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′
1 Like
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.
Answered By
If 2[130x]+[y012]=[5618]2\begin{bmatrix} 1 & 3 \ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \ 1 & 8 \end{bmatrix}2[103x]+[y102]=[5168],then the value of (x + y) is:
4
-4
6
8
If [a+b25ab]=[6258]\begin{bmatrix} a + b & 2 \ 5 & ab \end{bmatrix} = \begin{bmatrix} 6 & 2 \ 5 & 8 \end{bmatrix}[a+b52ab]=[6528], then the values of a and b are:
a = 4, b = 2
a = 2, b = 4
both (a) and (b)
none of these
If [xy+2z−3]+[y45]=[4912]\begin{bmatrix} x & y + 2 & z - 3 \end{bmatrix} + \begin{bmatrix} y & 4 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix}[xy+2z−3]+[y45]=[4912], then the value of yz is:
24
20
36
30
If x[21]+y[35]+[−8−11]x\begin{bmatrix} 2 \ 1 \end{bmatrix} + y\begin{bmatrix} 3 \ 5 \end{bmatrix} + \begin{bmatrix} -8 \ -11 \end{bmatrix}x[21]+y[35]+[−8−11] = 0, then the value of (2x – 3y) is:
1
3
-3
