78959
If the determinant of the adjoint of a (real) matrix of order 3 is 25 , then the determinant of the inverse of the matrix is
1 0.2
2 \(\pm 5\)
3 \(\frac{1}{\sqrt[5]{625}}\)
4 \(\pm 0.2\)
Explanation:
(D) : According to given summation, \(|\operatorname{adj} \mathrm{A}|=25\) and \(\mathrm{n}=3\) As we know that, \(|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}\) \(25=|\mathrm{A}|^{2}\) \(|\mathrm{~A}|= \pm 5\) Now, \(\quad\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}= \pm \frac{1}{5}= \pm 0.2\)
Karnataka CET-2013
Matrix and Determinant
78960
The characteristic equation of a matrix \(A\) is \(\lambda^{3}-5 \lambda^{2}-3 \lambda+2 I=0\), then \(|\operatorname{adj}(\mathbf{A})|=\)
1 4
2 9
3 25
4 \(\frac{1}{2}\)
Explanation:
(A) : Here, we have the characteristic equation of a matrix A is, \(\lambda^{3}-5 \lambda^{2}-3 \lambda+2 \mathrm{I}=0\) \(\because\) Characteristic equation, \(=(\mathrm{P} \lambda)=|\mathrm{A}-\lambda \mathrm{I}|=\lambda^{3}-5 \lambda^{2}-3 \lambda+2 \mathrm{I}\) So, \(\quad|\mathrm{A}|=\mathrm{P}(0)=2\) And, \(\quad A^{-1}=\frac{1}{|A|}\) adj A \(\left|A^{-1}\right|=\left|\frac{1}{|A|} \operatorname{adj} A\right|=\left[\frac{1}{|A|}\right]^{3}|\operatorname{adj} A|\) \(\because \quad\left|\mathrm{A}^{-1}\right|=|\mathrm{A}|^{-1}\) \(|\mathrm{A}|^{2}=|\operatorname{adj} \mathrm{A}|=4\)
Karnataka CET-2012
Matrix and Determinant
78962
If \(A\) is a \(3 \times 3\) nonsingular matrix and if \(|A|=3\), then \(\left|(\mathbf{2 A})^{-1}\right|=\)
1 3
2 24
3 \(\frac{1}{24}\)
4 \(\frac{1}{3}\)
Explanation:
(C) : It is given that, \(|A|=3 \text { and order of matrix } A=3 \times 3\) Then, \(\quad\left|(2 \mathrm{~A})^{-1}\right|=\frac{1}{|2 \mathrm{~A}|}=\frac{1}{2^{3}|\mathrm{~A}|}=\frac{1}{8 \times 3}=\frac{1}{24}\) \(\left(\because|\mathrm{aA}|=\mathrm{a}^{3}|\mathrm{~A}|\right)\)
78959
If the determinant of the adjoint of a (real) matrix of order 3 is 25 , then the determinant of the inverse of the matrix is
1 0.2
2 \(\pm 5\)
3 \(\frac{1}{\sqrt[5]{625}}\)
4 \(\pm 0.2\)
Explanation:
(D) : According to given summation, \(|\operatorname{adj} \mathrm{A}|=25\) and \(\mathrm{n}=3\) As we know that, \(|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}\) \(25=|\mathrm{A}|^{2}\) \(|\mathrm{~A}|= \pm 5\) Now, \(\quad\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}= \pm \frac{1}{5}= \pm 0.2\)
Karnataka CET-2013
Matrix and Determinant
78960
The characteristic equation of a matrix \(A\) is \(\lambda^{3}-5 \lambda^{2}-3 \lambda+2 I=0\), then \(|\operatorname{adj}(\mathbf{A})|=\)
1 4
2 9
3 25
4 \(\frac{1}{2}\)
Explanation:
(A) : Here, we have the characteristic equation of a matrix A is, \(\lambda^{3}-5 \lambda^{2}-3 \lambda+2 \mathrm{I}=0\) \(\because\) Characteristic equation, \(=(\mathrm{P} \lambda)=|\mathrm{A}-\lambda \mathrm{I}|=\lambda^{3}-5 \lambda^{2}-3 \lambda+2 \mathrm{I}\) So, \(\quad|\mathrm{A}|=\mathrm{P}(0)=2\) And, \(\quad A^{-1}=\frac{1}{|A|}\) adj A \(\left|A^{-1}\right|=\left|\frac{1}{|A|} \operatorname{adj} A\right|=\left[\frac{1}{|A|}\right]^{3}|\operatorname{adj} A|\) \(\because \quad\left|\mathrm{A}^{-1}\right|=|\mathrm{A}|^{-1}\) \(|\mathrm{A}|^{2}=|\operatorname{adj} \mathrm{A}|=4\)
Karnataka CET-2012
Matrix and Determinant
78962
If \(A\) is a \(3 \times 3\) nonsingular matrix and if \(|A|=3\), then \(\left|(\mathbf{2 A})^{-1}\right|=\)
1 3
2 24
3 \(\frac{1}{24}\)
4 \(\frac{1}{3}\)
Explanation:
(C) : It is given that, \(|A|=3 \text { and order of matrix } A=3 \times 3\) Then, \(\quad\left|(2 \mathrm{~A})^{-1}\right|=\frac{1}{|2 \mathrm{~A}|}=\frac{1}{2^{3}|\mathrm{~A}|}=\frac{1}{8 \times 3}=\frac{1}{24}\) \(\left(\because|\mathrm{aA}|=\mathrm{a}^{3}|\mathrm{~A}|\right)\)
