Let `x=[x_1x_2x_3],A=[1-1 2 2 0 1 3 2 1]a n dB=[3 2 1]dotIfA X=B ,` Then `X` is equal to `[1 2 3]` (b) `[-1-2-3]` (c) `[-1-2-3]` (d) `[-1 2 3]` (e) `[0 2 1]`

Explanation

we know that `AX=B`
`X=A^-1B`
`A^-1=1/abs(A)*adj(A)`
Given,`X=[(x_1),(x_2),(x_3)]`
`A=[(1,-1,2),(2,0,1),(3,2,1)]`
`B=[(3),(1),(4)]`
`abs(A)=|(1,-1,2),(2,0,1),(3,2,1)|`
`=1(0−2)+1(2−3)+2(4−0)=−2−1+8=5`
`adj(A)=[(-2,5,-1),(1,-5,3),(4,-5,2)]`
`A^-1=1/5[(-2,5,-1),(1,-5,3),(4,-5,2)]`
`X=A^-1B`
`=1/5[(-2,5,-1),(1,-5,3),(4,-5,2)][(3),(1),(4)]`
`=1/5[(-5),(10),(15)]`
`=[(-1),(2),(3)]`