Question

If A = `[(-1, 0, 0),(0, 1, 0),(0, 0, 1)]`, then A⁻¹ is ______.

Options

• `[(-1, 0, 0),(0, -1, 0),(0, 0, -1)]`

• `[(1, 0, 0),(0, -1, 0),(0, 0, -1)]`

• `[(-1, 0, 0),(0, -1, 0),(0, 0, 1)]`

• `[(-1, 0, 0),(0, 1, 0),(0, 0, 1)]`

Answer 1

If A = `[(-1, 0, 0),(0, 1, 0),(0, 0,1)]`, then A⁻¹ is `bbunderline([(-1, 0, 0),(0, 1, 0),(0, 0, 1)])`.

Explanation:

A = `[(-1, 0, 0),(0, 1, 0),(0, 0,1)]`

|A| = −1(1 − 0)

= −1

a₁₁ = 1, a₁₂ = 0, a₁₃ = 0, a₂₁ = 0, a₂₂ = −1, a₂₃ = 0, a₃₁ = 0, a₃₂ = 0, a₃₃ = −1

adj A = `[(1, 0, 0),(0, -1, 0),(0, 0,-1)]`

`A^-1 = 1/|A|` (adj A)

= `-1[(1, 0, 0),(0, -1, 0),(0, 0,-1)]`

= `[(-1, 0, 0),(0, 1, 0),(0, 0,1)]`