Question

Solve the following system of linear equations using matrices:

x – 2y = 10, 2x – у – z = 8, –2y + z = 7

Answer 1

Given equations are,

x – 2y = 10

2x – у – z = 8

And –2y + z = 7

The given equations can be arranged in matrix form as,

`[(1, -2, 0),(2, -1, -1),(0, -2, 1)] [(x),(y),(z)] = [(10),(8),(7)]`

Let A = `[(1, -2, 0),(2, -1, -1),(0, -2, 1)]`, X = `[(x),(y),(z)]` and B = `[(10),(8),(7)]`

∴ AX = B   ...(i)

Now, |A| = `|(1, -2, 0),(2, -1, -1),(0, -2, 1)|`

= 1(–1 – 2) + 2(2 + 0) + 0(– 4 + 0)

= –3 + 4

= 1 ≠ 0

So, A is invertible.

Now, the minors of the elements of |A| are

M₁₁ = – 3, M₁₂ = 2, M₁₃ = – 4

M₂₁ = – 2, M₂₂ = 1, M₂₃ = – 2

M₃₁ = 2, M₃₂ = –1, M₃₃ = 3

Co-factors of the elements of |A| are

A₁₁ = – 3, A₁₂ = – 2, A₁₃ = – 4

A₂₁ = 2, A₂₂ = 1, A₂₃ = 2

A₃₁ = 2, A₃₂ = 1, A₃₃ = 3

We know, `adj.A = [(A_11, A_12, A_13),(A_21, A_22, A_23),(A_31, A_32, A_33)]^T`

∴ `adj.A = [(-3, -2, -4),(2, 1, 2),(2, 1, 3)]^T`

= `[(-3, 2, 2),(-2, 1, 1),(-4, 2, 3)]`

∴ `A^-1 = 1/|A| (adj.A)`

= `1. [(-3, 2, 2),(-2, 1, 1),(-4, 2, 3)]`

From (i), we have

AX = B

⇒ A^–1 AX = A^–1 B   ...[Multiplying A^–1 on both sides]

⇒ IX = A^–1 B   ...[AA^–1 = I]

⇒ X = A^–1 B

⇒ `[(x),(y),(z)] = [(-3, 2, 2),(-2, 1, 1),(-4, 2, 3)] [(10),(8),(7)]`

⇒ `[(x),(y),(x)] = [(-30 + 16 + 14),(-20 + 8 + 7),(-40 + 16 + 21)]`

⇒ `[(x),(y),(z)] = [(0),(-5),(-3)]`

∴ x = 0, y = –5 and z = –3.