78871
If \(A\) is a matrix such \(\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] A\left[\begin{array}{ll}1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]\) then \(A\) is equal to
78873
If A is a zero square matrix of order \(n\) with \(\operatorname{det}(I+A) \neq 0\) and \(A^{3}=0\), where, \(I, O\) are unit and null matrices of order \(n \times n\) respectively, then \((I+A)^{-1}\) is equal to
1 \(\mathrm{I}-\mathrm{A}+\mathrm{A}^{2}\)
2 \(I+A+A^{2}\)
3 \(\mathrm{I}+\mathrm{A}^{-1}\)
4 \(\mathrm{I}+\mathrm{A}\)
Explanation:
(A) : Given, \(\operatorname{det}(\mathrm{I}+\mathrm{A}) \neq 0\) \(A^{3}=0\) where 0 is null matrix, \(I\) is the identity matrix. Here, \(\mathrm{A}^{3}+\mathrm{I}=\mathrm{I}\) [adding I on both sides] \((A+I)\left(A^{2}-I A+I^{2}\right)=I\left(\right.\) by formula of \(\left.a^{3}+b^{3}\right)\) \((\mathrm{A}+\mathrm{I})\left(\mathrm{A}^{2}-\mathrm{A}+\mathrm{I}\right)=\mathrm{I}\) \((\mathrm{I}+\mathrm{A})(\mathrm{I}+\mathrm{A})^{-2}=\mathrm{I}\) (by the rule of inverse matrix) Hence, \((\mathrm{I}+\mathrm{A})^{-1}=\left(\mathrm{A}^{2}-\mathrm{A}+\mathrm{I}\right)\)
AP EAMCET-2010
Matrix and Determinant
78874
If \(A\) is an invertible matrix of order \(n\), then the determinant of adj \(A\) is equal to:
1 \(|\mathrm{A}|^{\mathrm{n}}\)
2 \(|\mathrm{A}|^{\mathrm{n}+1}\)
3 \(|\mathrm{A}|^{\mathrm{n}-1}\)
4 \(|A|^{\mathrm{n}+2}\)
Explanation:
(C) : According to given summation, \(A(\operatorname{adj} A)=|A|(I)\) \(|A(\operatorname{adj} A)|=|(|A| I)|\) \(\left.|A| \operatorname{adj} A|=| A\right|^{n} \times|I|\) \(|A||\operatorname{arj} A|=|A|^{n}\) Case-I If \(|\mathrm{A}| \neq 0\) Then we get, \(|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}\) Case-II If \(|\mathrm{A}|=0\) Then, \(|\operatorname{adj} \mathrm{A}|=0\) And, we again get \(|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}\)
78871
If \(A\) is a matrix such \(\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] A\left[\begin{array}{ll}1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]\) then \(A\) is equal to
78873
If A is a zero square matrix of order \(n\) with \(\operatorname{det}(I+A) \neq 0\) and \(A^{3}=0\), where, \(I, O\) are unit and null matrices of order \(n \times n\) respectively, then \((I+A)^{-1}\) is equal to
1 \(\mathrm{I}-\mathrm{A}+\mathrm{A}^{2}\)
2 \(I+A+A^{2}\)
3 \(\mathrm{I}+\mathrm{A}^{-1}\)
4 \(\mathrm{I}+\mathrm{A}\)
Explanation:
(A) : Given, \(\operatorname{det}(\mathrm{I}+\mathrm{A}) \neq 0\) \(A^{3}=0\) where 0 is null matrix, \(I\) is the identity matrix. Here, \(\mathrm{A}^{3}+\mathrm{I}=\mathrm{I}\) [adding I on both sides] \((A+I)\left(A^{2}-I A+I^{2}\right)=I\left(\right.\) by formula of \(\left.a^{3}+b^{3}\right)\) \((\mathrm{A}+\mathrm{I})\left(\mathrm{A}^{2}-\mathrm{A}+\mathrm{I}\right)=\mathrm{I}\) \((\mathrm{I}+\mathrm{A})(\mathrm{I}+\mathrm{A})^{-2}=\mathrm{I}\) (by the rule of inverse matrix) Hence, \((\mathrm{I}+\mathrm{A})^{-1}=\left(\mathrm{A}^{2}-\mathrm{A}+\mathrm{I}\right)\)
AP EAMCET-2010
Matrix and Determinant
78874
If \(A\) is an invertible matrix of order \(n\), then the determinant of adj \(A\) is equal to:
1 \(|\mathrm{A}|^{\mathrm{n}}\)
2 \(|\mathrm{A}|^{\mathrm{n}+1}\)
3 \(|\mathrm{A}|^{\mathrm{n}-1}\)
4 \(|A|^{\mathrm{n}+2}\)
Explanation:
(C) : According to given summation, \(A(\operatorname{adj} A)=|A|(I)\) \(|A(\operatorname{adj} A)|=|(|A| I)|\) \(\left.|A| \operatorname{adj} A|=| A\right|^{n} \times|I|\) \(|A||\operatorname{arj} A|=|A|^{n}\) Case-I If \(|\mathrm{A}| \neq 0\) Then we get, \(|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}\) Case-II If \(|\mathrm{A}|=0\) Then, \(|\operatorname{adj} \mathrm{A}|=0\) And, we again get \(|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}\)
