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Mathematics Question on Matrices

If A'= \begin{bmatrix} 3 & 4 \\\ -1 & 2 \\\ 0 &1 \end{bmatrix}$$\begin{bmatrix} -1 & 2 & 1 \\\ 1 &2 & 3\end{bmatrix} , then verify that
(i) (A+B)′=A′+B′(A+B)'=A'+B'
(ii) (A−B)′=A′−B′(A-B)'=A'-B'

Answer

(i) It is known that A=(A')'
Therefore, we have:
A= [3−10 421]\begin{bmatrix} 3 & -1 & 0 \\\ 4 & 2 & 1 \end{bmatrix}

B'= [−11 22 13]\begin{bmatrix} -1 & 1 \\\ 2 & 2 \\\ 1 &3 \end{bmatrix}

A+BA+B = [3−10 421]\begin{bmatrix} 3 & -1 & 0 \\\ 4 & 2 & 1 \end{bmatrix} + [−121 123]\begin{bmatrix} -1 & 2 & 1 \\\ 1 &2 & 3\end{bmatrix}= [211 544]\begin{bmatrix} 2 & 1 & 1 \\\ 5 & 4 & 4 \end{bmatrix}

∴(A+B)′=\therefore (A+B)'= [25 14 14]\begin{bmatrix} 2 & 5 \\\ 1 & 4 \\\ 1 &4 \end{bmatrix}

A′+B′=A'+B'= [34 −12 01]\begin{bmatrix} 3 & 4 \\\ -1 & 2 \\\ 0 &1 \end{bmatrix}+ [−11 22 13]\begin{bmatrix} -1 & 1 \\\ 2 & 2 \\\ 1 &3 \end{bmatrix}= [25 14 14]\begin{bmatrix} 2 & 5 \\\ 1 & 4 \\\ 1 &4 \end{bmatrix}

Thus, we verified that:(A+B)'=A'+B'

(ii) A−BA-B= [3−10 421]\begin{bmatrix} 3 & -1 & 0 \\\ 4 & 2 & 1 \end{bmatrix}- [−121 123]\begin{bmatrix} -1 & 2 & 1 \\\ 1 &2 & 3\end{bmatrix} = [4−3−1 30−2]\begin{bmatrix} 4 & -3 & -1 \\\ 3 &0 & -2\end{bmatrix}

so(A−B)′ (A-B)' = [−43 −30 −1−2]\begin{bmatrix} -4 & 3 \\\ -3 & 0 \\\ -1 &-2 \end{bmatrix}

A'-B'= [34 −12 01]\begin{bmatrix} 3 & 4 \\\ -1 & 2 \\\ 0 &1 \end{bmatrix}- [−11 22 13]\begin{bmatrix} -1 & 1 \\\ 2 & 2 \\\ 1 &3 \end{bmatrix}= [−43 −30 −1−2]\begin{bmatrix} -4 & 3 \\\ -3 & 0 \\\ -1 &-2 \end{bmatrix}

Hence we verified that: (A−B)′=A′−B′(A-B)'=A'-B'