Given: - Equations are: –

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

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

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

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

Solving determinant, expanding along 1^st Row

⇒ D = 5[( – 8)( – 6) – ( – 1)(2)] – 7[( – 6)(6) – 3( – 1)] + 1[2(6) – 3( – 8)]

⇒ D = 5[48 + 2] – 7[ – 36 + 3] + 1[12 + 24]

⇒ D = 250 – 231 + 36

⇒ D = 55

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

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 11[( – 8)( – 6) – (2)( – 1)] – ( – 7)[(15)( – 6) – ( – 1)(7)] + 1[(15)2 – (7)( – 8)]

⇒ D₁ = 11[48 + 2] + 7[ – 90 + 7] + 1[30 + 56]

⇒ D₁ = 11[50] + 7[ – 83] + 86

⇒ D₁ = 550 – 581 + 86

⇒ D₁ = 55

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

Here

Solving determinant, expanding along 1^st Row

⇒ D₂ = 5[(15)( – 6) – (7)( – 1)] – 11[(6)( – 6) – ( – 1)(3)] + 1[(6)7 – (15)(3)]

⇒ D_{2 =} 5[ – 90 + 7] – 11[ – 36 + 3] + 1[42 – 45]

⇒ D₂ = 5[ – 83] – 11( – 33) – 3

⇒ D₂ = – 415 + 363 – 3

⇒ D₂ = – 55

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

Here

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 5[( – 8)(7) – (15)(2)] – ( – 7)[(6)(7) – (15)(3)] + 11[(6)2 – ( – 8)(3)]

⇒ D₃ = 5[ – 56 – 30] – ( – 7)[42 – 45] + 11[12 + 24]

⇒ D₃ = 5[ – 86] + 7[ – 3] + 11[36]

⇒ D₃ = – 430 – 21 + 396

⇒ D₃ = – 55

Thus by Cramer’s Rule, we have

⇒

⇒

⇒ x = 1

again,

⇒

⇒

⇒ y = – 1

and,

⇒

⇒

⇒ z = – 1