The adjoint matrix of $\left[ {\begin{array}{*{20}{c}} 3&{ - 3}&4\\ 2&{ - 3}&4\\ 0&{ - 1}&1 \end{array}} \right]$ is

$\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0\\ { - 2}&3&{ - 4}\\ { - 2}&3&{ - 3} \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 4&8&3\\ 2&1&6\\ 0&2&1 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} {11}&9&3\\ 1&2&8\\ 6&9&1 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 1&{ - 2}&1\\ { - 1}&3&3\\ { - 2}&3&{ - 3} \end{array}} \right]$

JEE MATHEMATICS MCQ: MATRICES

1 Easy

1 of 1

Application +4 | -1

The adjoint matrix of

\left[ {\begin{array}{cccccccccccccccccccc} 3&{ - 3}&4\\ 2&{ - 3}&4\\ 0&{ - 1}&1 \end{array}} \right]

View answer and explanation

Correct option: B

\left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 1}&0\\ { - 2}&3&{ - 4}\\ { - 2}&3&{ - 3} \end{array}} \right]

Option B is correct. The adjoint matrix is found by taking the transpose of the cofactor matrix.

Practice All Chapter MCQs →

\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|