If A = [1223], B =[2132] and C =[1331],\begin{bmatrix} 1 & 2 \ 2 & 3 \end{bmatrix}, \text{ B } = \begin{bmatrix} 2 & 1 \ 3 & 2 \end{bmatrix} \text{ and C } = \begin{bmatrix} 1 & 3 \ 3 & 1 \end{bmatrix},[1223], B =[2312] and C =[1331], find the matrix C(B - A).
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B−A=[2132]−[1223]=[2−11−23−22−3]=[1−11−1]∴C(B−A)=[1331][1−11−1]=[1×1+3×11×(−1)+3×(−1)3×1+1×13×(−1)+1×(−1)]=[1+3−1−33+1−3−1]=[4−44−4].B - A = \begin{bmatrix}[r] 2 & 1 \ 3 & 2 \end{bmatrix} - \begin{bmatrix}[r] 1 & 2 \ 2 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 - 1 & 1 - 2 \ 3 - 2 & 2 - 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & -1 \ 1 & -1 \end{bmatrix} \\[1.5em] \therefore C(B - A) = \begin{bmatrix}[r] 1 & 3 \ 3 & 1 \end{bmatrix} \begin{bmatrix}[r] 1 & -1 \ 1 & -1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 1 + 3 \times 1 & 1 \times (-1) + 3 \times (-1) \ 3 \times 1 + 1 \times 1 & 3 \times (-1) + 1 \times (-1) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 + 3 & -1 - 3 \ 3 + 1 & -3 - 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 & -4 \ 4 & -4 \end{bmatrix}.B−A=[2312]−[1223]=[2−13−21−22−3]=[11−1−1]∴C(B−A)=[1331][11−1−1]=[1×1+3×13×1+1×11×(−1)+3×(−1)3×(−1)+1×(−1)]=[1+33+1−1−3−3−1]=[44−4−4].
Hence, the matrix C(B - A) = [4−44−4]\begin{bmatrix} 4 & -4 \ 4 & -4 \end{bmatrix}[44−4−4].
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If A = [1−22−1]and B =[32−21],\begin{bmatrix}[r] 1 & -2 \ 2 & -1 \end{bmatrix} \text{and B } = \begin{bmatrix}[r] 3 & 2 \ -2 & 1 \end{bmatrix},[12−2−1]and B =[3−221], find 2B - A2.
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(ii) (B + C)A
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If A = [3051] and B[−4210]\begin{bmatrix}[r] 3 & 0 \ 5 & 1 \end{bmatrix} \text{ and B} \begin{bmatrix}[r] -4 & 2 \ 1 & 0 \end{bmatrix}[3501] and B[−4120], find A2 - 2AB + B2.
