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

2x – 3y – 4z = 29

– 2x + 5y – z = – 15

3x – y + 5z = – 11

Given: - Equations are: –

2x – 3y – 4z = 29

– 2x + 5y – z = – 15

3x – y + 5z = – 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

2x – 3y – 4z = 29

– 2x + 5y – z = – 15

3x – y + 5z = – 11

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

Solving determinant, expanding along 1^st Row

⇒ D = 2[(5)(5) – ( – 1)( – 1)] – ( – 3)[( – 2)(5) – 3( – 1)] + ( – 4)[( – 2)( – 1) – 3(5)]

⇒ D = 2[25 – 1] + 3[ – 10 + 3] – 4[2 – 15]

⇒ D = 48 – 21 + 52

⇒ D = 79

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

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 29[(5)(5) – ( – 1)( – 1)] – ( – 3)[( – 15)(5) – ( – 11)( – 1)] + ( – 4)[( – 15)( – 1) – ( – 11)(5)]

⇒ D₁ = 29[25 – 1] + 3[ – 75 – 11] – 4[15 + 55]

⇒ D₁ = 29[24] + 3[ – 86] – 4(70)

⇒ D₁ = 696 – 258 – 280

⇒ D₁ = 158

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

Here

Solving determinant, expanding along 1^st Row

⇒ D₂ = 2[( – 15)(5) – ( – 11)( – 1)] – 29[( – 2)(5) – 3( – 1)] + ( – 4)[( – 11)( – 2) – 3( – 15)]

⇒ D₂ = 2[ – 75 – 11] – 29( – 10 + 3) – 4(22 + 45)

⇒ D₂ = 2[ – 86] – 29( – 7) – 4(67)

⇒ D₂ = – 172 + 203 – 268

⇒ D₂ = – 237

And, Solve D₃ formed by replacing 1^st column by B matrices

Here

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 2[(5)( – 11) – ( – 15)( – 1)] – ( – 3)[( – 11)( – 2) – ( – 15)(3)] + 29[( – 2)( – 1) – (3)(5)]

⇒ D₃ = 2[ – 55 – 15] + 3(22 + 45) + 29(2 – 15)

⇒ D₃ = 2[ – 70] + 3[67] + 29[ – 13]

⇒ D₃ = – 140 + 201 – 377

⇒ D₃ = – 316

Thus by Cramer’s Rule, we have

⇒

⇒

⇒ x = 2

again,

⇒

⇒

⇒ y = – 3

and,

⇒

⇒

⇒ z = – 4