Adjoint and Inverse of Matrices
« Previous Topic: Product of Matrices and its Properties Next Topic: Solution of System of Linear Equation Using Matrix »
For More Updates join with Us
Matrix and Determinant

78867 If \(A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\) then \((A d j A)^{-1}=\)

1 \(\mathrm{A}+\mathrm{I}\)
2 \(\mathrm{A}-\mathrm{I}\)
3 \(\mathrm{A}\)
4 \(\operatorname{Adj}\left(\mathrm{A}^{-1}\right)\)
AP EAMCET-2019-20.04.2019
Matrix and Determinant

78868 Let \(M=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(N=\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]\). Then \(N M^{10} N^{-1}=\)

1 \(\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]\)
2 \(\left[\begin{array}{cc}1 & -5 \\ 0 & 1\end{array}\right]\)
3 \(\left[\begin{array}{cc}1 & -10 \\ 0 & 1\end{array}\right]\)
4 \(\left[\begin{array}{ll}1 & 10 \\ 0 & 1\end{array}\right]\)
AP EAMCET-2020-18.09.2020
Matrix and Determinant

78869 If \(A=\frac{1}{7}\left[\begin{array}{ccc}3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3\end{array}\right]\), then

1 \(\mathrm{A}^{-1}=\mathrm{A}\)
2 \(\mathrm{A}^{-1}=\mathrm{A}^{\mathrm{T}}\)
3 \(\mathrm{A}^{-1}\) does not exist
4 \(\mathrm{A}^{-1}=-\mathrm{A}\)
AP EAMCET-2021-04.07.2021
Matrix and Determinant

78870 \(\quad \mathbf{A}(\alpha, \beta)=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^{\beta}\end{array}\right]=[\mathbf{A}(\alpha, \beta)]^{-1}=\)

1 \(\mathrm{A}(-\alpha, \beta)\)
2 \(\mathrm{A}(-\alpha,-\beta)\)
3 \(\mathrm{A}(\alpha,-\beta)\)
4 \(\mathrm{A}(\alpha, \beta)\)
AP EAMCET-2011
NEET DIGITAL LIBRARY
Matrix and Determinant

78867 If \(A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\) then \((A d j A)^{-1}=\)

1 \(\mathrm{A}+\mathrm{I}\)
2 \(\mathrm{A}-\mathrm{I}\)
3 \(\mathrm{A}\)
4 \(\operatorname{Adj}\left(\mathrm{A}^{-1}\right)\)
AP EAMCET-2019-20.04.2019
Matrix and Determinant

78868 Let \(M=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(N=\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]\). Then \(N M^{10} N^{-1}=\)

1 \(\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]\)
2 \(\left[\begin{array}{cc}1 & -5 \\ 0 & 1\end{array}\right]\)
3 \(\left[\begin{array}{cc}1 & -10 \\ 0 & 1\end{array}\right]\)
4 \(\left[\begin{array}{ll}1 & 10 \\ 0 & 1\end{array}\right]\)
AP EAMCET-2020-18.09.2020
Matrix and Determinant

78869 If \(A=\frac{1}{7}\left[\begin{array}{ccc}3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3\end{array}\right]\), then

1 \(\mathrm{A}^{-1}=\mathrm{A}\)
2 \(\mathrm{A}^{-1}=\mathrm{A}^{\mathrm{T}}\)
3 \(\mathrm{A}^{-1}\) does not exist
4 \(\mathrm{A}^{-1}=-\mathrm{A}\)
AP EAMCET-2021-04.07.2021
Matrix and Determinant

78870 \(\quad \mathbf{A}(\alpha, \beta)=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^{\beta}\end{array}\right]=[\mathbf{A}(\alpha, \beta)]^{-1}=\)

1 \(\mathrm{A}(-\alpha, \beta)\)
2 \(\mathrm{A}(-\alpha,-\beta)\)
3 \(\mathrm{A}(\alpha,-\beta)\)
4 \(\mathrm{A}(\alpha, \beta)\)
AP EAMCET-2011
Join With Us for More Updates
Matrix and Determinant

