Question

Solve the following system of equations by matrix method:

3x + 4y + 7z = 14

2x − y + 3z = 4

x + 2y − 3z = 0

Answer 1

The given system can be written in matrix form as:

`[(3, 4, 7),(2, -1, 3),(1, 2, -3)][(x),(y),(z)] = [(14),(4),(0)]` or A X = B

A = `[(3, 4, 7),(2, -1, 3),(1, 2, -3)], X = [(x),(y),(z)] and B = [(14),(4),(0)]`

Now,

|A| = `3|(-1, 3),(2, -3)| -4|(2, 3),(1, -3)| + 7|(2, 3),(2, -3)|`

= 3(3 − 6) − 4(−6 − 3) + 7(4 + 1)

= −9 + 36 + 35

= 62

So, the above system has a unique solution, given by

X = A⁻¹ B

Cofactors of A are:

C₁₁ = (−1)^{1 + 1} 3 − 6 = −3

C₂₁ = (−1)^{2 + 1} − 12 − 14 = 26

C₃₁ = (−1)^{3 + 1} 12 + 7 = 19

C₁₂ = (−1)^{1 + 2} − 6 − 3 = 9

C₂₂ = (−1)^{2 + 1} − 3 − 7 = −10

C₃₂ = (−1)^{3 + 1} 9 − 14 = 5

C₁₃ = (−1)^{1 + 2} 4 + 1 = 5

C₂₃ = (−1)^{2 + 1} 6 − 4 = −2

C₃₃ = (−1)^{3 + 1} − 3 − 8 = −11

adj A = `[(-3, 9, 5),(26, -5, -2),(19, 5, -11)]^T`

= `[(-3, 26, 19),(9, -16, 5),(5, -2, -11)]`

A⁻¹ = `1/(|A|)`adj A

Now, X = A⁻¹ B = `1/62[(-3, 26, 19),(9, -16, 5),(5, -2, -11)][(14),(4),(0)]`

X = `1/62[(-42 + 104 + 0),(126 - 64 + 0),(70 - 8 + 0)]`

X = `1/62[(-42 + 104 + 0),(126 - 64 + 0),(70 - 8 + 0)]`

 X = `1/62 [(62),(62),(62)]`

C = `[(1),(1),(1)]`

Hence, X = 1, Y = 1 and Z = 1