. The second entry is:
    \[\begin{aligned}a_{12}&=0(2y)+x(y)+x(-y)\\&=0+xy-xy\\&=0\end{aligned}\]
  13. The third entry is:
    \[\begin{aligned}a_{13}&=0(z)+x(-z)+x(z)\\&=0-xz+xz\\&=0\end{aligned}\]
  14. Hence, the first row of \(A^TA\) is:
    \[\begin{bmatrix}2x^2 & 0 & 0\end{bmatrix}.\]
  15. Calculate the second row
  16. The second row of \(A^T\) is:
    \[\begin{bmatrix}2y & y & -y\end{bmatrix}\]
  17. The first entry is:
    \[\begin{aligned}a_{21}&=2y(0)+y(x)+(-y)(x)\\&=0+xy-xy\\&=0\end{aligned}\]
  18. The second entry is:
    \[\begin{aligned}a_{22}&=2y(2y)+y(y)+(-y)(-y)\\&=4y^2+y^2+y^2\\&=6y^2\end{aligned}\]
  19. The third entry is:
    \[\begin{aligned}a_{23}&=2y(z)+y(-z)+(-y)(z)\\&=2yz-yz-yz\\&=0\end{aligned}\]
  20. Hence, the second row of \(A^TA\) is:
    \[\begin{bmatrix}0 & 6y^2 & 0\end{bmatrix}\]
  21. Calculate the third row
  22. The third row of \(A^T\) is:
    \[\begin{bmatrix}z & -z & z\end{bmatrix}\]
  23. The first entry is:
    \[\begin{aligned}a_{31}&=z(0)+(-z)(x)+z(x)\\&=0-zx+zx\\&=0\end{aligned}\]
  24. The second entry is:
    \[\begin{aligned}a_{32}&=z(2y)+(-z)(y)+z(-y)\\&=2yz-yz-yz\\&=0\end{aligned}\]
  25. The third entry is:
    \[\begin{aligned}a_{33}&=z(z)+(-z)(-z)+z(z)\\&=z^2+z^2+z^2\\&=3z^2\end{aligned}\]
  26. Hence, the third row of \(A^TA\) is:
    \[\begin{bmatrix}0 & 0 & 3z^2\end{bmatrix}\]
  27. Combine all three rows
  28. Therefore:
    \[A^TA=\begin{bmatrix}2x^2 & 0 & 0\\0 & 6y^2 & 0\\0 & 0 & 3z^2\end{bmatrix}\]
  29. But the given condition is:
    \[A^TA=I\]
    Since
    \[I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\]
  30. we must have:
    \[\begin{bmatrix}2x^2 & 0 & 0\\0 & 6y^2 & 0\\0 & 0 & 3z^2\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\]
  31. Equate the corresponding diagonal entries
  32. From the \((1,1)\)-entries:
    \[2x^2=1\]
  33. Therefore:
    \[\begin{aligned} x^2&=\frac{1}{2}\\ x&=\pm\frac{1}{\sqrt{2}} \end{aligned}\]
  34. From the \((2,2)\)-entries:
    \[6y^2=1\]
  35. Therefore:
    \[\begin{aligned} y^2&=\frac{1}{6}\\ y&=\pm\frac{1}{\sqrt{6}} \end{aligned}\]
  36. From the \((3,3)\)-entries:
    \[3z^2=1\]
  37. Therefore:
    \[\begin{aligned} z^2&=\frac{1}{3}\\ z&=\pm\frac{1}{\sqrt{3}} \end{aligned}\]
💡 Answer
Final Answer
›

Therefore, the required values are:

\[\boxed{x=\pm\frac{1}{\sqrt{2}},\qquad y=\pm\frac{1}{\sqrt{6}},\qquad z=\pm\frac{1}{\sqrt{3}}}\]
🎯 Exam Significance
Exam Significance
›
  • This problem tests matrix multiplication, transpose, and the concept of an orthogonal matrix.
  • The condition
    \[ A^TA=I \]
    is a standard and important condition for identifying an orthogonal matrix.
  • Showing the multiplication row by row makes the solution complete and avoids losing marks for unexplained calculations.
  • The diagonal entries of \(A^TA\) provide the equations needed to determine \(x,\ y,\ z\).
  • Students should not omit the negative roots when solving equations such as
    \[ x^2=\frac{1}{2}. \]
For JEE and Competitive Entrance Examinations
  • The condition
    \[ A^TA=I \]
    immediately identifies \(A\) as an orthogonal matrix.
  • For an orthogonal matrix, the columns are mutually orthonormal. This provides a faster conceptual route to the equations.
  • The diagonal elements of \(A^TA\) are the squared lengths of the columns of \(A\), so each must equal \(1\).
  • The off-diagonal elements represent dot products between distinct columns and must therefore be zero.
  • Recognising this structure can significantly reduce calculation time in objective-type questions.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The notation \(A'\) represents the transpose \(A^T\).

  2. The defining condition for an orthogonal matrix is:

    \[ A^TA=I. \]

  3. For the given matrix:

    \[ A^T= \begin{bmatrix} 0 & x & x\\ 2y & y & -y\\ z & -z & z \end{bmatrix} \]

  4. Direct multiplication gives:

    \[ A^TA= \begin{bmatrix} 2x^2&0&0\\ 0&6y^2&0\\ 0&0&3z^2 \end{bmatrix} \]

  5. Comparing with \(I\) gives:

    \[ 2x^2=1,\qquad 6y^2=1,\qquad 3z^2=1. \]

  6. The required values are:

    \[ \boxed{ x=\pm\frac{1}{\sqrt2},\quad y=\pm\frac{1}{\sqrt6},\quad z=\pm\frac{1}{\sqrt3} }. \]

  7. There are \(8\) possible ordered triples because each of \(x,\ y,\ z\) has two independent choices.

← Q2
3 / 11  ·  27%
Q4 →
Q4
NUMERIC3 marks
For what value of \(x\) does \[\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1 & 0 & 2\end{bmatrix} \begin{bmatrix}0\\2\\x\end{bmatrix}=0?\]
📘 Concept & Theory
Concept/Theory
›
1. Compatibility of Matrix Multiplication

Matrix multiplication is defined when the number of columns of the first matrix is equal to the number of rows of the second matrix.

Here,

\[\begin{bmatrix}1&2&1\end{bmatrix}\]
is a \(1\times3\) matrix,
\[\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}\]
is a \(3\times3\) matrix, and
\[\begin{bmatrix}0\\2\\x\end{bmatrix}\]
is a \(3\times1\) matrix.

Therefore, the complete product has order:

\[ (1\times3)(3\times3)(3\times1)=1\times1. \]

Hence, the final result is a scalar.

2. Matrix Multiplication Rule

If

\[A=[a_1\ a_2\ a_3]\]
is a row matrix and
\[B=\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}\]
is a column matrix, then:

\[ AB=a_1b_1+a_2b_2+a_3b_3. \]

In this problem, it is convenient to first multiply the \(1\times3\) row matrix by the \(3\times3\) matrix. This produces another \(1\times3\) matrix, which can then be multiplied by the final \(3\times1\) column matrix.

3. Important Property

Matrix multiplication is associative. Therefore:

\[ ABC=(AB)C=A(BC). \]

Thus, we may calculate the product in either grouping, provided the matrix dimensions remain compatible.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the orders of the three matrices.

