Question

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

x – y + z = 3 

2x + y – z = 2 

– x – 2y + 2z = 1

Answer 1

Given : - 

Three equation

x + y + z = 3 

2x – y + z = 2 

– x – 2y + 2z = 1

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 given system of equation may or may not be consistent. 

However if consistent, then it has infinitely many solution.

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

Now, 

We have,

x + y + z = 3 

2x – y + z = 2 

– x – 2y + 2z = 1

Lets find D

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

Expanding along 1^st row 

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

⇒ D = 1[0] + 1[3] + [ – 3] 

⇒ D = 0

Again, 

D₁ by replacing 1^st column by B 

Here,

B = \(\begin{vmatrix} 3 \\[0.3em] 2\\[0.3em]1 \end{vmatrix}\)

⇒ D₁ = \(\begin{vmatrix} 3 & -1 & 1 \\[0.3em] 2 & 1 &-1 \\[0.3em] 1 & -2 & 2 \end{vmatrix}\)

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

⇒ D₁ = 3[2 – 2] + [4 + 1] + 1[ – 5] 

⇒ D₁ = 0 + 5 – 5 

⇒ D₁ = 0 

Also, 

D₂ by replacing 2^nd column by B 

Here,

B = \(\begin{vmatrix} 3 \\[0.3em] 2\\[0.3em]1 \end{vmatrix}\)

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

⇒ D₂ = 1[4 – ( – 1)(1)] – (3)[(2)2 – (1)] + 1[2 – ( – 2)] 

⇒ D₂ = 1[4 + 1] – 3[4 – 1] + 1[4] 

⇒ D₂ = 5 – 9 + 4 

⇒ D₂ = 0

Again, 

D₃ by replacing 3^rd column by B 

Here,

B = \(\begin{vmatrix} 3 \\[0.3em] 2\\[0.3em]1 \end{vmatrix}\)

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

⇒ D₃ = 1[1 – ( – 2)(2)] – ( – 1)[(2)1 – 2( – 1)] + 3[2( – 2) – 1( – 1)] 

⇒ D₃ = [1 + 4] + [2 + 2] + 3[ – 4 + 1] 

⇒ D₃ = 5 + 4 – 9 

⇒ 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 – y = 3 – z 

2x + y = 2 + z 

Now by applying Cramer’s rule to solve them, 

New D and D₁, D₂

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

⇒ D = 1 + 2 

⇒ D = 3 

Again, 

D₁ by replacing 1^st column with

⇒ B = \(\begin{vmatrix} 3-z \\[0.3em] 2+z \\[0.3em] \end{vmatrix}\)

⇒ D₁ = \(\begin{vmatrix} 3-z &-1 \\[0.3em] 2+z&1 \\[0.3em] \end{vmatrix}\) 

⇒ D₁ = 3 – z – ( – 1)(2 + z) 

⇒ D₁ = 5

Again, 

D₂ by replacing 2^nd column with

⇒ B = \(\begin{vmatrix} 3-z \\[0.3em] 2+z \\[0.3em] \end{vmatrix}\)

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

⇒ D₂ = 2 + z – 2 (3 – z) 

⇒ D₂ = – 4 + 3z 

Hence, 

Using Cramer’s rule

And z = k 

By changing value of k you may get infinite solutions.