If $A = \left[ {\begin{array}{*{20}{c}} {1\;\,\,\, - 3}\\ {2\;\;\;\;\;k} \end{array}} \right]\;$ and $\;{A^2} - 4A = 10I = A\;$ then $k$ is equal to

JEE MATHEMATICS MCQ: MATRICES

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Analyzing +4 | -1

A = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\,\,\, - 3}\\ {2\;\;\;\;\;k} \end{array}} \right]\;

and

\;{A^2} - 4A = 10I = A\;

then

k

is equal to

View answer and explanation

Correct option: C

4 and not 1

Option C (4 and not 1) is correct because substituting the matrix A into the given equation and solving for k yields k = 4, but k = 1 does not satisfy the initial equation.

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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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{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|

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

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|A|

is equal to