  2. Multiply the \(1\times3\) row matrix by the \(3\times3\) matrix.

  3. Calculate all three entries of the resulting row matrix explicitly.

  4. Multiply the resulting \(1\times3\) matrix by the \(3\times1\) column matrix.

  5. Set the resulting scalar equal to zero.

  6. Solve the resulting linear equation in \(x\).

  7. State the required value of \(x\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  8 steps
  1. Given —
    \[\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1 & 0 & 2\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=0\]
  2. Multiply the first two matrices
  3. First calculate:
    \[\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1 & 0 & 2\end{bmatrix}\]
    The resulting matrix will be of order \(1\times3\).
  4. Find the first entry
  5. The first entry is obtained by multiplying the first row of the first matrix by the first column of the second matrix:
    \[\begin{aligned}a_{11}&=1(1)+2(2)+1(1)\\&=1+4+1\\&=6\end{aligned}\]
  6. Find the second entry
  7. The second entry is obtained by multiplying the first row of the first matrix by the second column of the second matrix:
    \[\begin{aligned}a_{12}&=1(2)+2(0)+1(0)\\&=2+0+0\\&=2\end{aligned}\]
  8. Find the third entry
  9. The third entry is obtained by multiplying the first row of the first matrix by the third column of the second matrix:
    \[\begin{aligned}a_{13}&=1(0)+2(1)+1(2)\\&=0+2+2\\&=4\end{aligned}\]
  10. Write the resulting row matrix
    \[\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1 & 0 & 2\end{bmatrix}=\begin{bmatrix}6 & 2 & 4\end{bmatrix}\]
  11. Hence the original equation becomes:
    \[\begin{bmatrix}6 & 2 & 4\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=0\]
  12. Multiply the row matrix by the column matrix
  13. Using the rule for multiplying a row matrix by a column matrix:
    \[\begin{aligned}\begin{bmatrix}6 & 2 & 4\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}&=6(0)+2(2)+4(x)\\&=0+4+4x\\&=4+4x\end{aligned}\]
  14. Use the given condition
  15. The original expression is equal to zero. Therefore:
    \[ \begin{aligned} 4+4x&=0\\ 4x&=-4\\ x&=-\dfrac44\\ x&=-1 \end{aligned} \]
💡 Answer
Final Answer
›
\[\boxed{x=-1}\]
🎯 Exam Significance
Exam Significance
›
  • This question tests the basic but essential skill of multiplying matrices of different orders.
  • It checks whether the student can correctly calculate a row matrix multiplied by a square matrix and then by a column matrix.
  • Writing the intermediate matrix
    \[ \begin{bmatrix} 6&2&4 \end{bmatrix} \]
    makes the calculation transparent and helps avoid arithmetic errors.
  • The final matrix product is a \(1\times1\) matrix, which is treated as a scalar in the equation.
  • This is a good example of a question where correct matrix dimensions and multiplication order are important for obtaining full marks.
For JEE and Other Competitive Entrance Examinations
  • The problem tests speed and accuracy in matrix multiplication.
  • Recognising the dimensions
    \[ (1\times3)(3\times3)(3\times1) \]
    immediately shows that the final result is a scalar.
  • The associative property allows an efficient choice of multiplication order:
    \[ ABC=(AB)C=A(BC). \]
  • For objective questions, students can choose whichever grouping requires fewer calculations.
  • The question also reinforces the important fact that matrix multiplication is associative but generally not commutative.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Always check matrix dimensions before multiplying.

  2. Here:

    \[ (1\times3)(3\times3)(3\times1)=1\times1. \]

  3. The first two matrices give:

    \[\begin{bmatrix}1&2&1\end{bmatrix}\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}=\begin{bmatrix}6&2&4\end{bmatrix}.\]

  4. The complete expression therefore becomes:

    \[\begin{bmatrix}6&2&4\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=0.\]

  5. This gives:

    \[ 4+4x=0. \]

  6. Hence:

    \[ \boxed{x=-1}. \]
← Q3
4 / 11  ·  36%
Q5 →
Q5
NUMERIC3 marks
If \[A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\] show that \[A^2-5A+7I=0\]
📘 Concept & Theory
Concept/Theory
›

This problem illustrates an important application of matrix algebra: verifying that a given matrix satisfies a polynomial equation.

1. Matrix Polynomial

An expression such as

\[ A^2-5A+7I \]
is called a matrix polynomial. Here, \(A\) is a square matrix and \(I\) is the identity matrix of the same order as \(A\).

Since \(A\) is a \(2\times2\) matrix, the identity matrix required in the expression is:

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

2. Scalar Multiplication of a Matrix

Every entry of a matrix is multiplied by the scalar. Therefore:

\[5A=5\begin{bmatrix}3&1\\-1&2\end{bmatrix}=\begin{bmatrix}15&5\\-5&10\end{bmatrix}\]

3. Multiplication of Two Matrices

Matrix multiplication is performed row by column. For example, the \((1,1)\)-entry of \(A^2\) is obtained by multiplying the first row of \(A\) by the first column of \(A\).

4. Zero Matrix

The symbol \(0\) on the right-hand side represents the zero matrix of the same order as \(A\):

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

Therefore, to prove

\[ A^2-5A+7I=0, \]
we must calculate the left-hand side completely and show that every entry is zero.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given matrix \(A\).

