78985
If \(A=\left|\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right|\), then the value of \(|A \| \operatorname{adj}(\mathbf{A})|\) is
1 \(a^{3}\)
2 \(a^{6}\)
3 \(a^{9}\)
4 \(\mathrm{a}^{27}\)
Explanation:
(C): We have, \(A=\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{a} & 0 \\ 0 & 0 & \mathrm{a} \end{array}\right]\) Determinant of \(\mathrm{A}\) is - \(|\mathrm{A}|=\mathrm{a}\left[\mathrm{a}^{2}-0\right]-0+0\) \(|\mathrm{~A}|=\mathrm{a}^{3}\) Adjoint of given matrix is given as - \(\operatorname{Adj} \mathrm{A}=\left[\begin{array}{ccc} \mathrm{a}^{2} & 0 & 0 \\ 0 & \mathrm{a}^{2} & 0 \\ 0 & 0 & \mathrm{a}^{2} \end{array}\right]\) Now, we find the determinant of Adjoint A, \(|\operatorname{Adj} A|=\mathrm{a}^{2}\left[\left(\mathrm{a}^{2}\right)\left(\mathrm{a}^{2}\right)-(0)(0)\right]-0\left[(0)\left(\mathrm{a}^{2}\right)-(0)(0)\right]\) \(+0[(0)\) \(|\operatorname{Adj} A| =\mathrm{a}^{2}\left[\mathrm{a}^{4}\right]-0+0\) \(|\operatorname{Adj} A| =\mathrm{a}^{6}\) Finally, \(|\mathrm{A}||\operatorname{Adj} \mathrm{A}|=\mathrm{a}^{3}\left(\mathrm{a}^{6}\right)\) \(|\mathrm{A}||\operatorname{Adj} \mathrm{A}|=\mathrm{a}^{9}\)
BCECE-2011
Matrix and Determinant
78986
If \(A=\left[\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right]\), then \(A^{-1}\) is equal to
(B): It is given that, \(A=\left[\begin{array}{cc} 1 & 3 \\ 2 & -2 \end{array}\right]\) Here, for matrix \(A=\left[\begin{array}{cc}m & n \\ o & p\end{array}\right]\) \(\operatorname{Adj} A=\left[\begin{array}{cc} p & -n \\ -o & m \end{array}\right]\) Hence, adj \(A=\left[\begin{array}{cc}-2 & -3 \\ -2 & 1\end{array}\right]^{\mathrm{T}}\) \(=\left[\begin{array}{cc} -2 & -2 \\ -3 & 1 \end{array}\right]\) And, \(\quad|A|=\left|\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right|=-2-6=-8\) \(\therefore \quad \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \cdot \operatorname{adj}(\mathrm{A})\) \(\mathrm{A}^{-1}=-\frac{1}{8}\left[\begin{array}{cc} -2 & -3 \\ -2 & 1 \end{array}\right]\)
BCECE-2010
Matrix and Determinant
78988
For any \(2 \times 2\) matrix \(A\), if \(A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\), then \(|A|\) i.e., \(\operatorname{det} A\) is equal to :
1 20
2 100
3 10
4 0
Explanation:
(C) : We have, \(A(\operatorname{adj} A)=\left[\begin{array}{cc} 10 & 0 \tag{i}\\ 0 & 10 \end{array}\right]\) We know that, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\) \(\mathrm{AA}^{-1}|\mathrm{~A}| =\mathrm{A} \operatorname{adj} \mathrm{A}\) \(=\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}| \mathrm{I}\) From equation (i) and equation (ii), we get - \(|\mathrm{A}| \mathrm{I}=10\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) \(|\mathrm{A}| \mathrm{I}=10 \mathrm{I}\) \(|\mathrm{A}|=10\)
BCECE-2005
Matrix and Determinant
78989
If \(A=\left[\begin{array}{cc}1 & 3 \\ 3 & 10\end{array}\right]\) then adjoint of \(A\) is :
78985
If \(A=\left|\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right|\), then the value of \(|A \| \operatorname{adj}(\mathbf{A})|\) is
1 \(a^{3}\)
2 \(a^{6}\)
3 \(a^{9}\)
4 \(\mathrm{a}^{27}\)
Explanation:
