78807
If \(A\) is a non-singular matrix and \(\mathbf{A}^{2}-\mathbf{A}+\mathbf{I}=\) 0, then \(\mathbf{A}^{-1}=\)
1 \(\mathrm{A}\)
2 I - A
3 \(\mathrm{A}-\mathrm{I}\)
4 \(\mathrm{A}+\mathrm{I}\)
Explanation:
(B) : Given that, \(\mathrm{A}^{2}-\mathrm{A}+\mathrm{I}=0\) On multiplying by \(\mathrm{A}^{-1}\) we get - \(\left(A^{2}-A+I\right) A^{-1}=0\) \(\mathrm{A}-\mathrm{AA}^{-1}+\mathrm{A}^{-1}=0\) \(\mathrm{A}-\mathrm{I}+\mathrm{A}^{-1}=0\) \(\mathrm{A}^{-1}=\mathrm{I}-\mathrm{A}\)
MHT CET-2019
Matrix and Determinant
78808
The inverse of the matrix \(\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]\) is
78809
For a invertible matrix \(A\) if \(A(\operatorname{adj} A)=\) \(\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\), then \(|\mathbf{A}|=\)
1 100
2 -100
3 10
4 -10
Explanation:
(C) : Given that, \(A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\) \(A(\operatorname{adj} A)=10\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=10 I\) We know that, \(\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}\) \(A(\operatorname{adj} A)=|A| I \Rightarrow|A|=10\)
MHT CET-2017
Matrix and Determinant
78810
If the inverse of the matrix \(\left[\begin{array}{ccc}\alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3\end{array}\right]\) does not exist then the value of \(\alpha\) is
1 1
2 -1
3 0
4 -2
Explanation:
(D) : Let, \(A=\left[\begin{array}{ccc} \alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3 \end{array}\right]\) Now, \(|A|=7 \alpha+14\) According to question, \(\mathrm{A}^{-1}\) does not exists it means - \(|\mathrm{A}|=0\) \(7 \alpha+14=0\) \(7 \alpha=-14\) \(\alpha=-2\)
78807
If \(A\) is a non-singular matrix and \(\mathbf{A}^{2}-\mathbf{A}+\mathbf{I}=\) 0, then \(\mathbf{A}^{-1}=\)
1 \(\mathrm{A}\)
2 I - A
3 \(\mathrm{A}-\mathrm{I}\)
4 \(\mathrm{A}+\mathrm{I}\)
Explanation:
(B) : Given that, \(\mathrm{A}^{2}-\mathrm{A}+\mathrm{I}=0\) On multiplying by \(\mathrm{A}^{-1}\) we get - \(\left(A^{2}-A+I\right) A^{-1}=0\) \(\mathrm{A}-\mathrm{AA}^{-1}+\mathrm{A}^{-1}=0\) \(\mathrm{A}-\mathrm{I}+\mathrm{A}^{-1}=0\) \(\mathrm{A}^{-1}=\mathrm{I}-\mathrm{A}\)
MHT CET-2019
Matrix and Determinant
78808
The inverse of the matrix \(\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]\) is
78809
For a invertible matrix \(A\) if \(A(\operatorname{adj} A)=\) \(\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\), then \(|\mathbf{A}|=\)
1 100
2 -100
3 10
4 -10
Explanation:
(C) : Given that, \(A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\) \(A(\operatorname{adj} A)=10\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=10 I\) We know that, \(\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}\) \(A(\operatorname{adj} A)=|A| I \Rightarrow|A|=10\)
MHT CET-2017
Matrix and Determinant
78810
If the inverse of the matrix \(\left[\begin{array}{ccc}\alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3\end{array}\right]\) does not exist then the value of \(\alpha\) is
1 1
2 -1
3 0
4 -2
Explanation:
(D) : Let, \(A=\left[\begin{array}{ccc} \alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3 \end{array}\right]\) Now, \(|A|=7 \alpha+14\) According to question, \(\mathrm{A}^{-1}\) does not exists it means - \(|\mathrm{A}|=0\) \(7 \alpha+14=0\) \(7 \alpha=-14\) \(\alpha=-2\)