  2. Calculate \(A^2=A\cdot A\) using row-by-column multiplication.

  3. Calculate \(5A\) using scalar multiplication.

  4. Write the \(2\times2\) identity matrix \(I\).

  5. Calculate \(7I\).

  6. Substitute \(A^2\), \(5A\), and \(7I\) into \(A^2-5A+7I\).

  7. Perform the matrix subtraction and addition entry by entry.

  8. Show that the resulting matrix is the zero matrix.

  9. Conclude that the required matrix identity is proved.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  22 steps
  1. Given —
    \[A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\]
  2. To Prove —
    \[A^2-5A+7I=0\]
  3. Proof
  4. Calculate \(A^2\)
  5. By definition:
    \[A^2=A\cdot A\]
  6. Therefore:
    \[A^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}\]
  7. Calculate each entry of \(A^2\)
  8. The \((1,1)\)-entry is:
    \[\begin{aligned}a_{11}&=3(3)+1(-1)\\&=9-1\\&=8\end{aligned}\]
  9. The \((1,2)\)-entry is:
    \[\begin{aligned}a_{12}&=3(1)+1(2)\\&=3+2\\&=5\end{aligned}\]
  10. The \((2,1)\)-entry is:
    \[\begin{aligned}a_{21}&=(-1)(3)+2(-1)\\&=-3-2\\&=-5\end{aligned}\]
  11. The \((2,2)\)-entry is:
    \[\begin{aligned}a_{22}&=(-1)(1)+2(2)\\&=-1+4\\&=3\end{aligned}\]
  12. Hence:
    \[A^2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}\]
  13. Calculate \(5A\)
  14. Multiplying every entry of \(A\) by \(5\):
    \[\begin{aligned}5A&=5\begin{bmatrix}3&1\\-1&2\end{bmatrix}\\&=\begin{bmatrix}5(3)&5(1)\\5(-1)&5(2)\end{bmatrix}\\&=\begin{bmatrix}15&5\\-5&10\end{bmatrix}\end{aligned}\]
  15. Therefore:
    \[5A=\begin{bmatrix}15&5\\-5&10\end{bmatrix}\]
  16. Calculate \(7I\)
  17. Since \(A\) is a \(2\times2\) matrix, the corresponding identity matrix is:
    \[I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\]
  18. Therefore:
    \[\begin{aligned}7I&=7\begin{bmatrix}1&0\\0&1\end{bmatrix}\\&=\begin{bmatrix}7&0\\0&7\end{bmatrix}\end{aligned}\]
  19. Hence:
    \[7I=\begin{bmatrix}7&0\\0&7\end{bmatrix}\]
  20. Substitute the calculated matrices
  21. Now consider:
    \[A^2-5A+7I\]
  22. Substituting the values obtained above:
    \[A^2-5A+7I=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}\]
  23. Perform the matrix operations entry by entry
  24. The \((1,1)\)-entry is:
    \[8-15+7=0\]
  25. The \((1,2)\)-entry is:
    \[5-5+0=0\]
  26. The \((2,1)\)-entry is:
    \[-5-(-5)+0=-5+5=0\]
  27. The \((2,2)\)-entry is:
    \[3-10+7=0\]
  28. Therefore:
    \[\begin{aligned}A^2-5A+7I&=\begin{bmatrix}8-15+7&5-5+0\\-5-(-5)+0&3-10+7\end{bmatrix}\\&=\begin{bmatrix}0&0\\0&0\end{bmatrix}\end{aligned}\]
  29. Identify the zero matrix
  30. The matrix
    \[\begin{bmatrix}0&0\\0&0\end{bmatrix}\]
    is the \(2\times2\) zero matrix, denoted by \(0\).
  31. Hence:
    \[A^2-5A+7I=0\]
  32. Hence Proved
🎯 Exam Significance
Exam Significance
›
  • This problem tests the ability to calculate the square of a matrix accurately.
  • It reinforces scalar multiplication and multiplication by the identity matrix.
  • It is an important example of verifying a polynomial equation involving a matrix.
  • For full marks, each matrix operation should be performed in the correct order and the final result should explicitly be identified as the zero matrix.
  • The identity matrix must have the same order as \(A\). Since \(A\) is \(2\times2\), \(I\) must also be \(2\times2\).
For JEE and Competitive Entrance Examinations
  • Matrix polynomial identities are useful for reducing higher powers of a matrix.
  • From
    \[ A^2-5A+7I=0, \]
    we can immediately write:
    \[ A^2=5A-7I. \]
  • This relation can subsequently be used to express higher powers such as \(A^3,A^4,\) and so on in terms of \(A\) and \(I\).
  • Such reductions are useful in objective questions involving powers of matrices and matrix polynomials.
  • The result is also closely related to the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. For

    \[A=\begin{bmatrix}3&1\\-1&2\end{bmatrix},\]
    its square is:
    \[A^2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}.\]

  2. Scalar multiplication gives:

    \[5A=\begin{bmatrix}15&5\\-5&10\end{bmatrix}.\]

  3. Since

    \[I=\begin{bmatrix}1&0\\0&1\end{bmatrix},\]
    we have:
    \[7I=\begin{bmatrix}7&0\\0&7\end{bmatrix}.\]

  4. Substitution gives:

    \[A^2-5A+7I=\begin{bmatrix}0&0\\0&0\end{bmatrix}.\]

  5. Therefore:

    \[\boxed{A^2-5A+7I=0}.\]

  6. Equivalently, the matrix satisfies:

    \[\boxed{A^2=5A-7I}.\]
← Q4
5 / 11  ·  45%
Q6 →
Q6
NUMERIC3 marks

Find \(x\), if

\[\begin{bmatrix}x&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0\]

📘 Concept & Theory
Concept/Theory
›

This question involves the multiplication of a row matrix, a square matrix, and a column matrix. The product ultimately produces a \(1\times1\) matrix, which can be treated as a scalar.

1. Checking the Orders

The three matrices have orders:

\[\begin{bmatrix}x&-5&-1\end{bmatrix}\quad\text{is of order }1\times3,\]

\[\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\quad\text{is of order }3\times3,\]

and

\[\begin{bmatrix}x\\4\\1\end{bmatrix}\quad\text{is of order }3\times1\]

Therefore:

\[(1\times3)(3\times3)(3\times1)=1\times1.\]

Thus, the final product is a scalar.

2. Row-by-Column Multiplication

To multiply a row matrix by a square matrix, each entry of the resulting row is obtained by taking the dot product of the given row with the corresponding column of the square matrix.

For example, if

\[R=\begin{bmatrix}r_1&r_2&r_3\end{bmatrix},\]
then the first entry of \(RM\) is obtained from the first column of \(M\).

3. Efficient Strategy

By associativity of matrix multiplication:

\[ ABC=(AB)C=A(BC). \]

We can therefore choose a convenient multiplication order. Here, multiplying the first two matrices first produces a simple \(1\times3\) matrix, after which only one row-by-column multiplication remains.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Multiply the \(1\times3\) row matrix by the \(3\times3\) matrix.

  2. Calculate each of the three resulting entries carefully.

  3. Multiply the resulting \(1\times3\) matrix by the \(3\times1\) column matrix.

  4. Set the resulting scalar equal to zero.

  5. Simplify the resulting quadratic equation in \(x\).

  6. Solve the quadratic equation.

  7. Simplify the radical to obtain the exact values of \(x\).

  8. Verify the obtained values in the original equation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  14 steps
  1. Given —
    \[\begin{bmatrix}x&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0\]
  2. Multiply the first two matrices
  3. Consider:
    \[\begin{bmatrix}x&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\]
    The result will be a \(1\times3\) matrix.
  4. Calculate the first entry
  5. The first column of the square matrix is:
    \[\begin{bmatrix}1\\0\\2\end{bmatrix}\]
  6. Therefore:
    \[\begin{aligned}a_{11}&=x(1)+(-5)(0)+(-1)(2)\\&=x+0-2\\&=x-2\end{aligned}\]
  7. Calculate the second entry
  8. The second column of the square matrix is:
    \[\begin{bmatrix}0\\2\\0\end{bmatrix}\]
  9. Therefore:
    \[\begin{aligned}a_{12}&=x(0)+(-5)(2)+(-1)(0)\\&=0-10+0\\&=-10\end{aligned}\]
  10. Calculate the third entry
  11. The third column of the square matrix is:
    \[\begin{bmatrix}2\\1\\3\end{bmatrix}\]
  12. Therefore:
    \[\begin{aligned}a_{13}&=x(2)+(-5)(1)+(-1)(3)\\&=2x-5-3\\&=2x-8\end{aligned}\]
  13. Write the resulting matrix
  14. Hence
    \[\begin{aligned}\begin{bmatrix}x&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}&=\begin{bmatrix}x-2&-10&2x-8\end{bmatrix}\end{aligned}\]
  15. Therefore, the original equation becomes:
    \[\begin{bmatrix}x-2&-10&2x-8\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0\]
  16. Multiply the row matrix by the column matrix
  17. Using row-by-column multiplication:
    \[\begin{aligned}&(x-2)x+(-10)(4)+(2x-8)(1)=0\end{aligned}\]
  18. Expanding each term:
    \[\begin{aligned}x(x-2)-40+2x-8&=0\\x^2-2x-40+2x-8&=0\end{aligned}\]
  19. Simplify the equation
  20. The terms \(-2x\) and \(+2x\) cancel:
    \[x^2-48=0\]
  21. Therefore:
    \[ x^2=48\]
  22. Taking the square root of both sides:
    \[\begin{aligned}x&=\pm\sqrt{48}\\&=\pm\sqrt{16\times3}\\&=\pm\sqrt{16}\sqrt{3}\\&=\pm4\sqrt{3}\end{aligned}\]
💡 Answer
Final Answer
›

