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

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 $1 \leq j \leq n$.
(ii)(A-B)'=A'-B'
$Given A=\begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix}$ and $B= \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}$
LHS: $A-B=\begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix}+(-1)\begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}$
$A-B=\begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix}+\begin{bmatrix} 4 &- 1 & 5 \\ -1 & -2 & 0 \\ -1 & -3 & -1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} -1+4 & 2 -1& 3+5 \\ 5-1 & 7-2 & 9+0 \\ -2-1 & 1-3 & 1-1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix}$
$(A-B)'=\begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$
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'=A'+(-1)B'
$\;\;\;=\begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}+(-1)\begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}+\begin{bmatrix} 4 & -1 & -1 \\ -1 &- 2 & -3 \\ 5 & 0 &- 1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} -1+4 & 5-1 & -2-1 \\ 2-1 & 7-2 & 1-3 \\ 3+5 & 9+0 & 1-1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$
Hence LHS=RHS.

edited Mar 13, 2013