Show That \(AB \ne BA\)

Solution:

Step 1: Compute \(AB\)

\[ AB = \begin{bmatrix} 1 & 3 & -1 \\ 2 & -1 & -1 \\ 3 & 0 & -1 \end{bmatrix} \begin{bmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{bmatrix} \] \[ = \begin{bmatrix} 1(-2)+3(-1)+(-1)(-6) & 1(3)+3(2)+(-1)(9) & 1(-1)+3(-1)+(-1)(-4) \\ 2(-2)+(-1)(-1)+(-1)(-6) & 2(3)+(-1)(2)+(-1)(9) & 2(-1)+(-1)(-1)+(-1)(-4) \\ 3(-2)+0(-1)+(-1)(-6) & 3(3)+0(2)+(-1)(9) & 3(-1)+0(-1)+(-1)(-4) \end{bmatrix} \] \[ = \begin{bmatrix} 1 & 0 & 0 \\ 3 & -5 & 3 \\ 0 & 0 & 1 \end{bmatrix} \]

Step 2: Compute \(BA\)

\[ BA = \begin{bmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{bmatrix} \begin{bmatrix} 1 & 3 & -1 \\ 2 & -1 & -1 \\ 3 & 0 & -1 \end{bmatrix} \] \[ = \begin{bmatrix} -2(1)+3(2)+(-1)(3) & -2(3)+3(-1)+(-1)(0) & -2(-1)+3(-1)+(-1)(-1) \\ -1(1)+2(2)+(-1)(3) & -1(3)+2(-1)+(-1)(0) & -1(-1)+2(-1)+(-1)(-1) \\ -6(1)+9(2)+(-4)(3) & -6(3)+9(-1)+(-4)(0) & -6(-1)+9(-1)+(-4)(-1) \end{bmatrix} \] \[ = \begin{bmatrix} 1 & -9 & 0 \\ 0 & -5 & 0 \\ 0 & -27 & 1 \end{bmatrix} \]

Conclusion:

\[ AB = \begin{bmatrix} 1 & 0 & 0 \\ 3 & -5 & 3 \\ 0 & 0 & 1 \end{bmatrix} \ne BA = \begin{bmatrix} 1 & -9 & 0 \\ 0 & -5 & 0 \\ 0 & -27 & 1 \end{bmatrix} \]

Hence, matrix multiplication is not commutative.