A 2 × 2 matrix whose elements are given by aij = (i+2j)22\dfrac{(i + 2j)^2}{2}2(i+2j)2 is:
[3252 818]\begin{bmatrix} \dfrac{3}{2} & \dfrac{5}{2} \ \space & \space \ 8 & 18 \end{bmatrix}23 825 18
[5272 818]\begin{bmatrix} \dfrac{5}{2} & \dfrac{7}{2} \ \space & \space \ 8 & 18 \end{bmatrix}25 827 18
[92152 818]\begin{bmatrix} \dfrac{9}{2} & \dfrac{15}{2} \ \space & \space \ 8 & 18 \end{bmatrix}29 8215 18
[92252 818]\begin{bmatrix} \dfrac{9}{2} & \dfrac{25}{2} \ \space & \space \ 8 & 18 \end{bmatrix}29 8225 18
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Given,
aij = (i+2j)22\dfrac{(i + 2j)^2}{2}2(i+2j)2
a11=[1+2(1)]22=322=92,a12=[1+2(2)]22=522=252a21=[2+2]22=422=162=8,a22=[2+2(2)]22=622=362=18.a{11} = \dfrac{[1 + 2(1)]^2}{2} = \dfrac{3^2}{2} = \dfrac{9}{2}, a{12} = \dfrac{[1 + 2(2)]^2}{2} = \dfrac{5^2}{2} = \dfrac{25}{2} \\[1em] a{21} = \dfrac{[2 + 2]^2}{2} = \dfrac{4^2}{2} = \dfrac{16}{2} = 8, a{22} = \dfrac{[2 + 2(2)]^2}{2} = \dfrac{6^2}{2} = \dfrac{36}{2} = 18.a11=2[1+2(1)]2=232=29,a12=2[1+2(2)]2=252=225a21=2[2+2]2=242=216=8,a22=2[2+2(2)]2=262=236=18.
Matrix A = [92252818]\begin{bmatrix} \dfrac{9}{2} & \dfrac{25}{2} \\ 8 & 18 \end{bmatrix}29822518
Hence, option 4 is the correct option.
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A matrix of order 3 × 2 whose elements aij are given by aij = (i + j), is:
[234345]\begin{bmatrix} 2 & 3 & 4 \ 3 & 4 & 5 \end{bmatrix}[233445]
[234456]\begin{bmatrix} 2 & 3 & 4 \ 4 & 5 & 6 \end{bmatrix}[243546]
[233445]\begin{bmatrix} 2 & 3 \ 3 & 4 \ 4 & 5 \end{bmatrix}234345
[233456]\begin{bmatrix} 2 & 3 \ 3 & 4 \ 5 & 6 \end{bmatrix}235346
If a matrix has 12 elements, then the total number of possible orders it can have is:
4
6
8
cannot be determined
If [53x7]=[yz17]\begin{bmatrix} 5 & 3 \ x & 7 \end{bmatrix} = \begin{bmatrix} y & z \ 1 & 7 \end{bmatrix}[5x37]=[y1z7], then the value of (x + y + z) is:
9
11
10
If [x−y2x+z2x−y3z+w]=[−15013]\begin{bmatrix} x - y & 2x + z \ 2x - y & 3z + w \end{bmatrix} = \begin{bmatrix} -1 & 5 \ 0 & 13 \end{bmatrix}[x−y2x−y2x+z3z+w]=[−10513], then the value of (x + y + z + w) is:
12