(C): We have, \(A=\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{a} & 0 \\ 0 & 0 & \mathrm{a} \end{array}\right]\) Determinant of \(\mathrm{A}\) is - \(|\mathrm{A}|=\mathrm{a}\left[\mathrm{a}^{2}-0\right]-0+0\) \(|\mathrm{~A}|=\mathrm{a}^{3}\) Adjoint of given matrix is given as - \(\operatorname{Adj} \mathrm{A}=\left[\begin{array}{ccc} \mathrm{a}^{2} & 0 & 0 \\ 0 & \mathrm{a}^{2} & 0 \\ 0 & 0 & \mathrm{a}^{2} \end{array}\right]\) Now, we find the determinant of Adjoint A, \(|\operatorname{Adj} A|=\mathrm{a}^{2}\left[\left(\mathrm{a}^{2}\right)\left(\mathrm{a}^{2}\right)-(0)(0)\right]-0\left[(0)\left(\mathrm{a}^{2}\right)-(0)(0)\right]\) \(+0[(0)\) \(|\operatorname{Adj} A| =\mathrm{a}^{2}\left[\mathrm{a}^{4}\right]-0+0\) \(|\operatorname{Adj} A| =\mathrm{a}^{6}\) Finally, \(|\mathrm{A}||\operatorname{Adj} \mathrm{A}|=\mathrm{a}^{3}\left(\mathrm{a}^{6}\right)\) \(|\mathrm{A}||\operatorname{Adj} \mathrm{A}|=\mathrm{a}^{9}\)
BCECE-2011
Matrix and Determinant
78986
If \(A=\left[\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right]\), then \(A^{-1}\) is equal to
(B): It is given that, \(A=\left[\begin{array}{cc} 1 & 3 \\ 2 & -2 \end{array}\right]\) Here, for matrix \(A=\left[\begin{array}{cc}m & n \\ o & p\end{array}\right]\) \(\operatorname{Adj} A=\left[\begin{array}{cc} p & -n \\ -o & m \end{array}\right]\) Hence, adj \(A=\left[\begin{array}{cc}-2 & -3 \\ -2 & 1\end{array}\right]^{\mathrm{T}}\) \(=\left[\begin{array}{cc} -2 & -2 \\ -3 & 1 \end{array}\right]\) And, \(\quad|A|=\left|\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right|=-2-6=-8\) \(\therefore \quad \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \cdot \operatorname{adj}(\mathrm{A})\) \(\mathrm{A}^{-1}=-\frac{1}{8}\left[\begin{array}{cc} -2 & -3 \\ -2 & 1 \end{array}\right]\)
BCECE-2010
Matrix and Determinant
78988
For any \(2 \times 2\) matrix \(A\), if \(A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\), then \(|A|\) i.e., \(\operatorname{det} A\) is equal to :
1 20
2 100
3 10
4 0
Explanation:
(C) : We have, \(A(\operatorname{adj} A)=\left[\begin{array}{cc} 10 & 0 \tag{i}\\ 0 & 10 \end{array}\right]\) We know that, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\) \(\mathrm{AA}^{-1}|\mathrm{~A}| =\mathrm{A} \operatorname{adj} \mathrm{A}\) \(=\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}| \mathrm{I}\) From equation (i) and equation (ii), we get - \(|\mathrm{A}| \mathrm{I}=10\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) \(|\mathrm{A}| \mathrm{I}=10 \mathrm{I}\) \(|\mathrm{A}|=10\)
BCECE-2005
Matrix and Determinant
78989
If \(A=\left[\begin{array}{cc}1 & 3 \\ 3 & 10\end{array}\right]\) then adjoint of \(A\) is :
78985
If \(A=\left|\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right|\), then the value of \(|A \| \operatorname{adj}(\mathbf{A})|\) is
1 \(a^{3}\)
2 \(a^{6}\)
3 \(a^{9}\)
4 \(\mathrm{a}^{27}\)
Explanation:
(C): We have, \(A=\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{a} & 0 \\ 0 & 0 & \mathrm{a} \end{array}\right]\) Determinant of \(\mathrm{A}\) is - \(|\mathrm{A}|=\mathrm{a}\left[\mathrm{a}^{2}-0\right]-0+0\) \(|\mathrm{~A}|=\mathrm{a}^{3}\) Adjoint of given matrix is given as - \(\operatorname{Adj} \mathrm{A}=\left[\begin{array}{ccc} \mathrm{a}^{2} & 0 & 0 \\ 0 & \mathrm{a}^{2} & 0 \\ 0 & 0 & \mathrm{a}^{2} \end{array}\right]\) Now, we find the determinant of Adjoint A, \(|\operatorname{Adj} A|=\mathrm{a}^{2}\left[\left(\mathrm{a}^{2}\right)\left(\mathrm{a}^{2}\right)-(0)(0)\right]-0\left[(0)\left(\mathrm{a}^{2}\right)-(0)(0)\right]\) \(+0[(0)\) \(|\operatorname{Adj} A| =\mathrm{a}^{2}\left[\mathrm{a}^{4}\right]-0+0\) \(|\operatorname{Adj} A| =\mathrm{a}^{6}\) Finally, \(|\mathrm{A}||\operatorname{Adj} \mathrm{A}|=\mathrm{a}^{3}\left(\mathrm{a}^{6}\right)\) \(|\mathrm{A}||\operatorname{Adj} \mathrm{A}|=\mathrm{a}^{9}\)
BCECE-2011
