Show That \(AB \ne BA\)

Solution:

Step 1: Compute \(AB\)

\[ AB = \begin{bmatrix} 1 & 3 & 0 \\ 1 & 1 & 0 \\ 4 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 5 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 1(0)+3(1)+0(0) & 1(1)+3(0)+0(5) & 1(0)+3(0)+0(1) \\ 1(0)+1(1)+0(0) & 1(1)+1(0)+0(5) & 1(0)+1(0)+0(1) \\ 4(0)+1(1)+0(0) & 4(1)+1(0)+0(5) & 4(0)+1(0)+0(1) \end{bmatrix} \] \[ = \begin{bmatrix} 3 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 4 & 0 \end{bmatrix} \]

Step 2: Compute \(BA\)

\[ BA = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 5 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 0 \\ 1 & 1 & 0 \\ 4 & 1 & 0 \end{bmatrix} \] \[ = \begin{bmatrix} 0(1)+1(1)+0(4) & 0(3)+1(1)+0(1) & 0 \\ 1(1)+0(1)+0(4) & 1(3)+0(1)+0(1) & 0 \\ 0(1)+5(1)+1(4) & 0(3)+5(1)+1(1) & 0 \end{bmatrix} \] \[ = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 3 & 0 \\ 9 & 6 & 0 \end{bmatrix} \]

Conclusion:

\[ AB = \begin{bmatrix} 3 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 4 & 0 \end{bmatrix} \ne BA = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 3 & 0 \\ 9 & 6 & 0 \end{bmatrix} \]

Hence, matrix multiplication is not commutative.