#### Provide solution for RD Sharma maths class 12 chapter Determinants exercise multiple choise question 26

Correct option (d)

Hint:

Evaluate the given determinant by applying row or column operation.

Given:

$Let\; \; \; \; \Delta =\begin{vmatrix} b^{2}-ab &b-c &bc-ac \\ ab-a^{2} &a-b &b^{2}-ab \\ bc-ca &c-a &ab-a^{2} \end{vmatrix}$

We have to find the value of $\Delta$

Solution:

$Here\; \; \; \; \Delta =\begin{vmatrix} b^{2}-ab &b-c &bc-ac \\ ab-a^{2} &a-b &b^{2}-ab \\ bc-ca &c-a &ab-a^{2} \end{vmatrix}$

$\Rightarrow \Delta =\begin{vmatrix} b(b-a) &b-c &c(b-a) \\ a(b-a) &a-b &b(b-a) \\ c(b-a) &c-a &a(b-a) \end{vmatrix}$

Taking common (b-a) from column 1 and 3

$\Rightarrow \Delta=(b-a)^{2}\left|\begin{array}{lll} b & b-c & c \\ a & a-b & b \\ c & c-a & a \end{array}\right|$

Here, C1 and C2 are same and if any two row or column of a matrix is identical then determinant will be zero.

$\Rightarrow \Delta=\left|\begin{array}{lll} b & b & c \\ a & a & b \\ c & c & a \end{array}\right|$

$\Rightarrow \Delta=0$

$Hence,\; \; \; \; \begin{vmatrix} b^{2}-ab &b-c &bc-ac \\ ab-a^{2} &a-b &b^{2}-ab \\ bc-ca &c-a &ab-a^{2} \end{vmatrix}=0$