Matrix and Determinant
78986
If \(A=\left[\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right]\), then \(A^{-1}\) is equal to
(B): It is given that, \(A=\left[\begin{array}{cc} 1 & 3 \\ 2 & -2 \end{array}\right]\) Here, for matrix \(A=\left[\begin{array}{cc}m & n \\ o & p\end{array}\right]\) \(\operatorname{Adj} A=\left[\begin{array}{cc} p & -n \\ -o & m \end{array}\right]\) Hence, adj \(A=\left[\begin{array}{cc}-2 & -3 \\ -2 & 1\end{array}\right]^{\mathrm{T}}\) \(=\left[\begin{array}{cc} -2 & -2 \\ -3 & 1 \end{array}\right]\) And, \(\quad|A|=\left|\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right|=-2-6=-8\) \(\therefore \quad \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \cdot \operatorname{adj}(\mathrm{A})\) \(\mathrm{A}^{-1}=-\frac{1}{8}\left[\begin{array}{cc} -2 & -3 \\ -2 & 1 \end{array}\right]\)
BCECE-2010
Matrix and Determinant
78988
For any \(2 \times 2\) matrix \(A\), if \(A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\), then \(|A|\) i.e., \(\operatorname{det} A\) is equal to :
1 20
2 100
3 10
4 0
Explanation:
(C) : We have, \(A(\operatorname{adj} A)=\left[\begin{array}{cc} 10 & 0 \tag{i}\\ 0 & 10 \end{array}\right]\) We know that, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\) \(\mathrm{AA}^{-1}|\mathrm{~A}| =\mathrm{A} \operatorname{adj} \mathrm{A}\) \(=\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}| \mathrm{I}\) From equation (i) and equation (ii), we get - \(|\mathrm{A}| \mathrm{I}=10\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) \(|\mathrm{A}| \mathrm{I}=10 \mathrm{I}\) \(|\mathrm{A}|=10\)
BCECE-2005
Matrix and Determinant
78989
If \(A=\left[\begin{array}{cc}1 & 3 \\ 3 & 10\end{array}\right]\) then adjoint of \(A\) is :
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Matrix and Determinant
78985
If \(A=\left|\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right|\), then the value of \(|A \| \operatorname{adj}(\mathbf{A})|\) is
1 \(a^{3}\)
2 \(a^{6}\)
3 \(a^{9}\)
4 \(\mathrm{a}^{27}\)
Explanation:
(C): We have, \(A=\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{a} & 0 \\ 0 & 0 & \mathrm{a} \end{array}\right]\) Determinant of \(\mathrm{A}\) is - \(|\mathrm{A}|=\mathrm{a}\left[\mathrm{a}^{2}-0\right]-0+0\) \(|\mathrm{~A}|=\mathrm{a}^{3}\) Adjoint of given matrix is given as - \(\operatorname{Adj} \mathrm{A}=\left[\begin{array}{ccc} \mathrm{a}^{2} & 0 & 0 \\ 0 & \mathrm{a}^{2} & 0 \\ 0 & 0 & \mathrm{a}^{2} \end{array}\right]\) Now, we find the determinant of Adjoint A, \(|\operatorname{Adj} A|=\mathrm{a}^{2}\left[\left(\mathrm{a}^{2}\right)\left(\mathrm{a}^{2}\right)-(0)(0)\right]-0\left[(0)\left(\mathrm{a}^{2}\right)-(0)(0)\right]\) \(+0[(0)\) \(|\operatorname{Adj} A| =\mathrm{a}^{2}\left[\mathrm{a}^{4}\right]-0+0\) \(|\operatorname{Adj} A| =\mathrm{a}^{6}\) Finally, \(|\mathrm{A}||\operatorname{Adj} \mathrm{A}|=\mathrm{a}^{3}\left(\mathrm{a}^{6}\right)\) \(|\mathrm{A}||\operatorname{Adj} \mathrm{A}|=\mathrm{a}^{9}\)
BCECE-2011
Matrix and Determinant
78986
If \(A=\left[\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right]\), then \(A^{-1}\) is equal to
(B): It is given that, \(A=\left[\begin{array}{cc} 1 & 3 \\ 2 & -2 \end{array}\right]\) Here, for matrix \(A=\left[\begin{array}{cc}m & n \\ o & p\end{array}\right]\) \(\operatorname{Adj} A=\left[\begin{array}{cc} p & -n \\ -o & m \end{array}\right]\) Hence, adj \(A=\left[\begin{array}{cc}-2 & -3 \\ -2 & 1\end{array}\right]^{\mathrm{T}}\) \(=\left[\begin{array}{cc} -2 & -2 \\ -3 & 1 \end{array}\right]\) And, \(\quad|A|=\left|\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right|=-2-6=-8\) \(\therefore \quad \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \cdot \operatorname{adj}(\mathrm{A})\) \(\mathrm{A}^{-1}=-\frac{1}{8}\left[\begin{array}{cc} -2 & -3 \\ -2 & 1 \end{array}\right]\)
