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

$A^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$

since, $\left(A^{\prime}\right)^{\prime}=A$, we have

$A=\left[\begin{array}{ccc}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]$

$A+B=\left[\begin{array}{ccc}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]+\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$

$=\left[\begin{array}{lll}2 & 1 & 1 \\ 5 & 4 & 4\end{array}\right]$

LHS:

RHS:

$A^{\prime}+B^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]+\left[\begin{array}{rr}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]=\left[\begin{array}{ll}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right]$

$\mathrm{LHS}=\mathrm{RHS}$

(ii)

$A-B=\left[\begin{array}{ccc}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]-\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]=\left[\begin{array}{ccc}4 & -3 & -1 \\ 3 & 0 & -2\end{array}\right]$

$(A-B)^{\prime}=\left[\begin{array}{cr}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$

RHS:

$A^{\prime}-B^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]-\left[\begin{array}{rr}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]=\left[\begin{array}{cr}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$

$\mathrm{LHS}=\mathrm{RHS}$