Question

Solve the system of linear equations by Cramer’s rule 

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

Answer 1

Given as

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

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

Suppose D_j be determinant observe from D after replacing the j^th column 

So,

Provide that D ≠ 0

Here

5x – 7y + z = 11

6x – 8y – z = 15

3x + 2y – 6z = 7

On comparing with theorem, let's find D,D₁,D₂ and D₃

On 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

On solving D₁ formed by replacing 1^st column by B matrices

Now

On 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

On solving D₂ formed by replacing 1^st column by B matrices

Now

On solving determinant

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

⇒ D₂ = 5[– 90 + 7] – 11[– 36 + 3] + 1[42 – 45]

⇒ D₂ = 5[– 83] – 11(– 33) – 3

⇒ D₂ = – 415 + 363 – 3

⇒ D₂ = – 55

On solving D₃ formed by replacing 1^st column by B matrices

Now

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

So by Cramer’s Rule,