Evaluate : if possible :
[643−1][−13]\begin{bmatrix}[r] 6 & 4 \ 3 & -1 \end{bmatrix}\begin{bmatrix}[r] -1 \ 3 \end{bmatrix}[634−1][−13]
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For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.
⇒[643−1][−13]=[6×(−1)+4×33×(−1)+(−1)×3]=[−6+12−3+(−3)]=[6−6].\Rightarrow \begin{bmatrix}[r] 6 & 4 \ 3 & -1 \end{bmatrix}\begin{bmatrix}[r] -1 \ 3 \end{bmatrix} = \begin{bmatrix}[r] 6 \times (-1) + 4 \times 3 \ 3 \times (-1) + (-1) \times 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -6 + 12 \ -3 + (-3) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 6 \ -6 \end{bmatrix}.⇒[634−1][−13]=[6×(−1)+4×33×(−1)+(−1)×3]=[−6+12−3+(−3)]=[6−6].
Hence, [643−1][−13]=[6−6].\begin{bmatrix}[r] 6 & 4 \ 3 & -1 \end{bmatrix}\begin{bmatrix}[r] -1 \ 3 \end{bmatrix} = \begin{bmatrix}[r] 6 \ -6 \end{bmatrix}.[634−1][−13]=[6−6].
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[32][20]\begin{bmatrix}[r] 3 & 2 \ \end{bmatrix}\begin{bmatrix}[r] 2 \ 0 \end{bmatrix}[32][20]
[1−2][−23−14]\begin{bmatrix}[r] 1 & -2 \ \end{bmatrix}\begin{bmatrix}[r] -2 & 3 \ -1 & 4 \end{bmatrix}[1−2][−2−134]
[643−1][−13]\begin{bmatrix}[r] 6 & 4 \ 3 & -1 \end{bmatrix}\begin{bmatrix}[r] -1 & 3 \ \end{bmatrix}[634−1][−13]
If A = [025−2],B=[1−132]\begin{bmatrix}[r] 0 & 2 \ 5 & -2 \end{bmatrix}, B = \begin{bmatrix}[r] 1 & -1 \ 3 & 2 \end{bmatrix}[052−2],B=[13−12] and I is a unit matrix of order 2 × 2, find :
(i) AB
(ii) BA
(iii) AI
