Compute AB and BA
Question:
Given \[ A=\begin{bmatrix}1 & -2 \\ 2 & 3\end{bmatrix}, \quad B=\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1\end{bmatrix} \] Compute \(AB\) and \(BA\), wherever possible.
Given \[ A=\begin{bmatrix}1 & -2 \\ 2 & 3\end{bmatrix}, \quad B=\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1\end{bmatrix} \] Compute \(AB\) and \(BA\), wherever possible.
Solution:
Step 1: Check order
\(A\) is \(2 \times 2\), \(B\) is \(2 \times 3\)
- \(AB\): Possible (2×2 × 2×3 → 2×3)
- \(BA\): Not possible (2×3 × 2×2 ❌ mismatch)
Step 2: Compute \(AB\)
\[ AB = \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 1(1)+(-2)(2) & 1(2)+(-2)(3) & 1(3)+(-2)(1) \\ 2(1)+3(2) & 2(2)+3(3) & 2(3)+3(1) \end{bmatrix} \] \[ = \begin{bmatrix} 1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3 \end{bmatrix} = \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix} \]Step 3: Compute \(BA\)
Not defined because number of columns of \(B\) ≠ number of rows of \(A\).