78818
\(A\) and \(B\) are two matrices such that \(A B=B, B A\) \(=A\), then \(A^{2}+B^{2}=\)
1 \(2 \mathrm{AB}\)
2 \(2 \mathrm{BA}\)
3 \(\mathrm{A}+\mathrm{B}\)
4 \(\mathrm{AB}\)
Explanation:
(C) : Given that, \(\mathrm{AB}=\mathrm{B}\) Multiply by A on both sides we get - \(\mathrm{ABA}=\mathrm{BA} \tag{i}\) And, \(\quad \mathrm{BA}=\mathrm{A}\) Multiply by \(\mathrm{B}\) on both sides we get - \(\mathrm{BAB}=\mathrm{AB} \tag{ii}\) Adding (i) and (ii), we get- \(\mathrm{ABA}+\mathrm{BAB}=\mathrm{BA}+\mathrm{AB}\) \(\mathrm{A}(\mathrm{BA})+\mathrm{B}(\mathrm{AB})=\mathrm{BA}+\mathrm{AB}\) \(\mathrm{AA}+\mathrm{BB}=\mathrm{A}+\mathrm{B}\) \(\mathrm{A}^{2}+\mathrm{B}^{2}=\mathrm{A}+\mathrm{B}\)
COMEDK-2014
Matrix and Determinant
78820
If \(A, B, C\) are invertible matrices, then \((\mathrm{ABC})^{-1}\) equal to
(D) : According to question, A, B, C are invertible matrix. We know that, \((\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}\) Now, \((\mathrm{ABC})^{-1} =[\mathrm{A}(\mathrm{BC})]^{-1}\) \(=(\mathrm{BC})^{-1} \mathrm{~A}^{-1}\) \(=\mathrm{C}^{-1} \mathrm{~B}^{-1} \mathrm{~A}^{-1}\)
SRM JEE-2009
Matrix and Determinant
78821
If \(\left[\begin{array}{rr}\cos \frac{2 \pi}{5} & -\sin \frac{2 \pi}{5} \\ \sin \frac{2 \pi}{5} & \cos \frac{2 \pi}{5}\end{array}\right]^{n}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\), then the least positive integer \(\mathbf{n}\) is equal to
78818
\(A\) and \(B\) are two matrices such that \(A B=B, B A\) \(=A\), then \(A^{2}+B^{2}=\)
1 \(2 \mathrm{AB}\)
2 \(2 \mathrm{BA}\)
3 \(\mathrm{A}+\mathrm{B}\)
4 \(\mathrm{AB}\)
Explanation:
(C) : Given that, \(\mathrm{AB}=\mathrm{B}\) Multiply by A on both sides we get - \(\mathrm{ABA}=\mathrm{BA} \tag{i}\) And, \(\quad \mathrm{BA}=\mathrm{A}\) Multiply by \(\mathrm{B}\) on both sides we get - \(\mathrm{BAB}=\mathrm{AB} \tag{ii}\) Adding (i) and (ii), we get- \(\mathrm{ABA}+\mathrm{BAB}=\mathrm{BA}+\mathrm{AB}\) \(\mathrm{A}(\mathrm{BA})+\mathrm{B}(\mathrm{AB})=\mathrm{BA}+\mathrm{AB}\) \(\mathrm{AA}+\mathrm{BB}=\mathrm{A}+\mathrm{B}\) \(\mathrm{A}^{2}+\mathrm{B}^{2}=\mathrm{A}+\mathrm{B}\)
COMEDK-2014
Matrix and Determinant
78820
If \(A, B, C\) are invertible matrices, then \((\mathrm{ABC})^{-1}\) equal to
(D) : According to question, A, B, C are invertible matrix. We know that, \((\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}\) Now, \((\mathrm{ABC})^{-1} =[\mathrm{A}(\mathrm{BC})]^{-1}\) \(=(\mathrm{BC})^{-1} \mathrm{~A}^{-1}\) \(=\mathrm{C}^{-1} \mathrm{~B}^{-1} \mathrm{~A}^{-1}\)
SRM JEE-2009
Matrix and Determinant
78821
If \(\left[\begin{array}{rr}\cos \frac{2 \pi}{5} & -\sin \frac{2 \pi}{5} \\ \sin \frac{2 \pi}{5} & \cos \frac{2 \pi}{5}\end{array}\right]^{n}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\), then the least positive integer \(\mathbf{n}\) is equal to
