(D) matrix is not invertible if its determinant is equal to $0$.We set the determinant of the given matrix to $0$:$\begin{vmatrix} x & 2 & 3 \\ 4 & 5

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(D) matrix is not invertible if its determinant is equal to $0$.We set the determinant of the given matrix to $0$:$\begin{vmatrix} x & 2 & 3 \\ 4 & 5 & 6 \\ 2 & 3 & 5 \end{vmatrix} = 0$Expanding along the first row:$x(5 \times 5 - 6 \times 3) - 2(4 \times 5 - 6 \times 2) + 3(4 \times 3 - 5 \times 2) = 0$$x(25 - 18) - 2(20 - 12) + 3(12 - 10) = 0$$x(7) - 2(8) + 3(2) = 0$$7x - 16 + 6 = 0$$7x - 10 = 0$$7x = 10$$x = \frac{10}{7}$

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

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The value of $x$ such that the matrix $\begin{bmatrix} x & 2 & 3 \\ 4 & 5 & 6 \\ 2 & 3 & 5 \end{bmatrix}$ is not invertible is

MHT CET 2020 All MHT CET

A
$\frac{-10}{7}$
B
$\frac{7}{10}$
C
$\frac{-7}{10}$
D
$\frac{10}{7}$

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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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If ${\Delta _1} = \left| {\begin{array}{{20}{c}}1&0\\a&b\end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{{20}{c}}1&0\\c&d\end{array}} \right|$,then ${\Delta _2}{\Delta _1}$ is equal to

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If $k > 1$ and the determinant of the matrix $A^2$, where $A = \begin{bmatrix} k & k\alpha & \alpha \\ 0 & \alpha & k\alpha \\ 0 & 0 & k \end{bmatrix}$, is $k^2$, then $|\alpha|$ is equal to

MediumAP EAMCET 2014

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$f(x)$ is an $n^{\text{th}}$ degree polynomial satisfying $f(x) = \frac{1}{2} \begin{vmatrix} f(x) & f(\frac{1}{x}) - f(x) \\ 1 & f(\frac{1}{x}) \end{vmatrix}$. If $f(2) = 33$,then the value of $f(3)$ is

DifficultAP EAMCET 2025

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If $\left|\begin{array}{ccc}2 & 2k & 1 \\ 1 & k-1 & 1 \\ 2 & 1 & k+1\end{array}\right|=Ak^2+Bk+C$, then $A+B+C=$

EasyTS EAMCET 2021

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If $\left| \begin{array}{ccc} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{array} \right| = 0$,then $x =$

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The value of $x$ such that the matrix $\begin{bmatrix} x & 2 & 3 \\ 4 & 5 & 6 \\ 2 & 3 & 5 \end{bmatrix}$ is not invertible is

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If ${\Delta _1} = \left| {\begin{array}{*{20}{c}}1&0\\a&b\end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{*{20}{c}}1&0\\c&d\end{array}} \right|$,then ${\Delta _2}{\Delta _1}$ is equal to

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If ${\Delta _1} = \left| {\begin{array}{{20}{c}}1&0\\a&b\end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{{20}{c}}1&0\\c&d\end{array}} \right|$,then ${\Delta _2}{\Delta _1}$ is equal to