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A=[−6042] and B=[1013]A = \begin{bmatrix}[r] -6 & 0 \ 4 & 2 \end{bmatrix} \text{ and } B = \begin{bmatrix}[r] 1 & 0 \ 1 & 3 \end{bmatrix}.

Find matrix M, if M = 12A−2B+5l\dfrac{1}{2}A - 2B + 5l, where l is the identity matrix.

Matrices

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Answer

Given,

⇒M=12A−2B+5l=12[−6042]−2[1013]+5[1001]=[−3021]−[2026]+[5005]=[−3−2+50−0+02−2+01−6+5]=[0000].\Rightarrow M = \dfrac{1}{2}A - 2B + 5l \\[1em] = \dfrac{1}{2}\begin{bmatrix}[r] -6 & 0 \ 4 & 2 \end{bmatrix} - 2\begin{bmatrix}[r] 1 & 0 \ 1 & 3 \end{bmatrix} + 5\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -3 & 0 \ 2 & 1 \end{bmatrix} - \begin{bmatrix}[r] 2 & 0 \ 2 & 6 \end{bmatrix} + \begin{bmatrix}[r] 5 & 0 \ 0 & 5 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -3 - 2 + 5 & 0 - 0 + 0 \ 2 - 2 + 0 & 1 - 6 + 5 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix}.

Hence, M = [0000].\begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix}.

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