If $\begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{vmatrix} = \begin{vmatrix} A + Bx…
If $\begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{vmatrix} = \begin{vmatrix} A + Bx \end{vmatrix} (x - A)^{2}$, then the ordered pair $(A, B)$ is equal to.
Solution
$\begin{aligned}
\begin{vmatrix}
x-4 & 2x & 2x \\
2x & x-4 & 2x \\
2x & 2x & x-4
\end{vmatrix}
= (A + Bx)^2
\end{aligned}$
Put $x = 0 \Rightarrow
\begin{vmatrix}
-4 & 0 & 0 \\
0 & -4 & 0 \\
0 & 0 & -4
\end{vmatrix}
= A^3 \Rightarrow A = -4$
$\begin{aligned}
\begin{vmatrix}
x-4 & 2x & 2x \\
2x & x-4 & 2x \\
2x & 2x & x-4
\end{vmatrix}
= (Bx - 4)^2
\end{aligned}$
$\begin{aligned}
\begin{vmatrix}
1-\frac{4}{x} & 2 & 2 \\
2 & 1-\frac{4}{x} & 2 \\
2 & 2 & 1-\frac{4}{x}
\end{vmatrix}
= (B - \frac{4}{x})^2
\end{aligned}$
Put $x \rightarrow \infty \Rightarrow
\begin{vmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{vmatrix}
= B$
On expanding the determinant along the first row, we get $B = 5$.
Asked in: JEE Main 2018 (08 Apr)
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