Show that x = 2 is a root of the equation |(x,-6,-1)(2,-3x,x-3)(-3,2x,x+2)| = 0 and solve it completely. ​

Question

Show that x = 2 is a root of the equation \(\begin{vmatrix} x & -6& -1 \\[0.3em] 2 & -3x &x-3 \\[0.3em] -3 & 2x & x+2 \end{vmatrix}\) = 0 and solve it completely.

Answer 1

Let Δ = \(\begin{vmatrix} x & -6& -1 \\[0.3em] 2 & -3x &x-3 \\[0.3em] -3 & 2x & x+2 \end{vmatrix}\)

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₂→ R₂ – R₁, we get

Expanding the determinant along C₁, we have 

Δ = (x – 2)(x + 3)(x – 1)[(–3)(1) – (2)(1)] 

⇒ Δ = (x – 2)(x + 3)(x – 1)(–5) 

∴ Δ = –5(x – 2)(x + 3)(x – 1) 

The given equation is Δ = 0. 

⇒ –5(x – 2)(x + 3)(x – 1) = 0

⇒ (x – 2)(x + 3)(x – 1) = 0

Case – I : 

x – 2 = 0 

⇒ x = 2 

Case – II :

 x + 2 = 0 

⇒ x = –3 

Case – III : 

x – 1 = 0 

⇒ x = 1

Thus, 

2 is a root of the equation \(\begin{vmatrix} x & -6& -1 \\[0.3em] 2 & -3x &x-3 \\[0.3em] -3 & 2x & x+2 \end{vmatrix}\) = 0 and its other roots are –3 and 1.