Therefore, the required values of \(x\) are:

\[\boxed{x=\pm4\sqrt{3}}\]
🎯 Exam Significance
Exam Significance
›
  • This question tests the application of matrix multiplication to an algebraic equation involving an unknown \(x\).
  • It assesses whether students can correctly perform row-by-column multiplication without confusing matrix multiplication with ordinary multiplication.
  • Showing the intermediate matrix
    \[ \begin{bmatrix} x-2&-10&2x-8 \end{bmatrix} \]
    makes the solution systematic and reduces calculation errors.
  • The problem ultimately converts the matrix equation into a quadratic equation, requiring correct algebraic simplification.
  • Both roots must be retained because the equation
    \[ x^2=48 \]
    has two real solutions.
For JEE and Other Competitive Entrance Examinations
  • The question tests computational efficiency in matrix multiplication.
  • Recognising the dimensions
    \[ (1\times3)(3\times3)(3\times1) \]
    immediately confirms that the final result is a scalar.
  • The associativity property
    \[ ABC=(AB)C=A(BC) \]
    allows students to choose the multiplication order that minimises computation.
  • Competitive examinations frequently combine matrices with algebraic equations, so careful simplification after matrix multiplication is essential.
  • The cancellation of the linear terms
    \[ -2x+2x=0 \]
    is an important observation that reduces the equation directly to
    \[ x^2=48. \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The matrix orders are:

    \[ (1\times3)(3\times3)(3\times1)=1\times1. \]

  2. The first matrix product is:

    \[\begin{bmatrix}x&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}=\begin{bmatrix}x-2&-10&2x-8\end{bmatrix}\]

  3. Multiplying by the final column matrix gives:

    \[ (x-2)x-40+(2x-8)=0. \]

  4. Simplification gives:

    \[ x^2-48=0. \]

  5. Therefore:

    \[ x^2=48. \]

  6. Since

    \[ 48=16\times3, \]
    we obtain:
    \[ \boxed{x=\pm4\sqrt{3}}. \]

  7. Always consider both positive and negative roots when solving an equation of the form \(x^2=a\), provided both satisfy the original conditions.

← Q5
6 / 11  ·  55%
Q7 →
Q7
NUMERIC3 marks

A manufacturer produces three products \(x,y,z\), which he sells in two markets. Annual sales are indicated below:

Market Product \(x\) Product \(y\) Product \(z\)
Market I 10,000 2,000 18,000
Market II 6,000 20,000 8,000

(a) If the unit sale prices of \(x,y,z\) are ₹2.50, ₹1.50 and ₹1.00 respectively, find the total revenue in each market with the help of matrix algebra.

(b) If the unit costs of the above three commodities are ₹2.00, ₹1.00 and 50 paise respectively, find the gross profit.

📘 Concept & Theory
Concept/Theory
›

This problem demonstrates how matrix multiplication can be used to model and solve a real-world business problem involving sales, prices, costs and profit.

1. Sales Matrix

The annual quantities sold in the two markets can be represented by the \(2\times3\) sales matrix:

\[S=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}\]

Each row represents a market, while each column represents a commodity.

2. Price Matrix

The unit selling prices of \(x,y,z\) are represented by the column matrix:

\[P=\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}\]

The product \(SP\) gives the total revenue generated in each market.

3. Why Matrix Multiplication Works

The dimensions are:

\[(2\times3)(3\times1)=2\times1\]

Therefore, the result contains two entries, one for each market.

For Market I:

\[10000(2.50)+2000(1.50)+18000(1.00)\]

For Market II:

\[6000(2.50)+20000(1.50)+8000(1.00)\]

4. Cost Matrix

The unit costs are represented by:

\[C=\begin{bmatrix}2.00\\1.00\\0.50\end{bmatrix}\]

Therefore, the total cost in each market is:

\[SC\]

5. Gross Profit

Gross profit is calculated as:

\[\text{Gross Profit}=\text{Total Revenue}-\text{Total Cost}\]

Thus, the overall gross profit can be obtained by subtracting the total cost from the total revenue.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Construct the sales matrix from the annual sales data.