78959
If the determinant of the adjoint of a (real) matrix of order 3 is 25 , then the determinant of the inverse of the matrix is
1 0.2
2 \(\pm 5\)
3 \(\frac{1}{\sqrt[5]{625}}\)
4 \(\pm 0.2\)
Explanation:
(D) : According to given summation, \(|\operatorname{adj} \mathrm{A}|=25\) and \(\mathrm{n}=3\) As we know that, \(|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}\) \(25=|\mathrm{A}|^{2}\) \(|\mathrm{~A}|= \pm 5\) Now, \(\quad\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}= \pm \frac{1}{5}= \pm 0.2\)
Karnataka CET-2013
Matrix and Determinant
78960
The characteristic equation of a matrix \(A\) is \(\lambda^{3}-5 \lambda^{2}-3 \lambda+2 I=0\), then \(|\operatorname{adj}(\mathbf{A})|=\)
1 4
2 9
3 25
4 \(\frac{1}{2}\)
Explanation:
(A) : Here, we have the characteristic equation of a matrix A is, \(\lambda^{3}-5 \lambda^{2}-3 \lambda+2 \mathrm{I}=0\) \(\because\) Characteristic equation, \(=(\mathrm{P} \lambda)=|\mathrm{A}-\lambda \mathrm{I}|=\lambda^{3}-5 \lambda^{2}-3 \lambda+2 \mathrm{I}\) So, \(\quad|\mathrm{A}|=\mathrm{P}(0)=2\) And, \(\quad A^{-1}=\frac{1}{|A|}\) adj A \(\left|A^{-1}\right|=\left|\frac{1}{|A|} \operatorname{adj} A\right|=\left[\frac{1}{|A|}\right]^{3}|\operatorname{adj} A|\) \(\because \quad\left|\mathrm{A}^{-1}\right|=|\mathrm{A}|^{-1}\) \(|\mathrm{A}|^{2}=|\operatorname{adj} \mathrm{A}|=4\)
Karnataka CET-2012
Matrix and Determinant
78962
If \(A\) is a \(3 \times 3\) nonsingular matrix and if \(|A|=3\), then \(\left|(\mathbf{2 A})^{-1}\right|=\)
1 3
2 24
3 \(\frac{1}{24}\)
4 \(\frac{1}{3}\)
Explanation:
(C) : It is given that, \(|A|=3 \text { and order of matrix } A=3 \times 3\) Then, \(\quad\left|(2 \mathrm{~A})^{-1}\right|=\frac{1}{|2 \mathrm{~A}|}=\frac{1}{2^{3}|\mathrm{~A}|}=\frac{1}{8 \times 3}=\frac{1}{24}\) \(\left(\because|\mathrm{aA}|=\mathrm{a}^{3}|\mathrm{~A}|\right)\)
78959
If the determinant of the adjoint of a (real) matrix of order 3 is 25 , then the determinant of the inverse of the matrix is
1 0.2
2 \(\pm 5\)
3 \(\frac{1}{\sqrt[5]{625}}\)
4 \(\pm 0.2\)
Explanation:
(D) : According to given summation, \(|\operatorname{adj} \mathrm{A}|=25\) and \(\mathrm{n}=3\) As we know that, \(|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}\) \(25=|\mathrm{A}|^{2}\) \(|\mathrm{~A}|= \pm 5\) Now, \(\quad\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}= \pm \frac{1}{5}= \pm 0.2\)
Karnataka CET-2013
Matrix and Determinant
78960
The characteristic equation of a matrix \(A\) is \(\lambda^{3}-5 \lambda^{2}-3 \lambda+2 I=0\), then \(|\operatorname{adj}(\mathbf{A})|=\)
1 4
2 9
3 25
4 \(\frac{1}{2}\)
Explanation:
(A) : Here, we have the characteristic equation of a matrix A is, \(\lambda^{3}-5 \lambda^{2}-3 \lambda+2 \mathrm{I}=0\) \(\because\) Characteristic equation, \(=(\mathrm{P} \lambda)=|\mathrm{A}-\lambda \mathrm{I}|=\lambda^{3}-5 \lambda^{2}-3 \lambda+2 \mathrm{I}\) So, \(\quad|\mathrm{A}|=\mathrm{P}(0)=2\) And, \(\quad A^{-1}=\frac{1}{|A|}\) adj A \(\left|A^{-1}\right|=\left|\frac{1}{|A|} \operatorname{adj} A\right|=\left[\frac{1}{|A|}\right]^{3}|\operatorname{adj} A|\) \(\because \quad\left|\mathrm{A}^{-1}\right|=|\mathrm{A}|^{-1}\) \(|\mathrm{A}|^{2}=|\operatorname{adj} \mathrm{A}|=4\)
Karnataka CET-2012
Matrix and Determinant
78962
If \(A\) is a \(3 \times 3\) nonsingular matrix and if \(|A|=3\), then \(\left|(\mathbf{2 A})^{-1}\right|=\)
1 3
2 24
3 \(\frac{1}{24}\)
4 \(\frac{1}{3}\)
Explanation:
(C) : It is given that, \(|A|=3 \text { and order of matrix } A=3 \times 3\) Then, \(\quad\left|(2 \mathrm{~A})^{-1}\right|=\frac{1}{|2 \mathrm{~A}|}=\frac{1}{2^{3}|\mathrm{~A}|}=\frac{1}{8 \times 3}=\frac{1}{24}\) \(\left(\because|\mathrm{aA}|=\mathrm{a}^{3}|\mathrm{~A}|\right)\)