Question

The determinants:
$\left|\begin{array}{ccc}{b}^2-abb-cbc-ac& & \\ ab-a^{2}a-b{b}^2-ab& & \\ bc-acc-aab-a^2& & \end{array}\right|$ equals

Answer 1

Given:$\left|\begin{array}{ccc}{b}^2-abb-cbc-ac& & \\ ab-a^{2}a-b{b}^2-ab& & \\ bc-acc-aab-a^2& & \end{array}\right|$
Let $\Delta \left|\begin{array}{ccc}{b}^2-abb-cbc-ac& & \\ ab-a^{2}a-b{b}^2-ab& & \\ bc-acc-aab-a^2& & \end{array}\right|$
$\Rightarrow \Delta =\left|\begin{array}{ccc}b(b-a)b-cc(b-a)& & \\ a(b-a)a-bb(b-a)& & \\ c(b-a)c-aa(b-a)& & \end{array}\right|$
Taking $(b-a)$ common from $C_1$ and $C_2$ both
$\Rightarrow \Delta =\left|\begin{array}{ccc}bb-cc& & \\ aa-bb& & \\ cc-aa& & \end{array}\right|$
Applying $c_1\to c_1-c_3$
$\Rightarrow \Delta =(b-a{)}^2\left|\begin{array}{ccc}b-cb-cc& & \\ a-ba-bb& & \\ c-ac-aa& & \end{array}\right|$
If any two rows or columns are identical then value of determinant is zero.
Hence, $\Delta =0$ $(⸪C_1\text{and}{C}_2$ are identical)
Hence, correct option is 𝐷