If $\left|\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right|,$ then $x$ is equal to

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$\left|\begin{array}{ll}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{ll}6 & 2 \\ 18 & 6\end{array}\right|$

$\Rightarrow x^{2}-36=36-36$

$\Rightarrow x^{2}-36=0$

$\Rightarrow x^{2}=36$

$\Rightarrow x=\pm 6$

Hence, the correct answer is $\mathrm{B}$.

Similar Questions

For positive numbers $x,y$ and $z$  the numerical value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&{{{\log }_x}y}&{{{\log }_x}z}\\{{{\log }_y}x}&1&{{{\log }_y}z}\\{{{\log }_z}x}&{{{\log }_z}y}&1\end{array}\,} \right|$is

  • [IIT 1993]

$\left| {\,\begin{array}{*{20}{c}}{{{({a^x} + {a^{ - x}})}^2}}&{{{({a^x} - {a^{ - x}})}^2}}&1\\{{{({b^x} + {b^{ - x}})}^2}}&{{{({b^x} - {b^{ - x}})}^2}}&1\\{{{({c^x} + {c^{ - x}})}^2}}&{{{({c^x} - {c^{ - x}})}^2}}&1\end{array}\,} \right| = $

Evaluate the determinant $\Delta=\left|\begin{array}{rrr}1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0\end{array}\right|$

If $AB = A$ and $BA = B$, then

$\left| {\,\begin{array}{*{20}{c}}5&3&{ - 1}\\{ - 7}&x&{ - 3}\\9&6&{ - 2}\end{array}\,} \right| = 0$, then $ x$ is equal to