(B) matrix $A$ is singular if its determinant is zero,i.e.,$|A| = 0$.Given $A = \begin{bmatrix} a & 2 \\ 2 & 4 \end{bmatrix}$.The determinant $|A| = (

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(B) matrix $A$ is singular if its determinant is zero,i.e.,$|A| = 0$.Given $A = \begin{bmatrix} a & 2 \\ 2 & 4 \end{bmatrix}$.The determinant $|A| = (a \times 4) - (2 \times 2) = 4a - 4$.For the matrix to be singular,set $|A| = 0$:$4a - 4 = 0$$4a = 4$$a = 1$.Therefore,the value of $a$ is $1$.

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 $a$ for which the matrix $A = \begin{bmatrix} a & 2 \\ 2 & 4 \end{bmatrix}$ is singular is:

A
$a \ne 1$
B
$a = 1$
C
$a = 0$
D
$a = -1$

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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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EasyGSEB 2019

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If $\omega$ is a cube root of unity,then $\left| \begin{array}{ccc} x + 1 & \omega & \omega^2 \\ \omega & x + \omega^2 & 1 \\ \omega^2 & 1 & x + \omega \end{array} \right| = $

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The sum of the real roots of the equation $\left| \begin{matrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{matrix} \right| = 0$ is equal to

DifficultJEE MAIN 2019

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The value of $a$ for which the matrix $A = \begin{bmatrix} a & 2 \\ 2 & 4 \end{bmatrix}$ is singular is:

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$\left|\begin{array}{rr}\sin 35^{\circ} & -\cos 35^{\circ} \\ \sin 55^{\circ} & \cos 55^{\circ}\end{array}\right|=$ . . . . . .

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