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Mathematics Question on Applications of Determinants and Matrices

The value of the determinant ∣b2−abb−cbc−ac ab−a2a−bb2−ab bc−acc−aab−a2∣\begin{vmatrix}b^{2}-a b & b-c & b c-a c \\\ a b-a^{2} & a-b & b^{2}-a b \\\ b c-a c & c-a & a b-a^{2}\end{vmatrix} is

A

a b c

B

a+b+c

C

0

D

a b+b c+c a

Answer

0

Explanation

Solution

.Let Δ=∣b2−abb−cbc−ac ab−a2a−bb2−ab bc−acc−aab−a2∣\Delta =\begin{vmatrix} b^{2}-a b & b-c & b c-a c \\\ a b-a^{2} & a-b & b^{2}-a b \\\ b c-a c & c-a & a b-a^{2} \end{vmatrix}
=∣b(b−a)b−cc(b−a) a(b−a)a−bb(b−a) c(b−a)c−aa(b−a)∣=\begin{vmatrix} b(b-a) & b-c & c(b-a) \\\ a(b-a) & a-b & b(b-a) \\\ c(b-a) & c-a & a(b-a) \end{vmatrix}
Taking common (b−a)(b-a) from C1C_{1} and C3C_{3}, respectively
=(b−a)(b−a)∣bb−cc aa−bb cc−aa∣=(b-a)(b-a)\begin{vmatrix} b & b-c & c \\\ a & a-b & b \\\ c & c-a & a \end{vmatrix}
=(b−a)2∣b0c a0b c0a∣=(b-a)^{2}\begin{vmatrix} b & 0 & c \\\ a & 0 & b \\\ c & 0 & a \end{vmatrix}
[using C2→C2−(C1−C3C_{2} \rightarrow C_{2}-(C_{1}-C_{3})]
=0= 0