If $A = \left[ {\begin{array}{*{20}{c}} a&b&0\\ { - b}&a&0\\ 0&0&1 \end{array}} \right],\;{\rm{where}}\;{a^2} + {b^2} = 1,$ then $adj\left( A \right)$ is equal to (Here, ${A^T}{\rm{\;is\;the\;transpose\;of}}\;A)$

JEE MATHEMATICS MCQ: MATRICES

1 Easy

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Application +4 | -1

A = \left[ {\begin{array}{cccccccccccccccccccc} a&b&0\\ { - b}&a&0\\ 0&0&1 \end{array}} \right],\;\mathrm{where}\;{a^2} + {b^2} = 1,

then

adj\left( A \right)

is equal to
(Here,

{A^T}\mathrm{\;is\;the\;transpose\;of}\;A)

View answer and explanation

Correct option: A

{A^{ - 1}}

Option A (

{A^{ - 1}}

) is correct because the determinant of A is 1, making the adjugate of A equal to its inverse.

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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

a, b, c

a x+b y+c=0

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

2x + 3y + 1 = 0,3x + y - 2 = 0

ax + 2y - b = 0

\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|

\omega

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

A

2 \times 2

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

|A|