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# if $A = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} \text{ and } B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix} \text{, then verify that }$ $$\text{ (i) } (A+B)' = A' + B'$$

This is a multi-part questions that has been entered as multiple separate questiosn on Clay6.com

Toolbox:

• The transpose of a matrix can be obtained by interchanging the rows and the column => A=A'
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where $1 \leq i \leq m$ and $1 \leq j \leq n$.
$(A+B)'=A'+B'$.
Consider the LHS
$(A+B)'= \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} +\begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}$
$A+B=\begin{bmatrix} -1-4 & 2+1 &3 -5 \\ 5+1 &7+ 2 & 9+0 \\ -2+1 & 1+3 & 1+1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix}$
Transpose of $A+B$ can be obtained by changing the rows into column.
$(A+B)'=\begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$

Consider the RHS: A'+B'
Given $A=\begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix}$, $A'=\begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}$
Given $B=\begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}$, $B'=\begin{bmatrix} -4 & 1 & 1 \\ 1 & 2& 3 \\ -5 & 0 & 1 \end{bmatrix}$
$A'+B'=\begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}+\begin{bmatrix} -4 & 1 & 1 \\ 1 & 2& 3 \\ -5 & 0 & 1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} -1-4 & 5+1 & -2+1 \\ 2+1 & 7+2& 1+3 \\ 3-5 &9+ 0 & 1+1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} -5 & 6 & -1 \\ 3 & 9& 4 \\ -2 & 9 & 2 \end{bmatrix}$
Hence LHS=RHS.
(A+B)'=A'+B'.