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*}A=[−6402] and B=[1103].
Find matrix M, if M = 12A−2B+5l\dfrac{1}{2}A - 2B + 5l21A−2B+5l, where l is the identity matrix.
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*}.⇒M=21A−2B+5l=21[−6402]−2[1103]+5[1001]=[−3201]−[2206]+[5005]=[−3−2+52−2+00−0+01−6+5]=[0000].
Hence, M = [0000].\begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}.[0000].
