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Mathematics

If A=[12−34],B=[01−25] and C =[−20−11],A = \begin{bmatrix}[r] 1 & 2 \ -3 & 4 \end{bmatrix}, B = \begin{bmatrix}[r] 0 & 1 \ -2 & 5 \end{bmatrix} \text{ and C } = \begin{bmatrix}[r] -2 & 0 \ -1 & 1 \end{bmatrix}, find A(4B - 3C).

Matrices

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Answer

We need to find the value of A(4B - 3C).

A(4B - 3C)=[12−34](4[01−25]−3[−20−11])=[12−34]([04−820]−[−60−33])=[12−34][0−(−6)4−0−8−(−3)20−3]=[12−34][64−517]=[1×6+2×(−5)1×4+2×17−3×6+4×(−5)−3×4+4×17]=[6−104+34−18−20−12+68]=[−438−3856].\text{A(4B - 3C)} = \begin{bmatrix}[r] 1 & 2 \ -3 & 4 \end{bmatrix} \Big(4\begin{bmatrix}[r] 0 & 1 \ -2 & 5 \end{bmatrix} - 3\begin{bmatrix}[r] -2 & 0 \ -1 & 1 \end{bmatrix}\Big) \\[0.5em] = \begin{bmatrix}[r] 1 & 2 \ -3 & 4 \end{bmatrix} \Big(\begin{bmatrix}[r] 0 & 4 \ -8 & 20 \end{bmatrix} - \begin{bmatrix}[r] -6 & 0 \ -3 & 3 \end{bmatrix}\Big) \\[0.5em] = \begin{bmatrix}[r] 1 & 2 \ -3 & 4 \end{bmatrix} \begin{bmatrix}[r] 0 - (-6) & 4 - 0 \ -8 - (-3) & 20 - 3 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 1 & 2 \ -3 & 4 \end{bmatrix} \begin{bmatrix}[r] 6 & 4 \ -5 & 17 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 1 \times 6 + 2 \times (-5) & 1 \times 4 + 2 \times 17 \ -3 \times 6 + 4 \times (-5) & -3 \times 4 + 4 \times 17 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 6 - 10 & 4 + 34 \ -18 - 20 & -12 + 68 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] -4 & 38 \ -38 & 56 \end{bmatrix}.

Hence, the value of A(4B - 3C) = [−438−3856].\begin{bmatrix}[r] -4 & 38 \ -38 & 56 \end{bmatrix}.

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