  2. Construct the column matrix of unit selling prices.

  3. Multiply the sales matrix by the price matrix.

  4. Interpret the two resulting entries as the revenues in Market I and Market II.

  5. Add the two market revenues to obtain total revenue.

  6. Construct the column matrix of unit costs.

  7. Multiply the sales matrix by the cost matrix.

  8. Interpret the resulting entries as the costs in the two markets.

  9. Add the two market costs to obtain total cost.

  10. Subtract total cost from total revenue to obtain gross profit.

✏️ Solution
Part (a): Total Revenue in Each Market
›
Step-by-step Solution  ·  14 steps
  1. Form the Sales Matrix
  2. From the given annual sales data:
    \[S=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}\]
    The first row represents Market I and the second row represents Market II.
  3. Form the Unit Selling Price Matrix
  4. \[x=\text{₹}2.50,\quad y=\text{₹}1.50,\quad z=\text{₹}1.00.\]
  5. Hence:
    \[P=\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}\]
  6. Multiply the Sales Matrix by the Price Matrix
  7. The required revenue matrix is:
    \[R=SP\]
  8. Therefore:
    \[R=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}\]
  9. Calculate Revenue in Market I
  10. The first row of the sales matrix corresponds to Market I. Therefore:
    \[\begin{aligned}R_1&=10000(2.50)+2000(1.50)+18000(1.00)\\&=25000+3000+18000\\&=46000\end{aligned}\]
  11. Therefore, the total revenue in Market I is:
    \[\boxed{\text{₹}46,000}\]
  12. Calculate Revenue in Market II
  13. The second row corresponds to Market II. Hence:
    \[\begin{aligned}R_2&=6000(2.50)+20000(1.50)+8000(1.00)\\&=15000+30000+8000\\&=53000\end{aligned}\]
  14. Therefore, the total revenue in Market II is:
    \[\boxed{\text{₹}53,000}\]
  15. Write the Revenue Matrix
  16. \[\boxed{R=\begin{bmatrix}46000\\53000\end{bmatrix}}\]
  17. Thus:
    • Revenue in Market I = ₹46,000
    • Revenue in Market II = ₹53,000
  18. Calculate Total Revenue
  19. The total revenue from both markets is:
    \[\begin{aligned}\text{Total Revenue}&=46000+53000\\&=99000\end{aligned}\]
  20. Therefore:
    \[\boxed{\text{Total Revenue}=\text{₹}99,000}\]
✏️ Solution
Part (b): Gross Profit
›
Step-by-step Solution  ·  15 steps
  1. Form the Unit Cost Matrix
  2. The unit costs of \(x,y,z\) are:
    \[\text{₹}2.00,\quad\text{₹}1.00,\quad 50\text{ paise}\]
  3. Since:
    \[50\text{ paise}=\text{₹}0.50\]
  4. the unit cost matrix is:
    \[C=\begin{bmatrix}2.00\\1.00\\0.50\end{bmatrix}\]
  5. Calculate Total Cost in Each Market
  6. The cost matrix is obtained by:
    \[T=SC\]
  7. Therefore:
    \[T=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}\begin{bmatrix}2.00\\1.00\\0.50\end{bmatrix}\]
  8. Cost in Market I
    \[\begin{aligned}T_1&=10000(2.00)+2000(1.00)+18000(0.50)\\&=20000+2000+9000\\&=31000\end{aligned}\]
  9. Hence, the total cost in Market I is:
    \[\boxed{\text{₹}31,000}\]
  10. Cost in Market II
    \[\begin{aligned}T_2&=6000(2.00)+20000(1.00)+8000(0.50)\\&=12000+20000+4000\\&=36000\end{aligned}\]
  11. Hence, the total cost in Market II is:
    \[\boxed{\text{₹}36,000}\]
  12. Therefore, the Cost Matrix
    \[\boxed{T=\begin{bmatrix}31000\\36000\end{bmatrix}}\]
  13. The total cost in both markets is:
    \[\begin{aligned}\text{Total Cost}&=31000+36000\\&=67000\end{aligned}\]
  14. Hence:
    \[\boxed{\text{Total Cost}=\text{₹}67,000}\]
  15. Calculate Gross Profit
  16. Gross profit is:
    \[\text{Gross Profit}=\text{Total Revenue}-\text{Total Cost}\]
  17. Substituting the values:
    \[\begin{aligned}\text{Gross Profit}&=99000-67000\\&=32000\end{aligned}\]
  18. Therefore:
    \[\boxed{\text{Gross Profit}=\text{₹}32,000}\]
🎯 Exam Significance
Exam Significance
›
  • This is an important application-based question on matrix multiplication.
  • It tests whether students can translate a real-world situation into suitable matrices.
  • The question combines matrix multiplication with concepts of revenue, cost and gross profit.
  • Correctly identifying the dimensions of the matrices is important:
    \[ (2\times3)(3\times1)=2\times1. \]
  • Students should clearly distinguish between unit selling price and unit cost.
  • In a board examination, showing the matrix multiplication before interpreting the answer makes the solution logically complete.
For JEE and Competitive Entrance Examinations
  • The problem develops the ability to translate word problems into matrix notation.
  • Matrix multiplication provides a compact method of calculating multiple weighted sums simultaneously.
  • The same technique is applicable to problems involving production, transportation, inventories, prices, costs, revenue and profit.
  • The dimensions of the matrices provide an immediate check on whether a multiplication is mathematically defined.
  • Competitive questions may change the number of products or markets while preserving exactly the same matrix-algebra structure.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  11 points

  1. Sales data are represented by:

    \[ S= \begin{bmatrix} 10000&2000&18000\\ 6000&20000&8000 \end{bmatrix}. \]

  2. Unit selling prices are represented by:

    \[ P= \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}. \]

  3. Revenue is obtained by:

    \[ R=SP. \]

  4. The revenue matrix is:

    \[ R= \begin{bmatrix} 46000\\ 53000 \end{bmatrix}. \]

  5. Total revenue is:

    \[ \boxed{\text{₹}99,000}. \]

  6. Unit costs are represented by:

    \[ C= \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}. \]

  7. Total cost is obtained by:

    \[ T=SC. \]

  8. The cost matrix is:

    \[ T= \begin{bmatrix} 31000\\ 36000 \end{bmatrix}. \]

  9. Total cost is:

    \[ \boxed{\text{₹}67,000}. \]

  10. Gross profit is:

    \[ \text{Gross Profit} = \text{Revenue}-\text{Cost}. \]

  11. Therefore:

    \[ \boxed{\text{Gross Profit}=\text{₹}32,000}. \]
← Q6
7 / 11  ·  64%
Q8 →
Q8
NUMERIC3 marks
Find the matrix \(X\) such that \[X\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory
›

This problem requires us to determine an unknown matrix \(X\) from a matrix equation. The most direct method is to represent the unknown matrix by variables and then compare corresponding entries after multiplication.

1. Determine the Order of \(X\)

The given matrix

\[A=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}\]
is of order \(2\times3\).

The matrix on the right-hand side,

\[B=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}\]
is also of order \(2\times3\).

Since

\[ XA=B, \]
and \(A\) is \(2\times3\), \(X\) must have \(2\) columns so that the multiplication \(XA\) is defined.

Since the resulting matrix has \(2\) rows, \(X\) must have \(2\) rows. Therefore:

\[ \boxed{X\text{ is of order }2\times2}. \]

2. General Form of the Unknown Matrix

Let

\[X=\begin{bmatrix}a&b\\c&d\end{bmatrix}\]

Matrix multiplication is performed row by column. Thus:

\[\begin{aligned}XA&=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}\\&=\begin{bmatrix}a+4b&2a+5b&3a+6b\\c+4d&2c+5d&3c+6d\end{bmatrix}\end{aligned}\]

Notice that the entries of \(X\) can be found by comparing the corresponding entries of this product with the given matrix.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Determine the order of the unknown matrix \(X\).

