Question

Show that each of the following systems of linear equations is inconsistent : 

3x – y + 2z = 3 

2x + y + 3z = 5 

x – 2y – z = 1

Answer 1

Given : - Three equation

3x – y + 2z = 3 

2x + y + 3z = 5 

x – 2y – z = 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 solutions.

(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,

3x – y + 2z = 3 

2x + y + 3z = 5 

x – 2y – z = 1

Lets find D

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

Expanding along 1^st row,

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

⇒ D = 3[5] + 1[ – 5] + 2[ – 5] 

⇒ D = 0 

Again, 

D₁ by replacing 1^st column by B 

Here,

 B = \(\begin{vmatrix} 3 \\[0.3em] 5\\[0.3em]1 \end{vmatrix}\)

 ⇒ D₁ = \(\begin{vmatrix} 3 & -1 & 2 \\[0.3em] 5 & 1 &3 \\[0.3em] 1 & -2 & -1 \end{vmatrix}\)

⇒ D₁ = 3[ – 1 – 3( – 2)] – ( – 1)[( – 1)5 – 3] + 2[ – 10 – 1] 

⇒ D₁ = 3[5] + [ – 8] + 2[ – 11] 

⇒ D₁ = 15 – 8 – 22 

⇒ D₁ = – 15 

⇒ D₁ ≠ 0 

So, here we can see that 

D = 0 and D₁ is non – zero Hence the given system of equation is inconsistent. 

Hence Proved.