Verify \((AB)C = A(BC)\)
Question:
Given \[ A=\begin{bmatrix}1 & 2 & 0 \\ -1 & 0 & 1\end{bmatrix}, \quad B=\begin{bmatrix}1 & 0 \\ -1 & 2 \\ 0 & 3\end{bmatrix}, \quad C=\begin{bmatrix}1 \\ -1\end{bmatrix} \] verify that: \[ (AB)C = A(BC) \]
Given \[ A=\begin{bmatrix}1 & 2 & 0 \\ -1 & 0 & 1\end{bmatrix}, \quad B=\begin{bmatrix}1 & 0 \\ -1 & 2 \\ 0 & 3\end{bmatrix}, \quad C=\begin{bmatrix}1 \\ -1\end{bmatrix} \] verify that: \[ (AB)C = A(BC) \]
Solution:
Step 1: Compute \(AB\)
\[ AB = \begin{bmatrix} 1 & 2 & 0 \\ -1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 2 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 \\ -1 & 3 \end{bmatrix} \]Step 2: Compute \((AB)C\)
\[ (AB)C = \begin{bmatrix} -1 & 4 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} -1 – 4 \\ -1 – 3 \end{bmatrix} = \begin{bmatrix} -5 \\ -4 \end{bmatrix} \]Step 3: Compute \(BC\)
\[ BC = \begin{bmatrix} 1 & 0 \\ -1 & 2 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ -3 \\ -3 \end{bmatrix} \]Step 4: Compute \(A(BC)\)
\[ A(BC) = \begin{bmatrix} 1 & 2 & 0 \\ -1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -3 \\ -3 \end{bmatrix} = \begin{bmatrix} 1 – 6 \\ -1 – 3 \end{bmatrix} = \begin{bmatrix} -5 \\ -4 \end{bmatrix} \]Conclusion:
\[ (AB)C = A(BC) = \begin{bmatrix} -5 \\ -4 \end{bmatrix} \]Hence, associativity is verified.