78807
If \(A\) is a non-singular matrix and \(\mathbf{A}^{2}-\mathbf{A}+\mathbf{I}=\) 0, then \(\mathbf{A}^{-1}=\)
1 \(\mathrm{A}\)
2 I - A
3 \(\mathrm{A}-\mathrm{I}\)
4 \(\mathrm{A}+\mathrm{I}\)
Explanation:
(B) : Given that, \(\mathrm{A}^{2}-\mathrm{A}+\mathrm{I}=0\) On multiplying by \(\mathrm{A}^{-1}\) we get - \(\left(A^{2}-A+I\right) A^{-1}=0\) \(\mathrm{A}-\mathrm{AA}^{-1}+\mathrm{A}^{-1}=0\) \(\mathrm{A}-\mathrm{I}+\mathrm{A}^{-1}=0\) \(\mathrm{A}^{-1}=\mathrm{I}-\mathrm{A}\)
MHT CET-2019
Matrix and Determinant
78808
The inverse of the matrix \(\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]\) is
78809
For a invertible matrix \(A\) if \(A(\operatorname{adj} A)=\) \(\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\), then \(|\mathbf{A}|=\)
1 100
2 -100
3 10
4 -10
Explanation:
(C) : Given that, \(A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\) \(A(\operatorname{adj} A)=10\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=10 I\) We know that, \(\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}\) \(A(\operatorname{adj} A)=|A| I \Rightarrow|A|=10\)
MHT CET-2017
Matrix and Determinant
78810
If the inverse of the matrix \(\left[\begin{array}{ccc}\alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3\end{array}\right]\) does not exist then the value of \(\alpha\) is
1 1
2 -1
3 0
4 -2
Explanation:
(D) : Let, \(A=\left[\begin{array}{ccc} \alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3 \end{array}\right]\) Now, \(|A|=7 \alpha+14\) According to question, \(\mathrm{A}^{-1}\) does not exists it means - \(|\mathrm{A}|=0\) \(7 \alpha+14=0\) \(7 \alpha=-14\) \(\alpha=-2\)
78807
If \(A\) is a non-singular matrix and \(\mathbf{A}^{2}-\mathbf{A}+\mathbf{I}=\) 0, then \(\mathbf{A}^{-1}=\)
1 \(\mathrm{A}\)
2 I - A
3 \(\mathrm{A}-\mathrm{I}\)
4 \(\mathrm{A}+\mathrm{I}\)
Explanation:
(B) : Given that, \(\mathrm{A}^{2}-\mathrm{A}+\mathrm{I}=0\) On multiplying by \(\mathrm{A}^{-1}\) we get - \(\left(A^{2}-A+I\right) A^{-1}=0\) \(\mathrm{A}-\mathrm{AA}^{-1}+\mathrm{A}^{-1}=0\) \(\mathrm{A}-\mathrm{I}+\mathrm{A}^{-1}=0\) \(\mathrm{A}^{-1}=\mathrm{I}-\mathrm{A}\)
MHT CET-2019
Matrix and Determinant
78808
The inverse of the matrix \(\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]\) is
78809
For a invertible matrix \(A\) if \(A(\operatorname{adj} A)=\) \(\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\), then \(|\mathbf{A}|=\)
1 100
2 -100
3 10
4 -10
Explanation:
(C) : Given that, \(A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\) \(A(\operatorname{adj} A)=10\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=10 I\) We know that, \(\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}\) \(A(\operatorname{adj} A)=|A| I \Rightarrow|A|=10\)
MHT CET-2017
Matrix and Determinant
78810
If the inverse of the matrix \(\left[\begin{array}{ccc}\alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3\end{array}\right]\) does not exist then the value of \(\alpha\) is
1 1
2 -1
3 0
4 -2
Explanation:
(D) : Let, \(A=\left[\begin{array}{ccc} \alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3 \end{array}\right]\) Now, \(|A|=7 \alpha+14\) According to question, \(\mathrm{A}^{-1}\) does not exists it means - \(|\mathrm{A}|=0\) \(7 \alpha+14=0\) \(7 \alpha=-14\) \(\alpha=-2\)
