Solve the following system of the linear equations by Cramer’s rule:

3x + y + z = 2

2x – 4y + 3z = – 1

4x + y – 3z = – 11

Given: - Equations are: –

3x + y + z = 2

2x – 4y + 3z = – 1

4x + y – 3z = – 11

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

3x + y + z = 2

2x – 4y + 3z = – 1

4x + y – 3z = – 11

So by comparing with the theorem, let's find D, D₁, D₂ and D₃

Solving determinant, expanding along 1^st row

⇒ D = 3[( – 4)( – 3) – (3)(1)] – 1[(2)( – 3) – 12] + 1[2 – 4( – 4)]

⇒ D = 3[12 – 3] – [ – 6 – 12] + [2 + 16]

⇒ D = 27 + 18 + 18

⇒ D = 63

Again,

Solving determinant, expanding along 1^st row

⇒ D₁ = 2[( – 4)( – 3) – (3)(1)] – 1[( – 1)( – 3) – ( – 11)(3)] + 1[( – 1) – ( – 4)( – 11)]

⇒ D₁ = 2[12 – 3] – 1[3 + 33] + 1[ – 1 – 44]

⇒ D₁ = 2[9] – 36 – 45

⇒ D₁ = 18 – 36 – 45

⇒ D₁ = – 63

Again

Solving determinant, expanding along 1^st row

⇒ D₂ = 3[3 + 33] – 2[ – 6 – 12] + 1[ – 22 + 4]

⇒ D₂ = 3[36] – 2( – 18) – 18

⇒ D₂ = 126

And,

⇒

Solving determinant, expanding along 1^st row

⇒ D₃ = 3[44 + 1] – 1[ – 22 + 4] + 2[2 + 16]

⇒ D₃ = 3[45] – 1( – 18) + 2(18)

⇒ D₃ = 135 + 18 + 36

⇒ D₃ = 189

Thus by Cramer’s Rule, we have

⇒

⇒

⇒ x = – 1

again,

⇒

⇒

⇒ y = 2

and,

⇒

⇒

⇒ z = 3