Question

Solve the system of linear equations using the matrix method.

2x + 3y + 3z = 5

x − 2y + z = −4

3x − y − 2z = 3

Answer 1

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

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

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

= 2(4 + 1) − 3(−2 − 3) + 3(−1 + 6)

= 10 + 15 + 15

= 40 ≠ 0

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

A₁₁ = `(-1)^(1 + 1) |(-2,1),(-1,-2)|`

= 4 + 1

= 5

A₁₂ = `(-1)^(1 + 2) |(1,1),(3,-2)|`

= (−2 − 3)

= 5

A₁₃ = `(-1)^(1 + 3) |(1,-2),(3,-1)|`

= −1 + 6

= 5

A₂₁ = `(-1)^(2 + 1) |(3,3),(-1,-2)|`

= −(−6 + 3)

= 3

A₂₂ = `(-1)^(2 + 2) |(2,3),(3,-2)|`

= −4 − 9

= −13

A₂₃ = `(-1)^(2 + 3) |(2,3),(3,-1)|`

= −(−2 − 9)

= 11

A₃₁ = `(-1)^ (3 + 1) |(3,3),(-2,1)|`

= 3 + 6

= 9

A₃₂ = `(-1)^(3 + 2) |(2,3),(1,1)|`

= −(2 − 3)

= 1

A₃₃ = `(-1)^(3 + 3) |(2,3),(1, -2)|`

= −4 − 3

= −7

∴ A⁻¹ = `1/|A|` (Adj A)

= `1/40 [(5,5,5),(3,-13,11),(9,1,-7)]`

= `1/40 [(5,3,9),(5,-13,1),(5,11,-7)]`

X = A⁻¹B

⇒ `[(x),(y),(z)] = 1/40 [(5,3,9),(5,-13,1),(5,11,-7)] [(5),(-4),(3)]`

= `1/40 [(25 - 12 + 27),(25 + 52 + 3),(25 - 44 - 21)]`

= `1/40 [(40),(80),(-40)]`

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

∴ x = 1, y = 2 and z = −1