Show That \(AB \ne BA\)
Question:
Given \[ A=\begin{bmatrix}5 & -1 \\ 6 & 7\end{bmatrix}, \quad B=\begin{bmatrix}2 & 1 \\ 3 & 4\end{bmatrix} \] show that \(AB \ne BA\).
Given \[ A=\begin{bmatrix}5 & -1 \\ 6 & 7\end{bmatrix}, \quad B=\begin{bmatrix}2 & 1 \\ 3 & 4\end{bmatrix} \] show that \(AB \ne BA\).
Solution:
Step 1: Compute \(AB\)
\[ AB = \begin{bmatrix} 5(2)+(-1)(3) & 5(1)+(-1)(4) \\ 6(2)+7(3) & 6(1)+7(4) \end{bmatrix} \] \[ = \begin{bmatrix} 10-3 & 5-4 \\ 12+21 & 6+28 \end{bmatrix} = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix} \]Step 2: Compute \(BA\)
\[ BA = \begin{bmatrix} 2(5)+1(6) & 2(-1)+1(7) \\ 3(5)+4(6) & 3(-1)+4(7) \end{bmatrix} \] \[ = \begin{bmatrix} 10+6 & -2+7 \\ 15+24 & -3+28 \end{bmatrix} = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix} \]Conclusion:
\[ AB = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix} \ne BA = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix} \]Hence, matrix multiplication is not commutative.