78867 If \(A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\) then \((A d j A)^{-1}=\)

1 \(\mathrm{A}+\mathrm{I}\)
2 \(\mathrm{A}-\mathrm{I}\)
3 \(\mathrm{A}\)
4 \(\operatorname{Adj}\left(\mathrm{A}^{-1}\right)\)
AP EAMCET-2019-20.04.2019
Matrix and Determinant

78868 Let \(M=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(N=\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]\). Then \(N M^{10} N^{-1}=\)

1 \(\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]\)
2 \(\left[\begin{array}{cc}1 & -5 \\ 0 & 1\end{array}\right]\)
3 \(\left[\begin{array}{cc}1 & -10 \\ 0 & 1\end{array}\right]\)
4 \(\left[\begin{array}{ll}1 & 10 \\ 0 & 1\end{array}\right]\)
AP EAMCET-2020-18.09.2020
Matrix and Determinant

78869 If \(A=\frac{1}{7}\left[\begin{array}{ccc}3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3\end{array}\right]\), then

1 \(\mathrm{A}^{-1}=\mathrm{A}\)
2 \(\mathrm{A}^{-1}=\mathrm{A}^{\mathrm{T}}\)
3 \(\mathrm{A}^{-1}\) does not exist
4 \(\mathrm{A}^{-1}=-\mathrm{A}\)
AP EAMCET-2021-04.07.2021
Matrix and Determinant

78870 \(\quad \mathbf{A}(\alpha, \beta)=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^{\beta}\end{array}\right]=[\mathbf{A}(\alpha, \beta)]^{-1}=\)

1 \(\mathrm{A}(-\alpha, \beta)\)
2 \(\mathrm{A}(-\alpha,-\beta)\)
3 \(\mathrm{A}(\alpha,-\beta)\)
4 \(\mathrm{A}(\alpha, \beta)\)
AP EAMCET-2011
For More Updates join with Us
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Matrix and Determinant

78867 If \(A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\) then \((A d j A)^{-1}=\)

1 \(\mathrm{A}+\mathrm{I}\)
2 \(\mathrm{A}-\mathrm{I}\)
3 \(\mathrm{A}\)
4 \(\operatorname{Adj}\left(\mathrm{A}^{-1}\right)\)
AP EAMCET-2019-20.04.2019
Matrix and Determinant

78868 Let \(M=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(N=\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]\). Then \(N M^{10} N^{-1}=\)

1 \(\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]\)
2 \(\left[\begin{array}{cc}1 & -5 \\ 0 & 1\end{array}\right]\)
3 \(\left[\begin{array}{cc}1 & -10 \\ 0 & 1\end{array}\right]\)
4 \(\left[\begin{array}{ll}1 & 10 \\ 0 & 1\end{array}\right]\)
AP EAMCET-2020-18.09.2020
Matrix and Determinant

78869 If \(A=\frac{1}{7}\left[\begin{array}{ccc}3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3\end{array}\right]\), then

1 \(\mathrm{A}^{-1}=\mathrm{A}\)
2 \(\mathrm{A}^{-1}=\mathrm{A}^{\mathrm{T}}\)
3 \(\mathrm{A}^{-1}\) does not exist
4 \(\mathrm{A}^{-1}=-\mathrm{A}\)
AP EAMCET-2021-04.07.2021
Matrix and Determinant

78870 \(\quad \mathbf{A}(\alpha, \beta)=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^{\beta}\end{array}\right]=[\mathbf{A}(\alpha, \beta)]^{-1}=\)

1 \(\mathrm{A}(-\alpha, \beta)\)
2 \(\mathrm{A}(-\alpha,-\beta)\)
3 \(\mathrm{A}(\alpha,-\beta)\)
4 \(\mathrm{A}(\alpha, \beta)\)
AP EAMCET-2011