Evaluate : if possible :
[1−2][−23−14]\begin{bmatrix}[r] 1 & -2 \ \end{bmatrix}\begin{bmatrix}[r] -2 & 3 \ -1 & 4 \end{bmatrix}[1−2][−2−134]
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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.
⇒[1−2][−23−14]=[1×(−2)+(−2)×(−1)1×3+(−2)×4]=[−2+23+(−8)]=[0−5].\Rightarrow \begin{bmatrix}[r] 1 & -2 \ \end{bmatrix}\begin{bmatrix}[r] -2 & 3 \ -1 & 4 \end{bmatrix} = \begin{bmatrix}[r] 1 \times (-2) + (-2) \times (-1) & 1 \times 3 + (-2) \times 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -2 + 2 & 3 + (-8) \ \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 & -5 \ \end{bmatrix}.⇒[1−2][−2−134]=[1×(−2)+(−2)×(−1)1×3+(−2)×4]=[−2+23+(−8)]=[0−5].
Hence, [1−2][−23−14]=[0−5].\begin{bmatrix}[r] 1 & -2 \ \end{bmatrix}\begin{bmatrix}[r] -2 & 3 \ -1 & 4 \end{bmatrix} = \begin{bmatrix}[r] 0 & -5 \ \end{bmatrix}.[1−2][−2−134]=[0−5].
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If A = [1011],B=[0110] and C=[1100]\begin{bmatrix}[r] 1 & 0 \ 1 & 1 \end{bmatrix}, B = \begin{bmatrix}[r] 0 & 1 \ 1 & 0 \end{bmatrix}\text{ and C} = \begin{bmatrix}[r] 1 & 1 \ 0 & 0 \end{bmatrix}[1101],B=[0110] and C=[1010], the matrix A2 + 2B - 3C is :
[−2−141]\begin{bmatrix}[r] -2 & -1 \ 4 & 1 \end{bmatrix}[−24−11]
[2−141]\begin{bmatrix}[r] 2 & -1 \ 4 & 1 \end{bmatrix}[24−11]
[2141]\begin{bmatrix}[r] 2 & 1 \ 4 & 1 \end{bmatrix}[2411]
[21−4−1]\begin{bmatrix}[r] 2 & 1 \ -4 & -1 \end{bmatrix}[2−41−1]
[32][20]\begin{bmatrix}[r] 3 & 2 \ \end{bmatrix}\begin{bmatrix}[r] 2 \ 0 \end{bmatrix}[32][20]
[643−1][−13]\begin{bmatrix}[r] 6 & 4 \ 3 & -1 \end{bmatrix}\begin{bmatrix}[r] -1 \ 3 \end{bmatrix}[634−1][−13]
[643−1][−13]\begin{bmatrix}[r] 6 & 4 \ 3 & -1 \end{bmatrix}\begin{bmatrix}[r] -1 & 3 \ \end{bmatrix}[634−1][−13]