  2. Represent \(X\) as a general \(2\times2\) matrix.

  3. Multiply \(X\) by the given \(2\times3\) matrix.

  4. Compare corresponding entries of the resulting matrix with the given matrix.

  5. Use the first two equations to determine \(a\) and \(b\).

  6. Use the corresponding equations in the second row to determine \(c\) and \(d\).

  7. Substitute the values into \(X\).

  8. Verify the answer by multiplying \(X\) with the given matrix.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  27 steps
  1. Determine the order of \(X\)
  2. Given —
    \[X\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}\]
  3. The matrix being multiplied by \(X\) is of order \(2\times3\), and the resulting matrix is also of order \(2\times3\).
  4. Therefore, \(X\) must be a \(2\times2\) matrix.
  5. Let:
    \[X=\begin{bmatrix}a&b\\c&d\end{bmatrix}\]
  6. Perform the matrix multiplication
  7. Substitute \(X\) into the given equation:
    \[\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}\]
  8. Multiplying row by column:
    \[\begin{aligned}&\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}\\&=\begin{bmatrix}a(1)+b(4)&a(2)+b(5)&a(3)+b(6)\\ c(1)+d(4)&c(2)+d(5)&c(3)+d(6)\end{bmatrix}\\&=\begin{bmatrix}a+4b&2a+5b&3a+6b\\c+4d&2c+5d&3c+6d\end{bmatrix}\end{aligned}\]
  9. Hence:
    \[\begin{bmatrix}a+4b&2a+5b&3a+6b\\c+4d&2c+5d&3c+6d\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}\]
  10. Find \(a\) and \(b\)
  11. Comparing the first entry of both matrices:
    \[a+4b=-7\]
  12. Therefore:
    \[a=-7-4b\]
  13. Now compare the second entry of the first row:
    \[2a+5b=-8\]
  14. Substitute
    \[a=-7-4b:\]
    \[\begin{aligned}2(-7-4b)+5b&=-8\\-14-8b+5b&=-8\\-14-3b&=-8\end{aligned}\]
  15. Add \(14\) to both sides:
    \[-3b=6\]
  16. Hence:
    \[b=-2\]
  17. Substitute \(b=-2\) into
    \[a+4b=-7:\]
  18. \[\begin{aligned}a+4(-2)&=-7\\a-8&=-7\\a&=1\end{aligned}\]
  19. Therefore:
    \[\boxed{a=1,\qquad b=-2}\]
  20. Find \(c\) and \(d\)
  21. Compare the first entry of the second row:
    \[c+4d=2\]
  22. Therefore:
    \[c=2-4d\]
  23. Now compare the second entry of the second row:
    \[2c+5d=4\]
  24. Substitute
    \[c=2-4d:\]
  25. \[\begin{aligned}2(2-4d)+5d&=4\\4-8d+5d&=4\\4-3d&=4\end{aligned}\]
  26. Subtract \(4\) from both sides:
    \[-3d=0.\]
  27. Therefore:
    \[d=0\]
  28. Substitute \(d=0\) into:
    \[c+4d=2\]
  29. We get:
    \[\begin{aligned}c+4(0)&=2\\c&=2\end{aligned}\]
  30. Therefore:
    \[\boxed{c=2,\qquad d=0}\]
  31. Construct the matrix \(X\)
  32. We have obtained:
    \[a=1,\quad b=-2,\quad c=2,\quad d=0\]
  33. Therefore:
    \[X=\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}1&-2\\2&0\end{bmatrix}\]
  34. Hence:
    \[X=\begin{bmatrix}1&-2\\2&0\end{bmatrix}\]
🎯 Exam Significance
Exam Significance
›
  • This problem tests the concept of solving a matrix equation by comparing corresponding elements.
  • Determining the order of the unknown matrix is an important first step and demonstrates understanding of matrix multiplication.
  • Students must correctly apply row-by-column multiplication.
  • The question also tests simultaneous linear equations arising from matrix multiplication.
  • A complete board solution should show the formation of the equations, their solution and substitution back into the matrix.
For JEE and Competitive Entrance Examinations
  • This type of question tests speed and accuracy in matrix multiplication.
  • Matrix equations can often be converted into systems of linear equations by comparing corresponding entries.
  • The order argument
    \[ (2\times2)(2\times3)=2\times3 \]
    is an efficient way to determine the structure of the unknown matrix.
  • In objective examinations, the verification multiplication can also be used as a quick way to test a candidate answer.
  • The problem reinforces the fundamental principle that matrix multiplication is not commutative:
    \[ AB\ne BA \]
    in general.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. If

    \[ X(2\times3)=(2\times3), \]
    then \(X\) must be of order \(2\times2\).

  2. Represent the unknown matrix as:

    \[X=\begin{bmatrix}a&b\\c&d\end{bmatrix}\]

  3. Matrix multiplication gives:

    \[XA=\begin{bmatrix}a+4b&2a+5b&3a+6b\\c+4d&2c+5d&3c+6d\end{bmatrix}\]

  4. Comparing corresponding entries gives a system of linear equations.

  5. The first row gives:

    \[ a+4b=-7,\qquad 2a+5b=-8. \]

  6. Solving these equations gives:

    \[ a=1,\qquad b=-2. \]

  7. The second row gives:

    \[ c+4d=2,\qquad 2c+5d=4. \]

  8. Solving these equations gives:

    \[ c=2,\qquad d=0. \]

  9. Hence:

    \[\boxed{X=\begin{bmatrix}1&-2\\2&0\end{bmatrix}}\]
← Q7
8 / 11  ·  73%
Q9 →
Q9
NUMERIC3 marks
If \[A=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\] is such that \[A^2=I,\]
  1. \[1+\alpha^2+\beta\gamma=0\]
  2. \[1-\alpha^2+\beta\gamma=0\]
  3. \[1-\alpha^2-\beta\gamma=0\]
  4. \[1+\alpha^2-\beta\gamma=0\]
then which of the following is correct?
📘 Concept & Theory
Concept/Theory
›

The condition

\[A^2=I\]
means that multiplying the matrix \(A\) by itself produces the identity matrix.

For a \(2\times2\) matrix, matrix multiplication is performed row by column. Therefore, every entry of \(A^2\) must be calculated carefully.

Identity Matrix of Order 2

The identity matrix of order \(2\) is:

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

Since \(A^2=I\), corresponding entries of \(A^2\) and \(I\) must be equal. In particular, the diagonal entries of \(A^2\) must each be equal to \(1\), while the off-diagonal entries must be equal to \(0\).

Important Calculation

Given:

\[A=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\]

The first diagonal entry of \(A^2\) is:

\[\alpha^2+\beta\gamma\]

The second diagonal entry is also:

\[\gamma\beta+(-\alpha)^2=\beta\gamma+\alpha^2\]

Therefore:

\[\alpha^2+\beta\gamma=1\]

Rearranging:

\[1-\alpha^2-\beta\gamma=0\]

Hence option \(\mathrm{C}\) is expected to be correct.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given matrix \(A\).

