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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)]`
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 `underline([(x^(-1),0,0),(0, y^(-1),0),(0,0,z^(-1))]).`
Explnation:
Let, A = `[(x,0,0),(0,y,0),(0,0,z)]`
∴ |A| = x[yz - 0] = xyz
∴ `A_11 = (-1)^{1 + 1}[(y,0),(0,z)] = (-1)^2[yz - 0]`
= 1 × yz = yz
`A_12 = (-1)^{1 + 2}[(0,0),(0,z)] = (-1)^3[0 - 0] = 0`
`A_13 = (-1)^{1 + 3}[(0,y),(0,0)] = (-1)^4[0 - 0] = 0`
`A_21 = (-1)^{2 + 1}[(0,0),(0,z)] = (-1)^3[0 - 0] = 0`
`A_22 = (-1)^{2 + 2}[(x,0),(0,z)] = (-1)^4[xz - 0] = 0`
= 1 × zx = zx
`A_23 = (-1)^{2 + 3}[(x,0),(0,0)] = (-1)^5[0 - 0] = 0`
`A_31 = (-1)^{3 + 1}[(0,0),(0,z)] = (-1)^4[0 - 0] = 0`
`A_32 = (-1)^{3 + 2}[(x,0),(0,0)] = (-1)^5[0 - 0] = 0`
`A_33 = (-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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