Question

Choose the correct option:

If X, Y, Z are non zero real numbers, then the inverse of matrix A = `[(x, 0, 0),(0, y, 0),(0, 0, z)]`

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)]`

Answer 1

`[(x^-1, 0, 0),(0, y^-1, 0),(0, 0, z^-1)]`

Explanation:

Given A = `[(x, 0, 0),(0, y, 0),(0, 0, z)]`

∴ |A| = `x(yz - 0) = xyz ≠ 0`

Now, A₁₁ = yz, A₁₂ = 0, A₁₃ = 0

A₂₁ = 0, A₂₂ = xz, A₂₃ = 0

A₃₁ = 0, A₃₂ = 0, A₃₃ = xy

∴ adj A = `[(yz, 0, 0),(0, xz, 0),(0, 0, xy)]` 

A^–1 = `1/|A| (adj A) = 1/(xyz) [(yz, 0, 0),(0, xz, 0),(0, 0, xy)]` 

= `[((yz)/(xyz), 0, 0),(0, (xz)/(xyz), 0),(0, 0, (xy)/(xyz))] = [(1/x, 0, 0),(0, 1/y, 0),(0, 0, 1/z)]`

= `[(x^-1, 0, 0),(0, y^-1, 0),(0, 0, z^-1)]`