If [4−240]+3A=[−2−21−3]\begin{bmatrix}[r] 4 & -2 \ 4 & 0 \end{bmatrix} + 3A = \begin{bmatrix}[r] -2 & -2 \ 1 & -3 \end{bmatrix}[44−20]+3A=[−21−2−3]; find A.
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Given,
⇒[4−240]+3A=[−2−21−3]⇒3A=[−2−21−3]−[4−240]⇒3A=[−2−4−2−(−2)1−4−3−0]⇒3A=[−60−3−3]⇒A=13[−60−3−3]⇒A=[−20−1−1].\Rightarrow \begin{bmatrix}[r] 4 & -2 \ 4 & 0 \end{bmatrix} + 3A = \begin{bmatrix}[r] -2 & -2 \ 1 & -3 \end{bmatrix} \\[1em] \Rightarrow 3A = \begin{bmatrix}[r] -2 & -2 \ 1 & -3 \end{bmatrix} - \begin{bmatrix}[r] 4 & -2 \ 4 & 0 \end{bmatrix} \\[1em] \Rightarrow 3A = \begin{bmatrix}[r] -2 - 4 & -2 - (-2) \ 1 - 4 & -3 - 0 \end{bmatrix} \\[1em] \Rightarrow 3A = \begin{bmatrix}[r] -6 & 0 \ -3 & -3 \end{bmatrix} \\[1em] \Rightarrow A = \dfrac{1}{3}\begin{bmatrix}[r] -6 & 0 \ -3 & -3 \end{bmatrix} \\[1em] \Rightarrow A = \begin{bmatrix}[r] -2 & 0 \ -1 & -1 \end{bmatrix}.⇒[44−20]+3A=[−21−2−3]⇒3A=[−21−2−3]−[44−20]⇒3A=[−2−41−4−2−(−2)−3−0]⇒3A=[−6−30−3]⇒A=31[−6−30−3]⇒A=[−2−10−1].
Hence, A = [−20−1−1]\begin{bmatrix}[r] -2 & 0 \ -1 & -1 \end{bmatrix}[−2−10−1].
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Find x and y if :
(i) 3[4x]+2[y−3]=[100]3\begin{bmatrix}[r] 4 & x \end{bmatrix} + 2\begin{bmatrix}[r] y & -3 \end{bmatrix} = \begin{bmatrix}[r] 10 & 0 \end{bmatrix}3[4x]+2[y−3]=[100]
(ii) x[−12]−4[−2y]=[7−8]x\begin{bmatrix}[r] -1 \ 2 \end{bmatrix} - 4\begin{bmatrix}[r] -2 \ y \end{bmatrix} = \begin{bmatrix}[r] 7 \ -8 \end{bmatrix}x[−12]−4[−2y]=[7−8]
Given A = [2130],B=[1152] and C=[−3−100]\begin{bmatrix}[r] 2 & 1 \ 3 & 0 \end{bmatrix}, B = \begin{bmatrix}[r] 1 & 1 \ 5 & 2 \end{bmatrix} \text{ and } C = \begin{bmatrix}[r] -3 & -1 \ 0 & 0 \end{bmatrix}[2310],B=[1512] and C=[−30−10]; find :
(i) 2A - 3B + C
(ii) A + 2C - B
Given A = [1423] and B=[−4−1−3−2]\begin{bmatrix}[r] 1 & 4 \ 2 & 3 \end{bmatrix} \text{ and } B = \begin{bmatrix}[r] -4 & -1 \ -3 & -2 \end{bmatrix}[1243] and B=[−4−3−1−2]
(i) find the matrix 2A + B
(ii) find a matrix C such that :
C + B = [0000]\begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix}[0000]
If 2[3x01]+3[13y2]=[z−7158]2\begin{bmatrix}[r] 3 & x \ 0 & 1 \end{bmatrix} + 3\begin{bmatrix}[r] 1 & 3 \ y & 2 \end{bmatrix} = \begin{bmatrix}[r] z & -7 \ 15 & 8 \end{bmatrix}2[30x1]+3[1y32]=[z15−78]; find the values of x, y and z.
