Transpose of a Matrix

Transpose of matrix

A matrix can be transposed by interchanging its rows into columns or its columns into rows. The letter "T" in the superscript of the matrix is used to denote the transpose of the matrix.

In simple language, the transpose of a matrix is changing its rows into columns or columns into rows.

Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{\mathrm{m} \times \mathrm{n}}$ be a matrix, then matrix obtained by changing rows into columns or vice-versa will give transpose of $\mathrm{A}$ which is denoted as $\mathrm{A}^{\prime}$ or $\mathrm{A}^{\top}$. Hence $\mathrm{A}^{\prime}=\left[\mathrm{a}_{\mathrm{ji}}\right]_{\mathrm{n} \times \mathrm{m}}$

Steps to find Transpose of Matrix

For the time being, let's assume that a matrix is a 2×3 matrix. Its measurements are two rows by three columns. When a matrix is transposed, the elements in the first row of the original matrix are recorded in the first column of the new matrix. The entries from the second row of the original matrix are similarly contained in the second column of the new matrix. The new matrix's order is now 3×2, as it has two columns and three rows.

Example,

$
\mathrm{A}=\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right] \Rightarrow \mathrm{A}^{\prime}=\left[\begin{array}{lll}
a_{11} & a_{21} & a_{31} \\
a_{12} & a_{22} & a_{32} \\
a_{13} & a_{23} & a_{33}
\end{array}\right]
$

If, $A=\left[\begin{array}{ll}2 & 6 \\ 3 & 7 \\ 5 & 8\end{array}\right]$ then, $A^{\prime}=\left[\begin{array}{lll}2 & 3 & 5 \\ 6 & 7 & 8\end{array}\right]$

Properties of the transpose of a matrix:

If $A^{\prime}$ and $B^{\prime}$ denote the transpose of the matrices $A$ and $B$, then :

i) Transpose of the Transpose Matrix

The matrix that results from taking the transpose of the transpose matrix is equal to the original matrix. Hence $\left(A^{\prime}\right)^{\prime}=A$

ii) Addition of Transpose Matrix

The resultant transpose of the addition of two matrices $A$ and $B$ is precisely equal to the total of the transposes of $A$ and $B$ separately

Hence, $(A \pm B)^{\prime}=A^{\prime} \pm B^{\prime}$ (given that $A$ and $B$ are conformable for matrix addition)

iii) Multiplication by constant

The matrix acquired is identical to the transpose of the original matrix multiplied by the constant when a matrix is multiplied by a constant and its transpose is taken.

In other words, $(k A)^{\prime}=k A^{\prime}$

iv) Multiplication Properties of Transpose

The product of the transpose of the two matrices in reverse order equals the transpose of the product of two matrices.

That's $(A B)^{\prime}=B^{\prime} A^{\prime}($ given that $A$ and $B$ are conformable for matrix product $A B)$

Recommended Video Based on Transpose of Matrix:

Solved Example Based on Transpose of Matrix

Example 1: Let $
\mathrm{A}=\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{ccc}
9^2 & -10^2 & 11^2 \\
12^2 & 13^2 & -14^2 \\
-15^2 & 16^2 & 17^2
\end{array}\right]
$ then the value of $\mathrm{A}^{\prime} \mathrm{BA}$ is: [JEE Main 2022]

Solution:

Example 2: Let $\mathrm{S}$ be the set containing all $3 \times 3$ matrices with entries from $\{-1,0,1\}$. The total number of matrices $\mathrm{A} \in \mathrm{S}$ such that the sum of all the diagonal elements of $\mathrm{A}^{\mathrm{T}} \mathrm{A}$ is 6 is $\qquad$

[JEE Main 2022]
Solution:
$
\begin{aligned}
& \operatorname{Tr}_{\mathrm{r}}\left(\mathrm{AA}^{\top}\right)=6 \\
& \mathrm{AA}^{\top}=\left[\begin{array}{lll}
a & d & g \\
b & e & h \\
c & f & i
\end{array}\right]\left[\begin{array}{lll}
a & b & c \\
d & e & f \\
g & h & i
\end{array}\right]
\end{aligned}
$

Now given $\mathrm{a}^2+\mathrm{d}^2+\mathrm{y}^2+\mathrm{b}^2+\mathrm{e}^2+\mathrm{h}^2+\mathrm{c}^2+\mathrm{f}^2+\mathrm{i}^2=6$
$
\begin{aligned}
& ={ }^9 \mathrm{c}_3 \times 2^6 \\
& =5376
\end{aligned}
$

Hence, The total number of matrices $\mathrm{A} \in \mathrm{S}$ such that the sum of all the diagonal elements of $\mathrm{A}^{\mathrm{T}} \mathrm{A}$ is 6 is 5376.

