78783
If \(A=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]\) and \(A^{-1}=x A+y I\), where \(I\) is unit matrix of order 2 , then the values of \(x\) and \(y\) are respectively.
78784
The element in the third row and first column of the inverse of the matrix \(\left[\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right]\) is
1 4
2 2
3 3
4 -3
Explanation:
(D) : Given that, \(A=\left[\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right]\) \(|A|=\left|\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right|\) \(|A|=-1(0-1)+3(0+2)+2(3-6)\) \(|A|=1+6-6=1\) We have to find, \(a_{31}\) in \(A^{-1}\). So we will find \(\mathrm{A}_{13}\) in \(\mathrm{A}\). Co-factor of \(2=(-1)^{1+3}\left|\begin{array}{cc}-3 & 3 \\ 2 & -1\end{array}\right|=3-6=-3\) Hence, \(a_{31}\) in \(A^{-1}=\frac{-3}{|A|}=\frac{-3}{1}=-3\)
MHT CET-2020
Matrix and Determinant
78785
If \(A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]\) and \(A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2\end{array}\right]\), then the values of \(\alpha\) and \(\boldsymbol{\beta}\) are, respectively
1 15,5
2 \(-15,5\)
3 \(15,-5\)
4 \(-15,-5\)
Explanation:
(B) : We know that, A.A \(A^{-1}=I\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) \(\left[\begin{array}{ccc}6+0-\beta & -2+0+2 & 2+0-2 \\ 15+\alpha+0 & -5+6+0 & 5-5+0 \\ 0+\alpha+3 \beta & 0+6-6 & 0-5+6\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) \(\left[\begin{array}{ccc}6-\beta & 0 & 0 \\ 15+\alpha & 1 & 0 \\ \alpha+3 \beta & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) On comparing corresponding elements on both side, we get - OR \(6-\beta=1 \Rightarrow \beta=5 \text { and } 15+\alpha=0 \Rightarrow \alpha=-15\) We have, \(A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right] \Rightarrow|A|=2(3)-(5)=1\) \(\alpha\) is \(a_{21}\) in \(A^{-1}\). So, we will find cofactor of \(a_{12} x\) in \(A\). \(\therefore \alpha=\frac{(-1)^{1+2}\left|\begin{array}{ll}5 & 0 \\ 0 & 3\end{array}\right|}{|\mathrm{A}|} \Rightarrow \alpha=-15\) Similarly, \(\beta=\frac{(-1)^{1+3}\left|\begin{array}{ll}5 & 1 \\ 0 & 1\end{array}\right|}{|\mathrm{A}|} \Rightarrow \beta=5\) The value of \(\alpha\) and \(\beta\) are respectively \((-15,5)\).
MHT CET-2020
Matrix and Determinant
78786
If \(A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]\) then \(B^{-1} A^{-1}=\)
78783
If \(A=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]\) and \(A^{-1}=x A+y I\), where \(I\) is unit matrix of order 2 , then the values of \(x\) and \(y\) are respectively.
78784
The element in the third row and first column of the inverse of the matrix \(\left[\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right]\) is
1 4
2 2
3 3
4 -3
Explanation:
(D) : Given that, \(A=\left[\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right]\) \(|A|=\left|\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right|\) \(|A|=-1(0-1)+3(0+2)+2(3-6)\) \(|A|=1+6-6=1\) We have to find, \(a_{31}\) in \(A^{-1}\). So we will find \(\mathrm{A}_{13}\) in \(\mathrm{A}\). Co-factor of \(2=(-1)^{1+3}\left|\begin{array}{cc}-3 & 3 \\ 2 & -1\end{array}\right|=3-6=-3\) Hence, \(a_{31}\) in \(A^{-1}=\frac{-3}{|A|}=\frac{-3}{1}=-3\)
MHT CET-2020
Matrix and Determinant
78785
If \(A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]\) and \(A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2\end{array}\right]\), then the values of \(\alpha\) and \(\boldsymbol{\beta}\) are, respectively
1 15,5
2 \(-15,5\)
3 \(15,-5\)
4 \(-15,-5\)
Explanation:
(B) : We know that, A.A \(A^{-1}=I\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) \(\left[\begin{array}{ccc}6+0-\beta & -2+0+2 & 2+0-2 \\ 15+\alpha+0 & -5+6+0 & 5-5+0 \\ 0+\alpha+3 \beta & 0+6-6 & 0-5+6\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) \(\left[\begin{array}{ccc}6-\beta & 0 & 0 \\ 15+\alpha & 1 & 0 \\ \alpha+3 \beta & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) On comparing corresponding elements on both side, we get - OR \(6-\beta=1 \Rightarrow \beta=5 \text { and } 15+\alpha=0 \Rightarrow \alpha=-15\) We have, \(A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right] \Rightarrow|A|=2(3)-(5)=1\) \(\alpha\) is \(a_{21}\) in \(A^{-1}\). So, we will find cofactor of \(a_{12} x\) in \(A\). \(\therefore \alpha=\frac{(-1)^{1+2}\left|\begin{array}{ll}5 & 0 \\ 0 & 3\end{array}\right|}{|\mathrm{A}|} \Rightarrow \alpha=-15\) Similarly, \(\beta=\frac{(-1)^{1+3}\left|\begin{array}{ll}5 & 1 \\ 0 & 1\end{array}\right|}{|\mathrm{A}|} \Rightarrow \beta=5\) The value of \(\alpha\) and \(\beta\) are respectively \((-15,5)\).
MHT CET-2020
Matrix and Determinant
78786
If \(A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]\) then \(B^{-1} A^{-1}=\)
78783
If \(A=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]\) and \(A^{-1}=x A+y I\), where \(I\) is unit matrix of order 2 , then the values of \(x\) and \(y\) are respectively.
