Question

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

x + y – z = 0 

x – 2y + z = 0 

3x + 6y – 5z = 0

Answer 1

Given : - 

Three equation

x + y – z = 0 

x – 2y + z = 0 

3x + 6y – 5z = 0

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, 

x + y – z = 0 

x – 2y + z = 0 

3x + 6y – 5z = 0

Lets find D

⇒ D = \(\begin{vmatrix} 1 & 1 & -1 \\[0.3em] 1 & -2 & 1 \\[0.3em] 3 &6 &-5 \end{vmatrix}\)

Expanding along 1^st row 

⇒ D = 1[10 – (6)1] – (1)[( – 5)1 – (1)3] + ( – 1)[6 – ( – 2)3] 

⇒ D = 1[4] – 1[ – 8] – [12] 

⇒ D = 0 

Again, 

D₁ by replacing 1^st column by B 

Here,

B = \(\begin{vmatrix} 0 \\[0.3em] 0\\[0.3em]0 \end{vmatrix}\)

⇒ D₁ = \(\begin{vmatrix} 0 & 1 & -1 \\[0.3em] 0 & -2 & 1 \\[0.3em] 0 &6 &-5 \end{vmatrix}\)

As one column is zero its determinant is zero 

⇒ D₁ = 0 

Also, 

D₂ by replacing 2^nd column by B 

Here,

B = \(\begin{vmatrix} 0 \\[0.3em] 0\\[0.3em]0 \end{vmatrix}\)

⇒ D₂ = \(\begin{vmatrix} 1 & 0 & -1 \\[0.3em] 1 & 0 & 1 \\[0.3em] 3 &0 &-5 \end{vmatrix}\)

As one column is zero its determinant is zero 

⇒ D₂ = 0 

Again, 

D₃ by replacing 3^rd column by B 

Here,

B = \(\begin{vmatrix} 0 \\[0.3em] 0\\[0.3em]0 \end{vmatrix}\)

⇒ D₃ = \(\begin{vmatrix} 1 & 1 & 0 \\[0.3em] 1 & -2 & 0 \\[0.3em] 3 &6 &0 \end{vmatrix}\)

As one column is zero its determinant is zero 

⇒ 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 = z 

x – 2y = – z 

Now by applying Cramer’s rule to solve them, 

New D and D₁, D₂

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

⇒ D = – 2 – 1 

⇒ D = – 3 

Again, 

D₁ by replacing 1^st column with

And,

z = k 

By changing value of k you may get infinite solutions.