Q.If $A^T=\begin{pmatrix} 4 & 5 \\ -1 & 0 \\ 2 & 3\end{pmatrix}$ and $B=\begin{pmatrix} 2 & -1 & 1 \\ 7 & 5 & -2\end{pmatrix}$, verify the following

Step 1. Recover $A$. Transposing the given $A^T=\begin{pmatrix}4&5\\-1&0\\2&3\end{pmatrix}$ (swap rows and columns):

Step 2. (i) Compute $A+B$ and its transpose.

$A+B=\begin{pmatrix}4+2&-1-1&2+1\\5+7&0+5&3-2\end{pmatrix}=\begin{pmatrix}6&-2&3\\12&5&1\end{pmatrix}$

$(A+B)^T=\begin{pmatrix}6&12\\-2&5\\3&1\end{pmatrix}$

Step 3. (i) Compute $A^T+B^T$ and $B^T+A^T$.

$B^T=\begin{pmatrix}2&7\\-1&5\\1&-2\end{pmatrix}$

$A^T+B^T=\begin{pmatrix}4+2&5+7\\-1-1&0+5\\2+1&3-2\end{pmatrix}=\begin{pmatrix}6&12\\-2&5\\3&1\end{pmatrix}$

By commutativity of matrix addition, $B^T+A^T$ gives the identical matrix. All three expressions in (i) agree.

Step 4. (ii) Compute $A-B$ and its transpose.

$A-B=\begin{pmatrix}4-2&-1+1&2-1\\5-7&0-5&3+2\end{pmatrix}=\begin{pmatrix}2&0&1\\-2&-5&5\end{pmatrix}$

$(A-B)^T=\begin{pmatrix}2&-2\\0&-5\\1&5\end{pmatrix}$ …

• Forgetting to first recover $A$ from the given $A^T$ before computing $A+B$ or $A-B$ …