Question

If A = `[(1, 2, -3),(2, 0, -3),(1, 2, 0)]`, then find A⁻¹ and hence solve the following system of equations: 

x + 2y − 3z = 1

2x − 3z = 2

x + 2y = 3

Answer 1

Here, ` |A| = [(1, 2, -3),(2, 0, -3),(1, 2, 0)]`

= = 1(0 + 6) - 2(0 + 3) − 3(4)

= 6 − 6 − 12 = −12 ≠ 0

Thus, A is invertible.

Now, the cofactor of `|A|` are

C₁₁ = 6, C₁₂ = −3, C₁₃ = 4

C₂₁ = −6, C₂₂ = 3, C₂₃ = 0

C₃₁ = −6, C₃₂ = −3, C₃₃ = −4

∴ adj A = `[(6, -3, 4),(-6, 3, 0),(-6, -3, -4)]^T = [(6, -6, -6),(-3, 3, -3),(4, 0, -4)]`

So, A⁻¹ = `1/|A|  "adj"  A = 1/-12[(6, -6, -6),(-3, 3, -3),(4, 0, -4)]`

The given equations are 

x + 2y − 3z = 1, 2x − 3z = 2 and x + 2y = 3

which can be written in matrix form as AX = B, where

`A = [(1, 2, -3),(2, 0, -3),(1, 2, 0)], X =[(x),(y),(z)] and B = [(1),(2),(3)]`

=> `[(x),(y),(z)] = [(1, 2, -3),(2, 0, -3),(1, 2, 0)]^-1[(1),(2),(3)]`

= `-1/12 [(6, -6, -6),(-3, 3, -3),(4, 0, -4)][(1),(2),(3)]`

= `(-1)/12 [(6, -12, -18),(-3, +6, -9),(4, +0, -12)]`

= `(-1)/12[(-24),(-6),(-8)]`

∴ `[(x),(y),(z)] = [(2),(1/2),(2/3)]`

∴ x = 2, y = `1/2` and z = `2/3`