Question

Show that each of the following systems of linear equations has infinite number of solutions and solve : 

2x + y – 2z = 4 

x – 2y + z = – 2 

5x – 5y + z = – 2

Answer 1

Given : - 

Three equation

2x + y – 2z = 4 

x – 2y + z = – 2 

5x – 5y + z = – 2

Tip : - We know that 

For a system of 3 simultaneous linear equation with 3 unknowns 

(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

  x = \(\frac{D_1}{D}\), y = \(\frac{D_2}{D}\) and z = \(\frac{D_3}{D}\)

(ii) If D = 0 and D₁ = D₂ = D₃ = 0, then the system is consistent and has infinitely many solution. 

(iii) If D = 0 and one of D₁ , D₂ and D₃ is non – zero, then the system is inconsistent. 

Now, 

We have, 

2x + y – 2z = 4 

x – 2y + z = – 2 

5x – 5y + z = – 2

Lets find D

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

Expanding along 1^st row 

⇒ D = 2[ – 2 – ( – 5)(1)] – (1)[(1)1 – 5(1)] + ( – 2)[ – 5 – 5( – 2)] 

⇒ D = 2[3] – 1[ – 4] – 2[5] 

⇒ D = 0 

Again, 

D₁ by replacing 1^st column by B 

Here,

B = \(\begin{vmatrix} 4 \\[0.3em] -2\\[0.3em]-2 \end{vmatrix}\)

⇒ D₁ = \(\begin{vmatrix} 4 & 1 & -2 \\[0.3em]-2 & -2 & 1 \\[0.3em] -2 &-5 &1 \end{vmatrix}\)

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

⇒ D₁ = 4[ – 2 + 5] – [ – 2 + 2] – 2[6] 

⇒ D₁ = 12 + 0 – 12 

⇒ D₁ = 0

Also, 

D₂ by replacing 2^nd column by B 

Here,

 B = \(\begin{vmatrix} 4 \\[0.3em] -2\\[0.3em]-2 \end{vmatrix}\)

⇒ D₂ = \(\begin{vmatrix} 2 & 4 & -2 \\[0.3em]1 & -2 & 1 \\[0.3em] 5 &-2 &1 \end{vmatrix}\)

⇒ D₂ = 2[ – 2 – ( – 2)(1)] – (4)[(1)1 – (5)] + ( – 2)[ – 2 – 5( – 2)] 

⇒ D₂ = 2[ – 2 + 2] – 4[ – 4] + ( – 2)[8] 

⇒ D₂ = 0 + 16 – 16 

⇒ D₂ = 0

Again, 

D₃ by replacing 3^rd column by B 

Here,

B = \(\begin{vmatrix} 4 \\[0.3em] -2\\[0.3em]-2 \end{vmatrix}\)

⇒ D₃ = \(\begin{vmatrix} 2 & 1 & 4 \\[0.3em]1 & -2 & -2 \\[0.3em] 5 &-5 &-2 \end{vmatrix}\)

⇒ D₃ = 2[4 – ( – 2)( – 5)] – (1)[( – 2)1 – 5( – 2)] + 4[1( – 5) – 5( – 2)] 

⇒ D₃ = 2[ – 6] – [8] + 4[ – 5 + 10] 

⇒ D₃ = – 12 – 8 + 20 

⇒ D₃ = 0 

So, here we can see that 

D = D₁ = D₂ = D₃ = 0 

Thus, 

Either the system is consistent with infinitely many solutions or it is inconsistent.

Now, by 1^st two equations, written as 

x – 2y = – 2 – z 

5x – 5y = – 2 – z 

Now by applying Cramer’s rule to solve them, 

New D and D₁, D₂

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

⇒ D = – 5 + 10 

⇒ D = 5 

Again, 

D₁ by replacing 1^st column with

⇒ B = \(\begin{vmatrix} -2-z \\[0.3em] -2-z \\[0.3em] \end{vmatrix}\)

⇒ D₁ = \(\begin{vmatrix} -2-z &-2\\[0.3em] -2-z&-5 \\[0.3em] \end{vmatrix}\)

⇒ D₁ = 10 + 5z – ( – 2)( – 2 – z) 

⇒ D₁ = 6 + 3z 

Again, 

D₂ by replacing 2^nd column with

⇒ B = \(\begin{vmatrix} -2-z \\[0.3em] -2-z \\[0.3em] \end{vmatrix}\)

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

⇒ D₂ = – 2 – z – 5 ( – 2 – z) 

⇒ D₂ = 8 + 4z

Hence, using Cramer’s rule

And z = k 

By changing value of k you may get infinite solutions.