  2. Calculate \(A^2=A\cdot A\) using row-by-column multiplication.

  3. Observe that the off-diagonal entries become zero.

  4. Use the condition \(A^2=I\).

  5. Compare the diagonal entries with those of the identity matrix.

  6. Obtain the relation \(\alpha^2+\beta\gamma=1\).

  7. Rearrange it into the form given in the options.

  8. Identify the correct option.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. Given —
    \[A=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\]
  2. Also:
    \[A^2=I\]
  3. Therefore:
    \[A^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\]
  4. Calculate \(A^2\)
  5. By definition:
    \[A^2=A\cdot A\]
  6. Hence:
    \[A^2=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\]
  7. Calculate the first entry
  8. The first row of the first matrix is
    \[\begin{bmatrix}\alpha&\beta\end{bmatrix},\]
    and the first column of the second matrix is
    \[\begin{bmatrix}\alpha\\\gamma\end{bmatrix}\]
  9. Therefore:
    \[\begin{aligned}(A^2)_{11}&=\alpha(\alpha)+\beta(\gamma)\\&=\alpha^2+\beta\gamma\end{aligned}\]
  10. Calculate the second entry
  11. The first row of the first matrix is
    \[\begin{bmatrix}\alpha&\beta\end{bmatrix},\]
    and the second column of the second matrix is
    \[\begin{bmatrix}\beta\\-\alpha\end{bmatrix}\]
  12. Therefore:
    \[\begin{aligned}(A^2)_{12}&=\alpha(\beta)+\beta(-\alpha)\\&=\alpha\beta-\alpha\beta\\&=0\end{aligned}\]
  13. Calculate the third entry
  14. The second row of the first matrix is
    \[\begin{bmatrix}\gamma&-\alpha\end{bmatrix},\]
    and the first column of the second matrix is
    \[\begin{bmatrix}\alpha\\\gamma\end{bmatrix}\]
  15. Therefore:
    \[\begin{aligned}(A^2)_{21}&=\gamma(\alpha)+(-\alpha)(\gamma)\\&=\alpha\gamma-\alpha\gamma\\&=0\end{aligned}\]
  16. Calculate the fourth entry
  17. The second row of the first matrix is
    \[\begin{bmatrix}\gamma&-\alpha\end{bmatrix},\]
    and the second column of the second matrix is
    \[\begin{bmatrix}\beta\\-\alpha\end{bmatrix}\]
  18. Therefore:
    \[\begin{aligned}(A^2)_{22}&=\gamma(\beta)+(-\alpha)(-\alpha)\\&=\beta\gamma+\alpha^2\end{aligned}\]
  19. Write \(A^2\)
  20. Combining all four entries:
    \[A^2=\begin{bmatrix}\alpha^2+\beta\gamma&0\\0&\alpha^2+\beta\gamma\end{bmatrix}\]
  21. Since \(A^2=I\):
    \[\begin{bmatrix}\alpha^2+\beta\gamma&0\\0&\alpha^2+\beta\gamma\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\]
  22. Compare the diagonal entries
  23. Comparing the \((1,1)\) entries:
    \[\alpha^2+\beta\gamma=1\]
  24. Rearranging:
    \[1-\alpha^2-\beta\gamma=0\]
  25. This is exactly option \(\mathrm{C}\).
💡 Answer
Final Answer
›
Correct option is:
\[ \boxed{\mathrm{C}\;:\;1-\alpha^2-\beta\gamma=0}. \]
🎯 Exam Significance
Exam Significance
›
  • This question tests direct application of matrix multiplication and the identity matrix.
  • It is important to calculate every entry of \(A^2\) using the correct row-by-column rule.
  • The question reinforces the condition that if two matrices are equal, their corresponding entries are equal.
  • The zero off-diagonal entries demonstrate the cancellation:
    \[ \alpha\beta-\alpha\beta=0 \]
    and
    \[ \alpha\gamma-\alpha\gamma=0. \]
  • Such a multiple-choice question can be solved quickly once the structure of \(A^2\) is recognised.
For JEE and Competitive Entrance Examinations
  • This problem tests accuracy in symbolic matrix multiplication.
  • It provides a useful shortcut: once \(A^2\) is calculated, only one diagonal-entry equation is required to identify the correct option.
  • The form
    \[A=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\]
    produces equal diagonal entries in \(A^2\), which is a useful structural observation.
  • Competitive examinations frequently use such parameterised matrices to test conceptual understanding rather than lengthy calculations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Given:

    \[A=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\]

  2. Calculate:

    \[ A^2=A\cdot A. \]

  3. The off-diagonal entries are zero:

    \[ \alpha\beta-\alpha\beta=0, \]
    \[ \alpha\gamma-\alpha\gamma=0. \]

  4. Both diagonal entries are:

    \[ \alpha^2+\beta\gamma. \]

  5. Therefore:

    \[A^2=\begin{bmatrix}\alpha^2+\beta\gamma&0\\0&\alpha^2+\beta\gamma\end{bmatrix}\]

  6. Since \(A^2=I\):

    \[ \alpha^2+\beta\gamma=1. \]

  7. Rearranging:

    \[ \boxed{1-\alpha^2-\beta\gamma=0}. \]

  8. Hence the correct answer is:

    \[ \boxed{\mathrm{C}}. \]
← Q8
9 / 11  ·  82%
Q10 →
Q10
NUMERIC3 marks
If the matrix \(A\) is both symmetric and skew-symmetric, then:
  1. \(A\) is a diagonal matrix
  2. \(A\) is a zero matrix
  3. \(A\) is a square matrix
  4. None of these
📘 Concept & Theory
Concept/Theory
›
Symmetric Matrix

A square matrix \(A\) is called a symmetric matrix if:

\[ A^T=A \]

Skew-Symmetric Matrix

A square matrix \(A\) is called a skew-symmetric matrix if:

\[ A^T=-A \]

If the same matrix \(A\) is both symmetric and skew-symmetric, then both conditions must hold simultaneously:

\[ A^T=A \]
and
\[ A^T=-A. \]

Therefore:

\[ A=-A. \]

Adding \(A\) to both sides:

\[ A+A=0. \]

Hence:

\[ 2A=0. \]

Since the entries of the matrices are real numbers and \(2\neq0\), division by \(2\) gives:

\[ \boxed{A=0}.

Thus, the only matrix that can be both symmetric and skew-symmetric is the zero matrix.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Recall the definition of a symmetric matrix.

  2. Recall the definition of a skew-symmetric matrix.

  3. Apply both conditions to the same matrix \(A\).

  4. Equate \(A\) and \(-A\).

  5. Deduce \(2A=0\).

  6. Conclude that \(A=0\).

  7. Identify the correct option.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  7 steps
  1. Use the condition for a symmetric matrix
  2. Since \(A\) is symmetric:
    \[\boxed{A^T=A}\]
  3. Use the condition for a skew-symmetric matrix
  4. Since \(A\) is also skew-symmetric:
    \[\boxed{A^T=-A}\]
  5. Compare the two conditions
  6. From the two equations:
    \[A^T=A\]
    and
    \[A^T=-A\]
  7. we obtain:
    \[A=-A\]
  8. Add \(A\) to both sides
    \[A+A=-A+A\]
  9. Therefore:
    \[\begin{aligned}2A&=0\\\Rightarrow A&=0\end{aligned}\]
  10. Thus \(A\) must be the zero matrix.
💡 Answer
Final Answer
›
Correct answer is
\[\boxed{\text{(B) }A\text{ is a zero matrix}}\]
🎯 Exam Significance
Exam Significance
›
  • This question tests the fundamental definitions of symmetric and skew-symmetric matrices.
  • It is a typical conceptual multiple-choice question that can be solved without lengthy calculations.
  • The key equations to remember are:
    \[ A^T=A \]
    for symmetric matrices and
    \[ A^T=-A \]
    for skew-symmetric matrices.
  • Understanding why the zero matrix satisfies both conditions is important for short-answer and objective questions.
For JEE and Other Competitive Entrance Examinations
  • This is a high-value conceptual result that can often be used as a shortcut in objective questions.
  • Whenever a matrix is simultaneously symmetric and skew-symmetric, it must be the zero matrix, provided the underlying number system has characteristic different from \(2\), as is the case for real and complex matrices.
  • The result can also be recognised from:
    \[A^T=A\quad\text{and}\quad A^T=-A\Rightarrow A=-A\Rightarrow A=0\]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. A symmetric matrix satisfies:

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

  2. A skew-symmetric matrix satisfies:

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

  3. If a matrix is both, then:

    \[ A=-A. \]

  4. Therefore:

    \[ 2A=0. \]

  5. Hence:

    \[ \boxed{A=0}. \]

  6. The zero matrix is both symmetric and skew-symmetric.

  7. Therefore, the correct option is:

    \[ \boxed{\text{(B) }A\text{ is a zero matrix}}. \]
← Q9
10 / 11  ·  91%
Q11 →
Q11
NUMERIC3 marks
If \(A\) is a square matrix such that \[A^2=A,\] then \[(I+A)^3-7A\] is equal to:
  1. \(A\)
  2. \(I-A\)
  3. \(I\)
  4. \(3A\)
📘 Concept & Theory
Concept/Theory
›
Idempotent Matrix

A square matrix \(A\) satisfying

\[A^2=A\]
is called an idempotent matrix.

