Question
Let \[ A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 3 & 3 \end{bmatrix},\ B = \begin{bmatrix} 3 & 1 \\ 5 & 2 \\ -2 & 4 \end{bmatrix},\ C = \begin{bmatrix} 4 & 2 \\ -3 & 5 \\ 5 & 0 \end{bmatrix} \] Verify that \(AB = AC\) though \(B \ne C,\ A \ne O\).
Solution
Step 1: Compute \(AB\)
\[ AB = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 3 & 3 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 5 & 2 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} (3+5-2) & (1+2+4) \\ 3(3+5-2) & 3(1+2+4) \end{bmatrix} \] \[ = \begin{bmatrix} 6 & 7 \\ 18 & 21 \end{bmatrix} \]Step 2: Compute \(AC\)
\[ AC = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 3 & 3 \end{bmatrix} \begin{bmatrix} 4 & 2 \\ -3 & 5 \\ 5 & 0 \end{bmatrix} = \begin{bmatrix} (4-3+5) & (2+5+0) \\ 3(4-3+5) & 3(2+5+0) \end{bmatrix} \] \[ = \begin{bmatrix} 6 & 7 \\ 18 & 21 \end{bmatrix} \]Step 3: Compare
\[ AB = AC = \begin{bmatrix} 6 & 7 \\ 18 & 21 \end{bmatrix} \] \[ \text{But } B \ne C,\quad A \ne O \]Final Result
\[
AB = AC \quad \text{though} \quad B \ne C,\ A \ne O
\]
Hence verified.