Question:
Verify the associativity of matrix multiplication, i.e. \((AB)C = A(BC)\), where
\[ A = \begin{bmatrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \]
Solution:
Step 1: Compute AB
\[ AB = \begin{bmatrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix} \] \[ AB = \begin{bmatrix} 6 & -5 & 11 \\ 4 & -2 & 5 \\ 2 & -4 & 4 \end{bmatrix} \]Step 2: Compute (AB)C
\[ (AB)C = \begin{bmatrix} 6 & -5 & 11 \\ 4 & -2 & 5 \\ 2 & -4 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \] \[ (AB)C = \begin{bmatrix} -9 & 12 & 0 \\ -2 & 8 & -1 \\ -10 & 4 & -2 \end{bmatrix} \]Step 3: Compute BC
\[ BC = \begin{bmatrix} 0 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \] \[ BC = \begin{bmatrix} -3 & 0 & 0 \\ 3 & 0 & 3 \\ -1 & 4 & -2 \end{bmatrix} \]Step 4: Compute A(BC)
\[ A(BC) = \begin{bmatrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{bmatrix} \begin{bmatrix} -3 & 0 & 0 \\ 3 & 0 & 3 \\ -1 & 4 & -2 \end{bmatrix} \] \[ A(BC) = \begin{bmatrix} -9 & 12 & 0 \\ -2 & 8 & -1 \\ -10 & 4 & -2 \end{bmatrix} \]Final Result:
\[ (AB)C = A(BC) = \begin{bmatrix} -9 & 12 & 0 \\ -2 & 8 & -1 \\ -10 & 4 & -2 \end{bmatrix} \]Hence, associativity is verified.
Conclusion:
Thus, matrix multiplication is associative, i.e., \((AB)C = A(BC)\). This property holds for all matrices of compatible order.