78871
If \(A\) is a matrix such \(\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] A\left[\begin{array}{ll}1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]\) then \(A\) is equal to
78873
If A is a zero square matrix of order \(n\) with \(\operatorname{det}(I+A) \neq 0\) and \(A^{3}=0\), where, \(I, O\) are unit and null matrices of order \(n \times n\) respectively, then \((I+A)^{-1}\) is equal to
1 \(\mathrm{I}-\mathrm{A}+\mathrm{A}^{2}\)
2 \(I+A+A^{2}\)
3 \(\mathrm{I}+\mathrm{A}^{-1}\)
4 \(\mathrm{I}+\mathrm{A}\)
Explanation:
(A) : Given, \(\operatorname{det}(\mathrm{I}+\mathrm{A}) \neq 0\) \(A^{3}=0\) where 0 is null matrix, \(I\) is the identity matrix. Here, \(\mathrm{A}^{3}+\mathrm{I}=\mathrm{I}\) [adding I on both sides] \((A+I)\left(A^{2}-I A+I^{2}\right)=I\left(\right.\) by formula of \(\left.a^{3}+b^{3}\right)\) \((\mathrm{A}+\mathrm{I})\left(\mathrm{A}^{2}-\mathrm{A}+\mathrm{I}\right)=\mathrm{I}\) \((\mathrm{I}+\mathrm{A})(\mathrm{I}+\mathrm{A})^{-2}=\mathrm{I}\) (by the rule of inverse matrix) Hence, \((\mathrm{I}+\mathrm{A})^{-1}=\left(\mathrm{A}^{2}-\mathrm{A}+\mathrm{I}\right)\)
AP EAMCET-2010
Matrix and Determinant
78874
If \(A\) is an invertible matrix of order \(n\), then the determinant of adj \(A\) is equal to:
1 \(|\mathrm{A}|^{\mathrm{n}}\)
2 \(|\mathrm{A}|^{\mathrm{n}+1}\)
3 \(|\mathrm{A}|^{\mathrm{n}-1}\)
4 \(|A|^{\mathrm{n}+2}\)
Explanation:
(C) : According to given summation, \(A(\operatorname{adj} A)=|A|(I)\) \(|A(\operatorname{adj} A)|=|(|A| I)|\) \(\left.|A| \operatorname{adj} A|=| A\right|^{n} \times|I|\) \(|A||\operatorname{arj} A|=|A|^{n}\) Case-I If \(|\mathrm{A}| \neq 0\) Then we get, \(|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}\) Case-II If \(|\mathrm{A}|=0\) Then, \(|\operatorname{adj} \mathrm{A}|=0\) And, we again get \(|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}\)
78871
If \(A\) is a matrix such \(\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] A\left[\begin{array}{ll}1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]\) then \(A\) is equal to
78873
If A is a zero square matrix of order \(n\) with \(\operatorname{det}(I+A) \neq 0\) and \(A^{3}=0\), where, \(I, O\) are unit and null matrices of order \(n \times n\) respectively, then \((I+A)^{-1}\) is equal to
1 \(\mathrm{I}-\mathrm{A}+\mathrm{A}^{2}\)
2 \(I+A+A^{2}\)
3 \(\mathrm{I}+\mathrm{A}^{-1}\)
4 \(\mathrm{I}+\mathrm{A}\)
Explanation:
(A) : Given, \(\operatorname{det}(\mathrm{I}+\mathrm{A}) \neq 0\) \(A^{3}=0\) where 0 is null matrix, \(I\) is the identity matrix. Here, \(\mathrm{A}^{3}+\mathrm{I}=\mathrm{I}\) [adding I on both sides] \((A+I)\left(A^{2}-I A+I^{2}\right)=I\left(\right.\) by formula of \(\left.a^{3}+b^{3}\right)\) \((\mathrm{A}+\mathrm{I})\left(\mathrm{A}^{2}-\mathrm{A}+\mathrm{I}\right)=\mathrm{I}\) \((\mathrm{I}+\mathrm{A})(\mathrm{I}+\mathrm{A})^{-2}=\mathrm{I}\) (by the rule of inverse matrix) Hence, \((\mathrm{I}+\mathrm{A})^{-1}=\left(\mathrm{A}^{2}-\mathrm{A}+\mathrm{I}\right)\)
AP EAMCET-2010
Matrix and Determinant
78874
If \(A\) is an invertible matrix of order \(n\), then the determinant of adj \(A\) is equal to:
1 \(|\mathrm{A}|^{\mathrm{n}}\)
2 \(|\mathrm{A}|^{\mathrm{n}+1}\)
3 \(|\mathrm{A}|^{\mathrm{n}-1}\)
4 \(|A|^{\mathrm{n}+2}\)
Explanation:
(C) : According to given summation, \(A(\operatorname{adj} A)=|A|(I)\) \(|A(\operatorname{adj} A)|=|(|A| I)|\) \(\left.|A| \operatorname{adj} A|=| A\right|^{n} \times|I|\) \(|A||\operatorname{arj} A|=|A|^{n}\) Case-I If \(|\mathrm{A}| \neq 0\) Then we get, \(|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}\) Case-II If \(|\mathrm{A}|=0\) Then, \(|\operatorname{adj} \mathrm{A}|=0\) And, we again get \(|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}\)