78818
\(A\) and \(B\) are two matrices such that \(A B=B, B A\) \(=A\), then \(A^{2}+B^{2}=\)
1 \(2 \mathrm{AB}\)
2 \(2 \mathrm{BA}\)
3 \(\mathrm{A}+\mathrm{B}\)
4 \(\mathrm{AB}\)
Explanation:
(C) : Given that, \(\mathrm{AB}=\mathrm{B}\) Multiply by A on both sides we get - \(\mathrm{ABA}=\mathrm{BA} \tag{i}\) And, \(\quad \mathrm{BA}=\mathrm{A}\) Multiply by \(\mathrm{B}\) on both sides we get - \(\mathrm{BAB}=\mathrm{AB} \tag{ii}\) Adding (i) and (ii), we get- \(\mathrm{ABA}+\mathrm{BAB}=\mathrm{BA}+\mathrm{AB}\) \(\mathrm{A}(\mathrm{BA})+\mathrm{B}(\mathrm{AB})=\mathrm{BA}+\mathrm{AB}\) \(\mathrm{AA}+\mathrm{BB}=\mathrm{A}+\mathrm{B}\) \(\mathrm{A}^{2}+\mathrm{B}^{2}=\mathrm{A}+\mathrm{B}\)
COMEDK-2014
Matrix and Determinant
78820
If \(A, B, C\) are invertible matrices, then \((\mathrm{ABC})^{-1}\) equal to
(D) : According to question, A, B, C are invertible matrix. We know that, \((\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}\) Now, \((\mathrm{ABC})^{-1} =[\mathrm{A}(\mathrm{BC})]^{-1}\) \(=(\mathrm{BC})^{-1} \mathrm{~A}^{-1}\) \(=\mathrm{C}^{-1} \mathrm{~B}^{-1} \mathrm{~A}^{-1}\)
SRM JEE-2009
Matrix and Determinant
78821
If \(\left[\begin{array}{rr}\cos \frac{2 \pi}{5} & -\sin \frac{2 \pi}{5} \\ \sin \frac{2 \pi}{5} & \cos \frac{2 \pi}{5}\end{array}\right]^{n}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\), then the least positive integer \(\mathbf{n}\) is equal to
78818
\(A\) and \(B\) are two matrices such that \(A B=B, B A\) \(=A\), then \(A^{2}+B^{2}=\)
1 \(2 \mathrm{AB}\)
2 \(2 \mathrm{BA}\)
3 \(\mathrm{A}+\mathrm{B}\)
4 \(\mathrm{AB}\)
Explanation:
(C) : Given that, \(\mathrm{AB}=\mathrm{B}\) Multiply by A on both sides we get - \(\mathrm{ABA}=\mathrm{BA} \tag{i}\) And, \(\quad \mathrm{BA}=\mathrm{A}\) Multiply by \(\mathrm{B}\) on both sides we get - \(\mathrm{BAB}=\mathrm{AB} \tag{ii}\) Adding (i) and (ii), we get- \(\mathrm{ABA}+\mathrm{BAB}=\mathrm{BA}+\mathrm{AB}\) \(\mathrm{A}(\mathrm{BA})+\mathrm{B}(\mathrm{AB})=\mathrm{BA}+\mathrm{AB}\) \(\mathrm{AA}+\mathrm{BB}=\mathrm{A}+\mathrm{B}\) \(\mathrm{A}^{2}+\mathrm{B}^{2}=\mathrm{A}+\mathrm{B}\)
COMEDK-2014
Matrix and Determinant
78820
If \(A, B, C\) are invertible matrices, then \((\mathrm{ABC})^{-1}\) equal to
(D) : According to question, A, B, C are invertible matrix. We know that, \((\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}\) Now, \((\mathrm{ABC})^{-1} =[\mathrm{A}(\mathrm{BC})]^{-1}\) \(=(\mathrm{BC})^{-1} \mathrm{~A}^{-1}\) \(=\mathrm{C}^{-1} \mathrm{~B}^{-1} \mathrm{~A}^{-1}\)
SRM JEE-2009
Matrix and Determinant
78821
If \(\left[\begin{array}{rr}\cos \frac{2 \pi}{5} & -\sin \frac{2 \pi}{5} \\ \sin \frac{2 \pi}{5} & \cos \frac{2 \pi}{5}\end{array}\right]^{n}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\), then the least positive integer \(\mathbf{n}\) is equal to
