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
›
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
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
Also, if \(A\) is a square matrix, then \(A^T\) has the same order as \(A\).
Step-by-step Plan
›
Identify the order of each given matrix.
Interchange every row with the corresponding column.
Write the resulting matrix carefully, preserving every entry.
Check that the order changes from \(m\times n\) to \(n\times m\).
For a square matrix, verify that the entries across the principal diagonal have been reflected.
(i) Transpose of the Column Matrix<
›
-
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. - On interchanging rows and columns, the first column becomes the first row:\[ A^T= \begin{bmatrix} 5 & \dfrac{1}{2} & -1 \end{bmatrix} \]
- Hence,\[A^T= \begin{bmatrix} 5 & \dfrac{1}{2} & -1 \end{bmatrix}\]
- Order check:\[ 3\times1\longrightarrow1\times3 \]
(ii) Transpose of the \(2\times2\) Matrix
›
-
Given — matrix is \[ A= \begin{bmatrix} 1 & 2\\ -1 & 3 \end{bmatrix} \]The matrix \(A\) is of order \(2\times2\).
- Its rows are\[ R_1=\begin{bmatrix}1&2\end{bmatrix}, \quad R_2=\begin{bmatrix}-1&3\end{bmatrix} \]
- Therefore, the first row becomes the first column of \(A^T\):\[ \begin{bmatrix} 1\\ 2 \end{bmatrix} \]
- The second row becomes the second column of \(A^T\):\[ \begin{bmatrix} -1\\ 3 \end{bmatrix} \]
- Hence,\[ A^T= \begin{bmatrix} 1 & -1\\ 2 & 3 \end{bmatrix} \]
(iii) Transpose of the \(3\times3\) Matrix
›
-
Given — matrix is \[ B= \begin{bmatrix} -1 & 5 & 6\\ \sqrt{3} & 5 & 6\\ 2 & 3 & -1 \end{bmatrix} \]The matrix \(B\) is of order \(3\times3\).
- Write its rows as\[ R_1= \begin{bmatrix} -1&5&6 \end{bmatrix} \]\[ R_2= \begin{bmatrix} \sqrt{3}&5&6 \end{bmatrix} \]\[ R_3= \begin{bmatrix} 2&3&-1 \end{bmatrix} \]
- On taking the transpose, the first row becomes the first column:\[ \begin{bmatrix} -1\\ 5\\ 6 \end{bmatrix} \]
- The second row becomes the second column:\[ \begin{bmatrix} \sqrt{3}\\ 5\\ 3 \end{bmatrix} \]
- The third row becomes the third column:\[ \begin{bmatrix} 2\\ 6\\ -1 \end{bmatrix} \]
- Combining these columns, we obtain\[ B^T= \begin{bmatrix} -1 & \sqrt{3} & 2\\ 5 & 5 & 3\\ 6 & 6 & -1 \end{bmatrix} \]
- Order check:\[ 3\times3\longrightarrow3\times3 \]
- Since \(B\) is square, its transpose has the same order.
Final Answer
›
\( { (ii)\quad \begin{bmatrix} 1 & -1\\ 2 & 3 \end{bmatrix}} \)
\( { (iii)\quad \begin{bmatrix} -1 & \sqrt{3} & 2\\ 5 & 5 & 3\\ 6 & 6 & -1 \end{bmatrix}} \)
Exam Significance
›
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
›
-
The transpose of a matrix is obtained by interchanging its rows and columns.
-
If \(A\) is \(m\times n\), then \(A^T\) is \(n\times m\).
-
A column matrix becomes a row matrix after transposition.
-
A row matrix becomes a column matrix after transposition.
-
For a square matrix, transposition reflects the matrix about its principal diagonal.
-
Diagonal entries remain unchanged during transposition.
-
A matrix is symmetric if and only if \(A^T=A\).
-
Do not assume \(A^T=A\) merely because \(A\) is a square matrix.
-
For the second matrix in this question, the correct transpose is \(\begin{bmatrix}1&-1\2&3\end{bmatrix}\), not the original matrix.
-
Always perform an order check and an entry-by-entry check after transposition.