Question

Without expanding, show that the value of each of the following determinants is zero:

(i) \(\begin{vmatrix} sin^2 23° & sin^2 67° & cos180° \\[0.3em] -sin^267° &-sin^223° & cos^2180° \\[0.3em] cos180°& sin^223° &sin^267° \end{vmatrix}\)

(ii) \(\begin{vmatrix} √23 + √3 & √5& √5 \\[0.3em] √15 + √46&5 & √10\\[0.3em] 3 + √115& √15 &5\end{vmatrix}\)

(iii) \(\begin{vmatrix} sin^2 A & cot A & 1 \\[0.3em] sin^2 B&cot B & 1 \\[0.3em]sin^2 C & cot C &1 \end{vmatrix}\), where A, B and C are the angles of Δ ABC.

Answer 1

(i) Suppose 

On applying C₁ → C₁ + C₂

On using sin(90 - A) = cos A, sin² A + cos² A = 1 and cos 180° = -1

Take (-1) common from C₁ 

Here, C₁ = C₃ 

The value of determinant is zero.

(ii) Suppose 

On multiplying C₂ with √3 and C₃ with √23 

On taking common from C₂ and C₃ 

On applying  C₂ → C₂ + C₃

Here, C₁ = C₂ 

The value of determinant is zero.

(iii) Suppose 

Δ = sin² A(cot B - cot C) - cot A(sin² B - sin² C) + 1(sin² B cot C - cot B sin² C)

Here, A,B and C are angles of triangle

A + B + C = 180°

Δ = sin² A cot B - sin² A cot C - cot A sin² B + cot A sin² C + sin² B cot C - cot B sin² C

Using formula

Thus proved.