If A = [0−12103−2−30]\begin{bmatrix} 0 & -1 & 2 \ 1 & 0 & 3 \ -2 & -3 & 0 \end{bmatrix}01−2−10−3230, then (A + A′) is equal to :
0
2A
–A′
A′
3 Likes
Given,
A = [0−12103−2−30]\begin{bmatrix} 0 & -1 & 2 \ 1 & 0 & 3 \ -2 & -3 & 0 \end{bmatrix}01−2−10−3230
A′ = [01−2−10−3230]\begin{bmatrix} 0 & 1 & -2 \ -1 & 0 & -3 \ 2 & 3 & 0 \end{bmatrix}0−12103−2−30
(A + A′)
⇒[0−12103−2−30]+[01−2−10−3230]⇒[0+0−1+12+(−2)1+(−1)0+03+(−3)−2+2−3+30+0]⇒[000000000]\Rightarrow \begin{bmatrix} 0 & -1 & 2 \ 1 & 0 & 3 \ -2 & -3 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 1 & -2 \ -1 & 0 & -3 \ 2 & 3 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 + 0 & -1 + 1 & 2 + (-2) \ 1 + (-1) & 0 + 0 & 3 + (-3) \ -2 + 2 & -3 + 3 & 0 + 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}⇒01−2−10−3230+0−12103−2−30⇒0+01+(−1)−2+2−1+10+0−3+32+(−2)3+(−3)0+0⇒000000000
Hence, option 1 is the correct option.
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