The given relation allows us to simplify every higher power of \(A\). For example:

\[A^3=A^2A=AA=A^2=A.\]

Similarly:

\[A^4=A^2=A.\]

In fact, for every positive integer \(n\):

\[\boxed{A^n=A}.\]

Important Matrix Algebra Rules

Since \(I\) is the identity matrix:

\[IA=AI=A.\]

Therefore, when expanding

\[(I+A)^3,\]
the ordinary binomial expansion can be used because \(I\) commutes with every square matrix \(A\).

Thus:

\[(I+A)^3=I^3+3I^2A+3IA^2+A^3\]

Using

\[I^2=I,\qquad IA=A,\qquad A^2=A,\qquad A^3=A,\]
this expression can be simplified immediately.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Use the given condition \(A^2=A\).

  2. Deduce that \(A^3=A\).

  3. Expand \((I+A)^3\) using the binomial theorem.

  4. Replace \(A^2\) and \(A^3\) by \(A\).

  5. Combine the resulting terms involving \(A\).

  6. Subtract \(7A\).

  7. Verify that the remaining expression is \(I\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  18 steps
  1. Given —
    \[A^2=A\]
    Therefore, \(A\) is an idempotent matrix.
  2. Find \(A^3\)
  3. We have:
    \[A^3=A^2A\]
  4. Using \(A^2=A\):
  5. \[A^3=AA\]
  6. Therefore:
    \[A^3=A^2\]
  7. Again, since \(A^2=A\):
    \[\boxed{A^3=A}\]
  8. Expand \((I+A)^3\)
  9. Using the binomial expansion:
    \[(I+A)^3=I^3+3I^2A+3IA^2+A^3\]
  10. Simplify each term
  11. Since \(I\) is the identity matrix:
    \[I^3=I\]
  12. Also:
    \[I^2A=IA=A\]
  13. Hence:
    \[3I^2A=3A\]
  14. Next:
    \[IA^2=A^2\]
  15. Since \(A^2=A\):
    \[IA^2=A\]
  16. Therefore:
    \[3IA^2=3A\]
  17. Finally:
    \[A^3=A\]
  18. Substitute all simplified terms
    \[\begin{aligned}(I+A)^3&=I^3+3I^2A+3IA^2+A^3\\&=I+3A+3A+A\\&=I+7A\end{aligned}\]
  19. Subtract \(7A\),
  20. The required expression is:
    \[(I+A)^3-7A\]
  21. Substituting
    \[(I+A)^3=I+7A,\]
    we get:
    \[\begin{aligned}(I+A)^3-7A&=(I+7A)-7A\\&=I+7A-7A\\&=I\end{aligned}\]
  22. Hence:
    \[\boxed{(I+A)^3-7A=I}\]
💡 Answer
Final Answer
›
Correct option is:
\[\boxed{\text{(C) }I}\]
🎯 Exam Significance
Exam Significance
›
For CBSE Board Examinations
  • This question tests the concept of an idempotent matrix.
  • It requires correct use of matrix algebra and the binomial expansion.
  • It reinforces the important identity:
    \[ A^2=A\Rightarrow A^n=A,\quad n\geq1. \]
  • Writing the intermediate steps clearly helps avoid errors in expressions involving matrix powers.
For JEE and Other Competitive Entrance Examinations
  • The condition \(A^2=A\) should immediately suggest the substitution \(A^2\rightarrow A\) and \(A^3\rightarrow A\).
  • The problem is designed to test algebraic manipulation rather than numerical matrix multiplication.
  • Recognising that \(I\) commutes with \(A\) allows the binomial theorem to be applied directly.
  • The fastest route is:
    \[ (I+A)^3 = I+3A+3A^2+A^3 = I+7A. \]
  • Therefore:
    \[ (I+A)^3-7A=I. \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. A matrix satisfying

    \[ A^2=A \]
    is called an idempotent matrix.

  2. For an idempotent matrix:

    \[ \boxed{A^n=A,\quad n\geq1}. \]

  3. In particular:

    \[ A^2=A,\qquad A^3=A. \]

  4. Since \(I\) commutes with \(A\):

    \[ (I+A)^3=I+3A+3A^2+A^3. \]

  5. Substituting \(A^2=A\) and \(A^3=A\):

    \[ (I+A)^3=I+7A. \]

  6. Therefore:

    \[ \boxed{(I+A)^3-7A=I}. \]

  7. Hence the correct answer is:

    \[ \boxed{\text{Option (C) }I}. \]
← Q10
11 / 11  ·  100%
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NCERT Class 12 Matrices Miscellaneous Exercise Solutions — Complete Notes & Solutions · academia-aeternum.com
Master the NCERT Class 12 Mathematics Chapter 3 Matrices Miscellaneous Exercise with these step-by-step solutions designed for CBSE board examinations and competitive entrance exams. This comprehensive collection explains each problem through essential matrix concepts, solution roadmaps, detailed calculations, and logical reasoning, helping students understand not just what to calculate but why each step works. The exercise covers important topics such as symmetric and skew-symmetric matrices,…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Frequently Asked Questions

    The Miscellaneous Exercise covers important concepts such as symmetric and skew-symmetric matrices, matrix multiplication, transpose, identity matrices, matrix equations, idempotent matrices, and applications of matrix algebra.

    Yes. Each solution provides a detailed, step-by-step approach with the required matrix operations, calculations, reasoning, and final answer.

    The Miscellaneous Exercise combines concepts from the Matrices chapter and helps students practise higher-level problems that are useful for CBSE board examinations.

    Yes. The solutions strengthen matrix algebra, symbolic manipulation, identities, and conceptual shortcuts that are useful for JEE Main, JEE Advanced, and other competitive entrance examinations.

    A square matrix A is symmetric if its transpose is equal to the matrix itself, that is, A? = A.

    A square matrix A is skew-symmetric if its transpose is equal to the negative of the matrix, that is, A? = -A. Its diagonal entries are always zero.

    A matrix that is both symmetric and skew-symmetric must be the zero matrix, because A? = A and A? = -A imply A = -A and hence A = 0.

    A square matrix A is called idempotent if A² = A. For an idempotent matrix, every positive integral power satisfies An = A.

    Focus on definitions, properties, matrix multiplication, transpose, symmetric and skew-symmetric matrices, matrix equations, and NCERT exercises. Practise every step and review common mistakes.

    Yes. The key takeaways, solution roadmaps, important concepts, exam significance, and quick-revision sections make these solutions useful for last-minute CBSE and competitive exam revision.

    Matrices MCQs are multiple-choice questions designed to test concepts such as matrix operations, transpose, determinants, inverse, rank, and special types of matrices.

    Yes. These questions cover concepts and problem patterns that are relevant to JEE Main preparation and other engineering entrance examinations.

    Yes. Several questions focus on conceptual properties, matrix identities, inverse matrices, determinants, and special matrices that are useful for JEE Advanced-level preparation.

    The questions cover types of matrices, matrix operations, transpose, symmetric and skew-symmetric matrices, determinants, inverse, singular matrices, orthogonal matrices, rank, trace, and matrix identities.

    Yes. Every MCQ includes the correct answer along with a concise explanation to help students understand the underlying concept.

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