(D) Given the determinant equation: $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2x & 4 \\ 6 & x\end{array}\right|$

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(D) Given the determinant equation: $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2x & 4 \\ 6 & x\end{array}\right|$.Expanding both sides:$(2 \times 1) - (5 \times 4) = (2x \times x) - (6 \times 4)$.Simplify the expressions:$2 - 20 = 2x^2 - 24$.$-18 = 2x^2 - 24$.Rearranging the terms to solve for $x^2$:$2x^2 = 24 - 18$.$2x^2 = 6$.$x^2 = 3$.Taking the square root on both sides:$x = \pm \sqrt{3}$.

Class 12 Mathematics 3 and 4 .Determinants and Matrices Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

Medium

Find the values of $x$,if $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2x & 4 \\ 6 & x\end{array}\right|$.

A
$x=\pm \sqrt{2}$
B
$x=\pm \sqrt{7}$
C
$x=\pm \sqrt{6}$
D
$x=\pm \sqrt{3}$

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3 and 4 .Determinants and Matrices

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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DifficultJEE MAIN 2024

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Medium

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Find the values of $x$,if $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2x & 4 \\ 6 & x\end{array}\right|$.

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If $a, b, c$ are non-zero real numbers and the system of equations $(a - 1)x = y + z,$ $(b - 1)y = z + x,$ $(c - 1)z = x + y$ has a non-trivial solution,then $ab + bc + ca$ equals

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If the system of equations
$x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0$
$x+(\cos \alpha) y+(\sin \alpha) z=0$
$x+(\sin \alpha) y-(\cos \alpha) z=0$
has a non-trivial solution,then $\alpha \in \left(0, \frac{\pi}{2}\right)$ is equal to :