78784
The element in the third row and first column of the inverse of the matrix \(\left[\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right]\) is
1 4
2 2
3 3
4 -3
Explanation:
(D) : Given that, \(A=\left[\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right]\) \(|A|=\left|\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right|\) \(|A|=-1(0-1)+3(0+2)+2(3-6)\) \(|A|=1+6-6=1\) We have to find, \(a_{31}\) in \(A^{-1}\). So we will find \(\mathrm{A}_{13}\) in \(\mathrm{A}\). Co-factor of \(2=(-1)^{1+3}\left|\begin{array}{cc}-3 & 3 \\ 2 & -1\end{array}\right|=3-6=-3\) Hence, \(a_{31}\) in \(A^{-1}=\frac{-3}{|A|}=\frac{-3}{1}=-3\)
MHT CET-2020
Matrix and Determinant
78785
If \(A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]\) and \(A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2\end{array}\right]\), then the values of \(\alpha\) and \(\boldsymbol{\beta}\) are, respectively
1 15,5
2 \(-15,5\)
3 \(15,-5\)
4 \(-15,-5\)
Explanation:
(B) : We know that, A.A \(A^{-1}=I\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) \(\left[\begin{array}{ccc}6+0-\beta & -2+0+2 & 2+0-2 \\ 15+\alpha+0 & -5+6+0 & 5-5+0 \\ 0+\alpha+3 \beta & 0+6-6 & 0-5+6\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) \(\left[\begin{array}{ccc}6-\beta & 0 & 0 \\ 15+\alpha & 1 & 0 \\ \alpha+3 \beta & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) On comparing corresponding elements on both side, we get - OR \(6-\beta=1 \Rightarrow \beta=5 \text { and } 15+\alpha=0 \Rightarrow \alpha=-15\) We have, \(A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right] \Rightarrow|A|=2(3)-(5)=1\) \(\alpha\) is \(a_{21}\) in \(A^{-1}\). So, we will find cofactor of \(a_{12} x\) in \(A\). \(\therefore \alpha=\frac{(-1)^{1+2}\left|\begin{array}{ll}5 & 0 \\ 0 & 3\end{array}\right|}{|\mathrm{A}|} \Rightarrow \alpha=-15\) Similarly, \(\beta=\frac{(-1)^{1+3}\left|\begin{array}{ll}5 & 1 \\ 0 & 1\end{array}\right|}{|\mathrm{A}|} \Rightarrow \beta=5\) The value of \(\alpha\) and \(\beta\) are respectively \((-15,5)\).
MHT CET-2020
Matrix and Determinant
78786
If \(A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]\) then \(B^{-1} A^{-1}=\)
78783
If \(A=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]\) and \(A^{-1}=x A+y I\), where \(I\) is unit matrix of order 2 , then the values of \(x\) and \(y\) are respectively.
78784
The element in the third row and first column of the inverse of the matrix \(\left[\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right]\) is
1 4
2 2
3 3
4 -3
Explanation:
(D) : Given that, \(A=\left[\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right]\) \(|A|=\left|\begin{array}{ccc}-1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0\end{array}\right|\) \(|A|=-1(0-1)+3(0+2)+2(3-6)\) \(|A|=1+6-6=1\) We have to find, \(a_{31}\) in \(A^{-1}\). So we will find \(\mathrm{A}_{13}\) in \(\mathrm{A}\). Co-factor of \(2=(-1)^{1+3}\left|\begin{array}{cc}-3 & 3 \\ 2 & -1\end{array}\right|=3-6=-3\) Hence, \(a_{31}\) in \(A^{-1}=\frac{-3}{|A|}=\frac{-3}{1}=-3\)
MHT CET-2020
Matrix and Determinant
78785
If \(A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]\) and \(A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2\end{array}\right]\), then the values of \(\alpha\) and \(\boldsymbol{\beta}\) are, respectively
1 15,5
2 \(-15,5\)
3 \(15,-5\)
4 \(-15,-5\)
Explanation:
(B) : We know that, A.A \(A^{-1}=I\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) \(\left[\begin{array}{ccc}6+0-\beta & -2+0+2 & 2+0-2 \\ 15+\alpha+0 & -5+6+0 & 5-5+0 \\ 0+\alpha+3 \beta & 0+6-6 & 0-5+6\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) \(\left[\begin{array}{ccc}6-\beta & 0 & 0 \\ 15+\alpha & 1 & 0 \\ \alpha+3 \beta & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\) On comparing corresponding elements on both side, we get - OR \(6-\beta=1 \Rightarrow \beta=5 \text { and } 15+\alpha=0 \Rightarrow \alpha=-15\) We have, \(A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right] \Rightarrow|A|=2(3)-(5)=1\) \(\alpha\) is \(a_{21}\) in \(A^{-1}\). So, we will find cofactor of \(a_{12} x\) in \(A\). \(\therefore \alpha=\frac{(-1)^{1+2}\left|\begin{array}{ll}5 & 0 \\ 0 & 3\end{array}\right|}{|\mathrm{A}|} \Rightarrow \alpha=-15\) Similarly, \(\beta=\frac{(-1)^{1+3}\left|\begin{array}{ll}5 & 1 \\ 0 & 1\end{array}\right|}{|\mathrm{A}|} \Rightarrow \beta=5\) The value of \(\alpha\) and \(\beta\) are respectively \((-15,5)\).
MHT CET-2020
Matrix and Determinant
78786
If \(A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]\) then \(B^{-1} A^{-1}=\)
