Question

If $A^{\prime }=\left[\begin{array}{cc}3& 4\\ -1& 2\\ 0& 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right],$ then verify that:
$\left(i\right)$ $(A+B{)}^{\prime }=A^{\prime }+B^{\prime }$
$\left(ii\right)$ $(A-B{)}^{\prime }=A^{\prime }-B^{\prime }$

Answer 1

$\left(i\right)$ Given: $A^{\prime }=\left[\begin{array}{cc}3& 4\\ -1& 2\\ 0& 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right]$
To verify: $(A+B{)}^{\prime }=A^{\prime }+B^{\prime }$
$A=(A^{\prime }{)}^{\prime }={\left[\begin{array}{cc}3& 4\\ -1& 2\\ 0& 1\end{array}\right]}^{\prime }=\left[\begin{array}{ccc}3& -1& 0\\ 4& 2& 1\end{array}\right]$
Solving $L.H.S.$
$(A+B{)}^{\prime }={(\left[\begin{array}{ccc}3& -1& 0\\ 4& 2& 1\end{array}\right]+\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right])}^{\prime }$
$\Rightarrow (A+B{)}^{\prime }={\left(\left[\begin{array}{ccc}3+(-1)& -1+2& 0+1\\ 4+1& 2+2& 1+3\end{array}\right]\right)}^{\prime }$
$\Rightarrow (A+B{)}^{\prime }={\left(\left[\begin{array}{ccc}2& 1& 1\\ 5& 4& 4\end{array}\right]\right)}^{\prime }$
$\Rightarrow (A+B{)}^{\prime }=\left[\begin{array}{cc}2& 5\\ 1& 4\\ 1& 4\end{array}\right]$

Solving $R.H.S$
$A^{\prime }+B^{\prime }=\left[\begin{array}{cc}3& 4\\ -1& 2\\ 0& 1\end{array}\right]+{\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right]}^{\prime }$
$\Rightarrow A^{\prime }+B^{\prime }=\left[\begin{array}{cc}3& 4\\ -1& 2\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}-1& 1\\ 2& 2\\ 1& 3\end{array}\right]$
$\Rightarrow A^{\prime }+B^{\prime }=\left[\begin{array}{cc}3+(-1)& 4+1\\ -1+2& 2+2\\ 0+1& 1+3\end{array}\right]$
$\Rightarrow A^{\prime }+B^{\prime }=\left[\begin{array}{cc}2& 5\\ 1& 4\\ 1& 4\end{array}\right]$
So, $L.H.S.=R.H.S.$
Hence verified.

$\left(ii\right)$ Given: $A^{\prime }=\left[\begin{array}{cc}3& 4\\ -1& 2\\ 0& 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right]$
To verify: $(A-B{)}^{\prime }=A^{\prime }-B^{\prime }$
$A=(A^{\prime }{)}^{\prime }={\left[\begin{array}{cc}3& 4\\ -1& 2\\ 0& 1\end{array}\right]}^{\prime }=\left[\begin{array}{ccc}3& -1& 0\\ 4& 2& 1\end{array}\right]$
Solving $L.H.S.$
$(A-B{)}^{\prime }={(\left[\begin{array}{ccc}3& -1& 0\\ 4& 2& 1\end{array}\right]-\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right])}^{\prime }$
$\Rightarrow (A-B{)}^{\prime }={\left(\left[\begin{array}{ccc}3-(-1)& -1-2& 0-1\\ 4-1& 2-2& 1-3\end{array}\right]\right)}^{\prime }$
$\Rightarrow (A-B{)}^{\prime }={\left(\left[\begin{array}{ccc}4& -3& -1\\ 3& 0& -2\end{array}\right]\right)}^{\prime }$
$\Rightarrow (A-B{)}^{\prime }=\left[\begin{array}{cc}4& 3\\ -3& 0\\ -1& -2\end{array}\right]$

Solving $R.H.S$
$A^{\prime }-B^{\prime }=\left[\begin{array}{cc}3& 4\\ -1& 2\\ 0& 1\end{array}\right]-{\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right]}^{\prime }$
$\Rightarrow A^{\prime }-B^{\prime }=\left[\begin{array}{cc}3& 4\\ -1& 2\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}-1& 1\\ 2& 2\\ 1& 3\end{array}\right]$
$\Rightarrow A^{\prime }-B^{\prime }=\left[\begin{array}{cc}3-(-1)& 4-1\\ -1-2& 2-2\\ 0-1& 1-3\end{array}\right]$
$\Rightarrow A^{\prime }-B^{\prime }=\left[\begin{array}{cc}4& 3\\ -3& 0\\ -1& -2\end{array}\right]$
So, $L.H.S.=R.H.S.$
Hence verified.