Example 3: Let $\mathrm{X}=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$ and $\mathrm{A}=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]$. For $\mathrm{k} \in \mathbb{N}$, if $\mathrm{X}^{\prime} \mathrm{A}^{\mathrm{k}} \mathrm{X}=33$, then $\mathrm{k}$ is equal to $\qquad$

Solution:

$
\begin{aligned}
& \mathrm{A}^2=\left[\begin{array}{ccc}
-1 & 2 & 3 \\
0 & 1 & 6 \\
0 & 0 & -1
\end{array}\right]\left[\begin{array}{ccc}
-1 & 2 & 3 \\
0 & 1 & 6 \\
0 & 0 & -1
\end{array}\right] \\
& \quad=\left[\begin{array}{lll}
1 & 0 & 6 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\mathrm{I}+\left[\begin{array}{lll}
0 & 0 & 6 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] \\
& \text { Let }\left[\begin{array}{lll}
0 & 0 & 6 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=\mathrm{B} \\
& \mathrm{B}^2=0, \\
& \therefore \mathrm{B}^3=0=\mathrm{B}^4=\mathrm{B}^5=\cdots \\
& \text { Now }\left(\mathrm{A}^2\right)^{\mathrm{P}}=(\mathrm{I}+\mathrm{B})^{\mathrm{P}} . \\
& \quad=\mathrm{I}+\mathrm{pB} \\
& \quad=\left[\begin{array}{lll}
1 & 0 & 6 \mathrm{p} \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}
$

Now $x^{\prime} A^k x$
Let $\mathrm{k}$ be even and $\mathrm{k}=2 \mathrm{p}$
$
\begin{aligned}
& {\left[\begin{array}{lll}
1 & 1 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 6 \mathrm{p} \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]} \\
& =\left[\begin{array}{lll}
1 & 1 & 6 \mathrm{p}+1
\end{array}\right]\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right] \\
& =[6 \mathrm{p}+3]
\end{aligned}
$

Now $6 \mathrm{p}+3=33$
$
\begin{aligned}
& \Rightarrow \mathrm{p}=5 \\
& \mathrm{k}=2 \mathrm{p}=10
\end{aligned}
$

Hence, the answer is 10.

Example 4: Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{\prime}=\alpha A+I$, where a $\epsilon \mathbb{R}-\{-1,1\}$. If $\left|\mathrm{A}^2-\mathrm{A}\right|=4$, then the sum of all possible values of $\alpha$ is equal to :

Solution:

$
\begin{aligned}
& \mathrm{A}^{\mathrm{T}}=\alpha \mathrm{A}+\mathrm{I} \\
& \mathrm{A}=\alpha \mathrm{A}^{\mathrm{T}}+\mathrm{I} \\
& \mathrm{A}=\alpha(\alpha \mathrm{A}+\mathrm{I})+\mathrm{I} \\
& \mathrm{A}=\alpha^2 \mathrm{~A}+(\alpha+1) \mathrm{I} \\
& \mathrm{A}\left(1-\alpha^2\right)=(\alpha+1) \mathrm{I} \\
& \mathrm{A}=\frac{\mathrm{I}}{1-\alpha} \ldots(1) \\
& |\mathrm{A}|=\frac{1}{(1-\alpha)^2} \ldots(2) \\
& \left|\mathrm{A}^2-\mathrm{A}\right|=|\mathrm{A}||\mathrm{A}-\mathrm{I}| \ldots(3) \\
& \mathrm{A}-\mathrm{I}=\frac{\mathrm{I}}{\mathrm{I}-\alpha}-\mathrm{I}=\frac{\alpha}{1-\alpha} \mathrm{I} \\
& |\mathrm{A}-\mathrm{I}|=\left(\frac{\alpha}{1-\alpha}\right)^2 \ldots(4)
\end{aligned}
$

Now $\left|\mathrm{A}^2-\mathrm{A}\right|=4$
$
|\mathrm{A}||\mathrm{A}-\mathrm{I}|=4
$

$
\begin{aligned}
& \Rightarrow \frac{1}{(1-\alpha)^2} \frac{\alpha^2}{\left(1-\alpha^2\right)}=4 \\
& \Rightarrow \frac{\alpha}{(1-\alpha)^2}= \pm 2 \\
& \Rightarrow 2(1-\alpha)^2= \pm \alpha \\
& \left(C_1\right) 2(1-\alpha)^2=\alpha \\
& \left(C_2\right) 2(1-\alpha)^3=-\alpha \\
& 2 \alpha^2-5 \alpha+2=0 \alpha_1 \\
& 2 \alpha^2-3 \alpha+2=0 \\
& \alpha_1+\alpha_2=\frac{5}{2} \\
& \alpha \notin \mathrm{R}
\end{aligned}
$

Hence, the sum of all possible values of $\alpha$ is equal to $\frac{5}{2}$