Question

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

6x + y – 3z = 5 

x + 3y – 2z = 5 

2x + y + 4z = 8

Answer 1

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

\(\begin{vmatrix} b_1 \\[0.3em] b_2 \\[0.3em]: \\[0.3em] b_n \end{vmatrix}\)

Then,

x₁ = \(\frac{D_1}{D},\) x₂ =\(\frac{D_2}{D},\)....,x_n = \(\frac{D_n}{D}\)

provided that D ≠ 0

Now, here we have

6x + y – 3z = 5 

x + 3y – 2z = 5 

2x + y + 4z = 8

So by comparing with the theorem, let's find D, D_1,D₂ and D₃

⇒ D = \(\begin{vmatrix} 6 &1 & -3 \\[0.3em] 1 & 3 &-2 \\[0.3em] 2 &1 &4 \end{vmatrix}\)

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