Question

Find the determinant of A=\(\begin{bmatrix}c^2&cb&ca\\ab&a^2&-ac\\ab&bc&-b^2\end{bmatrix}\)

(a) abc(a^3+b^3+c^3+abc)

(b) abc(a^3+b^3+c^3-abc)

(c) abc(a^3+b^3+c^3+abc)

(d) (a^3-b^3+c^3-abc)

I have been asked this question in an international level competition.

Question is taken from Properties of Determinants topic in chapter Determinants of Mathematics – Class 12

Answer 1

The correct option is (b) abc(a^3+b^3+c^3-abc)

For explanation I would say: Given that, A=\(\begin{bmatrix}c^2&cb&ca\\ab&a^2&-ac\\ab&bc&-b^2\end{bmatrix}\)

Taking c a, b common from R1, R2, R3 respectively, we get

Δ=abc\(\begin{bmatrix}c&b&a\\b&a&-c\\a&c&-b\end{bmatrix}\)

Δ=abc{(c(-ab+c^2)-b(-b^2+ac)+a(bc-a^2)

Δ=abc(-abc+c^3+b^3-abc+abc-a^3)

Δ=abc(a^3+b^3+c^3-abc).