Let $A$ be a non-singular square matrix. Then, $\left| {{\rm{adj\;}}A} \right|$ is equal to

JEE MATHEMATICS MCQ: MATRICES

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Let

A

be a non-singular square matrix. Then,

\left| {\mathrm{adj\;}A} \right|

is equal to

View answer and explanation

Correct option: B

{\left| A \right|^{n - 1}}

Option B (

\left| A \right|^{n - 1}

) is correct because the determinant of the adjoint of a matrix A is equal to the determinant of A raised to the power of (n-1), where n is the order of the matrix.

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\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

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a x+b y+c=0

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{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

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\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

, is If

\omega

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\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

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, if

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

, then

|A|

is equal to