Question

Solve the system of linear equations using the matrix method.

x − y + 2z = 7

3x + 4y − 5z = −5

2x − y + 3z = 12

Answer 1

`[(1,-1,2),(3,4,-5),(2,-1,3)] [(x),(y),(z)] = [(7),(-5),(12)]` AX = B

A = `[(1,-1,2),(3,4,-5),(2,-1,3)]` X = `[(x),(y),(z)]` or B = `[(7),(-5),(12)]`

Now, |A| = `|(1,-1,2),(3,4,-5),(2,-1,3)|`

= 1(12 − 5) + 1(9 + 10) + 2(−3 − 8)

= 7 + 19 − 22

= 4 ≠ 0

⇒ A⁻¹ exists and hence the given equation has a unique solution.

A₁₁ = `(-1)^(1 + 1) |(4,-5),(-1,3)|`

= 12 − 5

= 7

A₁₂ = `(-1)^(1 + 2) |(3,-5),(2,3)|`

= −(9 + 10)

= −19

A₁₃ = `(-1)^(1 + 3) |(3,4),(2,-1)|`

= −3 − 8

= −11

A₂₁ = `(-1)^(2 + 1) |(-1,2),(-1,3)|`

= −(−3 + 2)

= 1

A₂₂ = `(-1)^(2 + 2) |(1,2),(1,3)|`

= 3 − 4

= −1

A₂₃ = `(-1)^(2 + 3) |(1,-1),(2,-1)|`

= −1(−1 + 2)

= −1

A₃₁ = `(-1)^ (3 + 1) |(-1,2),(4,-5)|`

= 5 − 8

= −3

A₃₂ = `(-1)^(3 + 2) |(1,2),(3,-5)|`

= −(−5 − 6)

= 11

A₃₃ = `(-1)^(3 + 3) |(1,-1),(3, 4)|`

= 4 + 3

= 7

∴ A⁻¹ = `1/|A|` (Adj A)

= `1/4 [(7,-19,-11),(1,1,-1),(-3,11,7)]`

= `1/4 [(7,1,-3),(-19,-1,11),(-11,-1,7)]`

X = A⁻¹B

⇒ `[(x),(y),(z)] = 1/4 [(7,1,-3),(-19,-1,11),(-11,-1,7)] [(7),(-5),(12)]`

= `1/4 [(49 - 5 - 36),(-133 + 5 + 132),(-77 + 5 + 84)]`

= `1/4 [(8),(4),(12)]`

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

So, x = 2, y = 1 and z = 3