Question

What is the value of \(\begin{vmatrix}sin^2 ⁡a & sina\, cosa & cos^2 ⁡a \\sin^2 ⁡b & sinb\, cosb & cos^2 ⁡b \\sin^2⁡ c & sinc\, cosc & cos^2⁡ c \end {vmatrix}\)?

(a) -sin(a – b) sin(b – c) sin(c – a)

(b) sin(a – b) sin(b – c) sin(c – a)

(c) -sin(a + b) sin(b + c) sin(c + a)

(d) sin(a + b) sin(b + c) sin(c + a)

This question was addressed to me during an interview for a job.

This is a very interesting question from Determinant in section Determinants of Mathematics – Class 12

Answer 1

The correct answer is (a) -sin(a – b) sin(b – c) sin(c – a)

Easiest explanation: We have, \(\begin{vmatrix}sin^2 a & sina \,cosa & cos^2 a\\sin^2 b & sinb\, cosb & cos^2 b\\sin^2 c & sinc\, cosc & cos^2 c \end {vmatrix}\)

Now, multiplying by 2 in both numerator and denominator of column 2 and C1 = C1 + C3 we get,    

1/2 \(\begin{vmatrix}sin^2 a + cos^2 a & 2sina\, cosa & cos^2 a\\sin^2 b + cos^2 b & 2sinb\, cosb & cos^2 b\\sin^2 c + cos^2 c & 2sinc\, cosc & cos^2 c \end {vmatrix}\)

= 1/2 \(\begin{vmatrix}sin^2 a + cos^2 a & sin2a & cos^2 a\\sin^2 b + cos^2 b & sin2b & cos^2 b\\sin^2 c + cos^2 c & sin2c & cos^2 c \end {vmatrix}\)

= 1/2 \(\begin{vmatrix}1 & sin2a & cos^2 a\\1 & sin2b & cos^2 b\\1 & sin2c & cos^2 c \end {vmatrix}\)

Solving further,

= 1/2 \(\begin{vmatrix}1 & sin2a & cos^2⁡a \\0 & sin2b-sin2a & cos^2⁡ b-cos^2 ⁡a \\0 & sin2c-sin2a & cos^2 ⁡c-cos^2 ⁡a \end {vmatrix}\)

= 1/2 [(sin2b – sin2a)(cos^2⁡c – cos^2⁡a) – (cos^2 b – cos^2a)(sin2c – sin2a)]

Now, since, [cos^2 ⁡A + cos^2 B = sin(A + B) * sin(B – A)]

So, 1/2 [2 cos(a + b) sin(b – a) * sin(c + a)sin(a – c) – sin(a + b)sin(a – b) * 2 cos(a + c)sin(c – a)]

= sin(a – b)sin(c – a)[sin(c + a)cos(a + b) – cos(c + a) sin(a + b)]

= sin(a – b) sin(c – a) sin(c + a – a – b)

= -sin(a – b) sin(b – c) sin(c – a)