Q.If $A=\begin{bmatrix}1 & 0 & 1\\3 & 1 & 2\end{bmatrix}$, $B=\begin{bmatrix}2 & 1 & -4\\3 & 5 & -2\end{bmatrix}$ and $C=\begin{bmatrix}0 & 2 & 3\\-1 & -1 & 0\end{bmatrix}$, verify that $(A+2B+2C)^T=A^T+2B^T+3C^T$

$A=\begin{bmatrix}1 & 0 & 1\\3 & 1 & 2\end{bmatrix}$, $B=\begin{bmatrix}2 & 1 & -4\\3 & 5 & -2\end{bmatrix}$, $C=\begin{bmatrix}0 & 2 & 3\\-1 & -1 & 0\end{bmatrix}$.

LHS: $A+2B+2C=\begin{bmatrix}1+4+0 & 0+2+4 & 1-8+6\\3+6-2 & 1+10-2 & 2-4+0\end{bmatrix}=\begin{bmatrix}5 & 6 & -1\\7 & 9 & -2\end{bmatrix}$.

So $(A+2B+2C)^T=\begin{bmatrix}5 & 7\\6 & 9\\-1 & -2\end{bmatrix}$.

Now check the RHS exactly as printed in the book, $A^T+2B^T+3C^T$:

$A^T=\begin{bmatrix}1 & 3\\0 & 1\\1 & 2\end{bmatrix}$, $2B^T=\begin{bmatrix}4 & 6\\2 & 10\\-8 & -4\end{bmatrix}$, $3C^T=\begin{bmatrix}0 & -3\\6 & -3\\9 & 0\end{bmatrix}$.

$A^T+2B^T+3C^T=\begin{bmatrix}5 & 6\\8 & 8\\2 & -2\end{bmatrix}$ — this does NOT equal $(A+2B+2C)^T=\begin{bmatrix}5 & 7\\6 & 9\\-1 & -2\end{bmatrix}$. …

Compute $(A+2B+2C)^T$ directly and compare with the printed RHS $A^T+2B^T+3C^T$; also check the c …

• Blindly forcing the printed RHS coefficient (3) to match without checking, which produces a false 'verification'. …