Show That \(AB = BA = O_{3\times3}\)
Question:
If \[ A=\begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix}, \quad B=\begin{bmatrix} -1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5 \end{bmatrix} \] show that: \[ AB = BA = O_{3\times3} \]
If \[ A=\begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix}, \quad B=\begin{bmatrix} -1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5 \end{bmatrix} \] show that: \[ AB = BA = O_{3\times3} \]
Solution:
Step 1: Compute \(AB\)
Multiply rows of \(A\) with columns of \(B\): \[ AB = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] (Each row of \(A\) is orthogonal to each column of \(B\), giving zero.)Step 2: Compute \(BA\)
\[ BA = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] (Similarly, rows of \(B\) give zero when multiplied with columns of \(A\).)Conclusion:
\[ AB = BA = O_{3\times3} \]Hence proved.