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

x + y = 1

x + z = – 6

x – y – 2z = 3

Given: - Equations are: –

x + y = 1

x + z = – 6

x – y – 2z = 3

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

x + y = 1

x + z = – 6

x – y – 2z = 3

So by comparing with theorem, lets find D , D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 1[(0)( – 2) – (1)( – 1)] – 1[( – 2)(1) – 1] + 0[ – 1 – 0]

⇒ D = 1[0 + 1] – 1[ – 3] – 0[ – 2]

⇒ D = 1 + 3 + 0

⇒ D = 4

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 1[(0)( – 2) – (1)( – 1)] – 1[( – 2)( – 6) – 3] + 0[6 – 0]

⇒ D₁ = 1[0 + 1] – 1[12 – 3] + 0[6]

⇒ D₁ = 1[1] – 9 + 0

⇒ D₁ = 1 – 9 + 0

⇒ D₁ = – 8

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₂ = 1[( – 6)( – 2) – (1)(3)] – 1[( – 2)(1) – 1] + 0[3 + 6]

⇒ D₂ = 1[12 – 3] – 1( – 2 – 1) + 0(9)

⇒ D₂ = 9 + 3

⇒ D₂ = 12

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 1[(0)(3) – ( – 1)( – 6)] – 1[(3)(1) – 1( – 6)] + 1[ – 1 + 0]

⇒ D₃ = 1[0 – 6] – 1(3 + 6) + 1( – 1)

⇒ D₃ = – 6 – 9 – 1

⇒ D₃ = – 16

Thus by Cramer’s Rule, we have

⇒

⇒

⇒ x = – 2

again,

⇒

⇒

⇒ y = 3

and,

⇒

⇒

⇒ z = – 4