Question

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

2x – 3y – 4z = 29 

– 2x + 5y – z = – 15 

3x – y + 5z = – 11

Answer 1

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

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

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

2x – 3y – 4z = 29 

– 2x + 5y – z = – 15 

3x – y + 5z = – 11

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

⇒ D = \(\begin{vmatrix} 2 &-3 &-4 \\[0.3em] -2 &5 &-1 \\[0.3em] 3 &-1 &5\end{vmatrix}\)

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,

 B = \(\begin{vmatrix} 29 \\[0.3em] -15 \\[0.3em] -11 \end{vmatrix}\) 

⇒ D₁ = \(\begin{vmatrix} 29&-3 & -4 \\[0.3em] -15& 5 &-1 \\[0.3em] -11 & -1 & 5 \end{vmatrix}\)

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,

B = \(\begin{vmatrix} 29 \\[0.3em] -15 \\[0.3em] -11 \end{vmatrix}\) 

⇒ D₂ = \(\begin{vmatrix} 2&-2 & 3 \\[0.3em]-2& -15 &-1 \\[0.3em]3 & -11 & 5 \end{vmatrix}\)

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,

 B = \(\begin{vmatrix} 29 \\[0.3em] -15 \\[0.3em] -11 \end{vmatrix}\) 

⇒ D₃ = \(\begin{vmatrix} 2&-3 & 29 \\[0.3em]-2& 5 &-15 \\[0.3em]3 & -1 & -11 \end{vmatrix}\)

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