(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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Medium

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MediumIIT 1978

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MediumAP EAMCET 2024

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If $-9$ is a root of the equation $\left| \begin{array}{ccc} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array} \right| = 0$,then the other two roots are:

MediumIIT 1983

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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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The solution of the equation $\left| \begin{array}{ccc} \cos \theta & \sin \theta & \cos \theta \\ -\sin \theta & \cos \theta & \sin \theta \\ -\cos \theta & -\sin \theta & \cos \theta \end{array} \right| = 0$ is:

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