Q.If $A = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix}$, $B = \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix}$ and $C = \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix}$, show that $(A+B)+C = A+(B+C)$.

Given $A = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix}$, $B = \begin{bmatrix} -1 & 2 \\ 2 & 2 \\ 0 & 3 \end{bmatrix}$, $C = \begin{bmatrix} 4 & 3 \\ -1 & 4 \\ -2 & 1 \end{bmatrix}$.

From part (i), $A+B = \begin{bmatrix} 1 & -1 \\ 7 & -2 \\ -6 & 4 \end{bmatrix}$. So

$(A+B)+C = \begin{bmatrix} 1+4 & -1+3 \\ 7+(-1) & -2+4 \\ -6+(-2) & 4+1 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 6 & 2 \\ -8 & 5 \end{bmatrix}$

Now compute $B+C = \begin{bmatrix} -1+4 & 2+3 \\ 2+(-1) & 2+4 \\ 0+(-2) & 3+1 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\ 1 & 6 \\ -2 & 4 \end{bmatrix}$, so

$A+(B+C) = \begin{bmatrix} 2+3 & -3+5 \\ 5+1 & -4+6 \\ -6+(-2) & 1+4 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 6 & 2 \\ -8 & 5 \end{bmatrix}$

Both groupings give the same matrix, confirming $(A+B)+C=A+(B+C)$.

Compute (A+B)+C and A+(B+C) separately, entry-by-entry, and show both give the same matrix.

• Re-deriving A+B instead of reusing the already-verified result from part (i), inviting fresh arithmetic errors.
• Adding entries out of position when combining three matrices.