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]
22 Likes
As, I is a unit matrix of order 2.
∴ I = [1001]\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix}[1001]
Given,
⇒M+4I=[8−342]⇒M+4[1001]=[8−342]⇒M+[4004]=[8−342]⇒M=[8−342]−[4004]⇒M=[8−4−3−04−02−4]⇒M=[4−34−2].\Rightarrow M + 4I = \begin{bmatrix}[r] 8 & -3 \ 4 & 2 \end{bmatrix} \\[1em] \Rightarrow M + 4\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix}[r] 8 & -3 \ 4 & 2 \end{bmatrix} \\[1em] \Rightarrow M + \begin{bmatrix}[r] 4 & 0 \ 0 & 4 \end{bmatrix} = \begin{bmatrix}[r] 8 & -3 \ 4 & 2 \end{bmatrix} \\[1em] \Rightarrow M = \begin{bmatrix}[r] 8 & -3 \ 4 & 2 \end{bmatrix} - \begin{bmatrix}[r] 4 & 0 \ 0 & 4 \end{bmatrix} \\[1em] \Rightarrow M = \begin{bmatrix}[r] 8 - 4 & -3 - 0 \ 4 - 0 & 2 - 4 \end{bmatrix} \\[1em] \Rightarrow M = \begin{bmatrix}[r] 4 & -3 \ 4 & -2 \end{bmatrix}.⇒M+4I=[84−32]⇒M+4[1001]=[84−32]⇒M+[4004]=[84−32]⇒M=[84−32]−[4004]⇒M=[8−44−0−3−02−4]⇒M=[44−3−2].
Hence, Option 4 is the correct option.
Answered By
8 Likes
If 4[5x]−5[y−2]=[1022]4\begin{bmatrix}[r] 5 & x \end{bmatrix} - 5\begin{bmatrix}[r] y & -2 \end{bmatrix} = \begin{bmatrix}[r] 10 & 22 \end{bmatrix}4[5x]−5[y−2]=[1022], the values of x and y are :
x = 2 and y = 3
x = 3 and y = 2
x = -3 and y = 2
x = 3 and y = -2
If A = [−3−70−8] and A−B=[64−30]\begin{bmatrix}[r] -3 & -7 \ 0 & -8 \end{bmatrix}\text{ and } A - B = \begin{bmatrix}[r] 6 & 4 \ -3 & 0 \end{bmatrix}[−30−7−8] and A−B=[6−340], then matrix B is :
[911−318]\begin{bmatrix}[r] 9 & 11 \ -3 & 18 \end{bmatrix}[9−31118]
[−9−1138]\begin{bmatrix}[r] -9 & -11 \ 3 & 8 \end{bmatrix}[−93−118]
[9−11−38]\begin{bmatrix}[r] 9 & -11 \ -3 & 8 \end{bmatrix}[9−3−118]
[−9−11−3−8]\begin{bmatrix}[r] -9 & -11 \ -3 & -8 \end{bmatrix}[−9−3−11−8]
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
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]
