Given A = [473−2] and B=[12−14]\begin{bmatrix}[r] 4 & 7 \ 3 & -2 \end{bmatrix} \text{ and } B = \begin{bmatrix}[r] 1 & 2 \ -1 & 4 \end{bmatrix}[437−2] and B=[1−124], then A - 2B is :
[−235−10]\begin{bmatrix}[r] -2 & 3 \ 5 & -10 \end{bmatrix}[−253−10]
[−2−3−510]\begin{bmatrix}[r] -2 & -3 \ -5 & 10 \end{bmatrix}[−2−5−310]
[235−10]\begin{bmatrix}[r] 2 & 3 \ 5 & -10 \end{bmatrix}[253−10]
[23510]\begin{bmatrix}[r] 2 & 3 \ 5 & 10 \end{bmatrix}[25310]
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Substituting values of A and B in A - 2B, we get :
⇒A−2B=[473−2]−2[12−14]=[473−2]−[24−28]=[4−27−43−(−2)−2−8]=[233+2−10]=[235−10].\Rightarrow A - 2B = \begin{bmatrix}[r] 4 & 7 \ 3 & -2 \end{bmatrix} - 2\begin{bmatrix}[r] 1 & 2 \ -1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 & 7 \ 3 & -2 \end{bmatrix} - \begin{bmatrix}[r] 2 & 4 \ -2 & 8 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 - 2 & 7 - 4 \ 3 - (-2) & -2 - 8 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 & 3 \ 3 + 2 & -10 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 & 3 \ 5 & -10 \end{bmatrix}.⇒A−2B=[437−2]−2[1−124]=[437−2]−[2−248]=[4−23−(−2)7−4−2−8]=[23+23−10]=[253−10].
Hence, Option 3 is the correct option.
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If I is a unit matrix of order 2 and M + 4I = [8−342]\begin{bmatrix}[r] 8 & -3 \ 4 & 2 \end{bmatrix}[84−32], then matrix M is :
[434−2]\begin{bmatrix}[r] 4 & 3 \ 4 & -2 \end{bmatrix}[443−2]
[4342]\begin{bmatrix}[r] 4 & 3 \ 4 & 2 \end{bmatrix}[4432]
[4−3−42]\begin{bmatrix}[r] 4 & -3 \ -4 & 2 \end{bmatrix}[4−4−32]
[4−34−2]\begin{bmatrix}[r] 4 & -3 \ 4 & -2 \end{bmatrix}[44−3−2]
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], the values of x, y and z are :
x = 8, y = -5 and z = 9
x = -8, y = 5 and z = 9
x = -8, y = -5 and z = -9
x = -8, y = 5 and z = -9
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
