# Show that x = 2 i

Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation Ri Ri + kRj or Ci Ci + kCj.

Applying R2 R2 – R1, we get

Taking the term (x – 2) common from R2, we get

Applying R3 R3 – R1, we get

Taking the term (x + 3) common from R3, we get

Applying C1 C1 + C3, 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 and its other roots are –3 and 1.

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