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Question
If x, y, z are nonzero real numbers, then the inverse of matrix A = `[(x,0,0),(0,y,0),(0,0,z)]` is ______.
Options
`[(x^(-1),0,0),(0, y^(-1),0),(0,0,z^(-1))]`
`xyz[(x^(-1),0,0),(0,y^(-1),0),(0,0,z^(-1))]`
`1/(xyz)[(x,0,0),(0,y,0),(0,0,z)]`
`1/(xyz)[(1,0,0),(0,1,0),(0,0,1)]`
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Solution
If x, y, z are nonzero real numbers, then the inverse of matrix A = `[(x,0,0),(0,y,0),(0,0,z)]` is `underlinebb([(x^(-1),0,0),(0, y^(-1),0),(0,0,z^(-1))])`.
Explanation:
Let, A = `[(x,0,0),(0,y,0),(0,0,z)]`
∴ |A| = x[yz − 0] = xyz
∴ A11 = `(-1)^(1 + 1)[(y,0),(0,z)]`
= (−1)2 [yz − 0]
= 1 × yz
= yz
A12 = `(-1)^(1 + 2)[(0,0),(0,z)]`
= (−1)3 [0 − 0]
= 0
A13 = `(-1)^(1 + 3)[(0,y),(0,0)]`
= (−1)4 [0 − 0]
= 0
A21 = `(-1)^(2 + 1)[(0,0),(0,z)]`
= (−1)3 [0 − 0]
= 0
A22 = `(-1)^(2 + 2)[(x,0),(0,z)]`
= (−1)4 [xz − 0]
= 1 × zx
= zx
A23 = `(-1)^(2 + 3)[(x,0),(0,0)]`
= (−1)5 [0 − 0]
= 0
A31 = `(-1)^(3 + 1)[(0,0),(0,z)]`
= (−1)4 [0 − 0]
= 0
A32 = `(-1)^(3 + 2)[(x,0),(0,0)]`
= (−1)5 [0 − 0]
= 0
A33 = `(-1)^(3 + 3)[(x,0),(0,y)]`
= (−1)6 [xy − 0]
= xy
∴ adj A = `[(yz,0,0),(0,zx,0),(0,0,xy)] = [(yz,0,0),(0,zx,0),(0,0,xy)]`
A−1 = `1/|A|` (adj A)
= `1/(xyz)[(yz,0,0),(0,zx,0),(0,0,xy)]`
= `[(1/x,0,0),(0,1/y,0),(0,0,1/z)]`
= `[(x^-1,0,0),(0,y^-1,0),(0,0,z^-1)]`
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