Question

If $\left|\begin{array}{ccc}x-4& 2x& 2x\\ 2x& x-4& 2x\\ 2x& 2x& x-4\end{array}\right|=\left(A+Bx\right){\left(x-A\right)}^2,$ then the ordered pair $(A,B)$ is equal to

• A. $\left(4,5\right)$
• B. $\left(-4,-5\right)$
• C. $\left(-4,3\right)$
• D. $\left(-4,5\right)$

Answer 1

Answer: (D)

$\left|\begin{array}{ccc}x-4& 2x& 2x\\ 2x& x-4& 2x\\ 2x& 2x& x-4\end{array}\right|=\left(A+Bx\right){\left(x-A\right)}^2$
Put $x=0\Rightarrow \left|\begin{array}{ccc}-4& 0& 0\\ 0& -4& 0\\ 0& 0& -4\end{array}\right|=A^3\Rightarrow A=-4$
$\left|\begin{array}{ccc}x-4& 2x& 2x\\ 2x& x-4& 2x\\ 2x& 2x& x-4\end{array}\right|=\left(Bx-4\right){\left(x+4\right)}^2$
$\left|\begin{array}{ccc}1-\frac{4}{x}& 2& 2\\ 2& 1-\frac{4}{x}& 2\\ 2& 2& 1-\frac{4}{x}\end{array}\right|=\left(B-\frac{4}{x}\right){\left(1+\frac{4}{x}\right)}^2$
Put $x\to \in fty\Rightarrow \left|\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right|=B$
On expanding the determinant along the first row, we get $B=5$ .