Given: - Equations are: –

6x + y – 3z = 5

X + 3y – 2z = 5

2x + y + 4z = 8

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

6x + y – 3z = 5

x + 3y – 2z = 5

2x + y + 4z = 8

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

Solving determinant, expanding along 1^st Row

⇒ D = 6[(4)(3) – (1)( – 2)] – 1[(4)(1) + 4] – 3[1 – 3(2)]

⇒ D = 6[12 + 2] – [8] – 3[ – 5]

⇒ D = 84 – 8 + 15

⇒ D = 91

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

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 5[(4)(3) – ( – 2)(1)] – 1[(5)(4) – ( – 2)(8)] – 3[(5) – (3)(8)]

⇒ D₁ = 5[12 + 2] – 1[20 + 16] – 3[5 – 24]

⇒ D₁ = 5[14] – 36 – 3( – 19)

⇒ D₁ = 70 – 36 + 57

⇒ D₁ = 91

Again, Solve D₂ formed by replacing 1^st column by B matrices

Here

Solving determinant

⇒ D₂ = 6[20 + 16] – 5[4 – 2( – 2)] + ( – 3)[8 – 10]

⇒ D₂ = 6[36] – 5(8) + ( – 3)( – 2)

⇒ D₂ = 182

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

Here

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 6[24 – 5] – 1[8 – 10] + 5[1 – 6]

⇒ D₃ = 6[19] – 1( – 2) + 5( – 5)

⇒ D₃ = 114 + 2 – 25

⇒ D₃ = 91

Thus by Cramer’s Rule, we have

⇒

⇒

⇒ x = 1

again,

⇒

⇒

⇒ y = 2

and,

⇒

⇒

⇒ z = 1