BCECE-2010
Matrix and Determinant
78988
For any \(2 \times 2\) matrix \(A\), if \(A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\), then \(|A|\) i.e., \(\operatorname{det} A\) is equal to :
1 20
2 100
3 10
4 0
Explanation:
(C) : We have, \(A(\operatorname{adj} A)=\left[\begin{array}{cc} 10 & 0 \tag{i}\\ 0 & 10 \end{array}\right]\) We know that, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\) \(\mathrm{AA}^{-1}|\mathrm{~A}| =\mathrm{A} \operatorname{adj} \mathrm{A}\) \(=\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}| \mathrm{I}\) From equation (i) and equation (ii), we get - \(|\mathrm{A}| \mathrm{I}=10\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) \(|\mathrm{A}| \mathrm{I}=10 \mathrm{I}\) \(|\mathrm{A}|=10\)
BCECE-2005
Matrix and Determinant
78989
If \(A=\left[\begin{array}{cc}1 & 3 \\ 3 & 10\end{array}\right]\) then adjoint of \(A\) is :
78985
If \(A=\left|\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right|\), then the value of \(|A \| \operatorname{adj}(\mathbf{A})|\) is
1 \(a^{3}\)
2 \(a^{6}\)
3 \(a^{9}\)
4 \(\mathrm{a}^{27}\)
Explanation:
(C): We have, \(A=\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{a} & 0 \\ 0 & 0 & \mathrm{a} \end{array}\right]\) Determinant of \(\mathrm{A}\) is - \(|\mathrm{A}|=\mathrm{a}\left[\mathrm{a}^{2}-0\right]-0+0\) \(|\mathrm{~A}|=\mathrm{a}^{3}\) Adjoint of given matrix is given as - \(\operatorname{Adj} \mathrm{A}=\left[\begin{array}{ccc} \mathrm{a}^{2} & 0 & 0 \\ 0 & \mathrm{a}^{2} & 0 \\ 0 & 0 & \mathrm{a}^{2} \end{array}\right]\) Now, we find the determinant of Adjoint A, \(|\operatorname{Adj} A|=\mathrm{a}^{2}\left[\left(\mathrm{a}^{2}\right)\left(\mathrm{a}^{2}\right)-(0)(0)\right]-0\left[(0)\left(\mathrm{a}^{2}\right)-(0)(0)\right]\) \(+0[(0)\) \(|\operatorname{Adj} A| =\mathrm{a}^{2}\left[\mathrm{a}^{4}\right]-0+0\) \(|\operatorname{Adj} A| =\mathrm{a}^{6}\) Finally, \(|\mathrm{A}||\operatorname{Adj} \mathrm{A}|=\mathrm{a}^{3}\left(\mathrm{a}^{6}\right)\) \(|\mathrm{A}||\operatorname{Adj} \mathrm{A}|=\mathrm{a}^{9}\)
BCECE-2011
Matrix and Determinant
78986
If \(A=\left[\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right]\), then \(A^{-1}\) is equal to
(B): It is given that, \(A=\left[\begin{array}{cc} 1 & 3 \\ 2 & -2 \end{array}\right]\) Here, for matrix \(A=\left[\begin{array}{cc}m & n \\ o & p\end{array}\right]\) \(\operatorname{Adj} A=\left[\begin{array}{cc} p & -n \\ -o & m \end{array}\right]\) Hence, adj \(A=\left[\begin{array}{cc}-2 & -3 \\ -2 & 1\end{array}\right]^{\mathrm{T}}\) \(=\left[\begin{array}{cc} -2 & -2 \\ -3 & 1 \end{array}\right]\) And, \(\quad|A|=\left|\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right|=-2-6=-8\) \(\therefore \quad \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \cdot \operatorname{adj}(\mathrm{A})\) \(\mathrm{A}^{-1}=-\frac{1}{8}\left[\begin{array}{cc} -2 & -3 \\ -2 & 1 \end{array}\right]\)
BCECE-2010
Matrix and Determinant
78988
For any \(2 \times 2\) matrix \(A\), if \(A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]\), then \(|A|\) i.e., \(\operatorname{det} A\) is equal to :
1 20
2 100
3 10
4 0
Explanation:
(C) : We have, \(A(\operatorname{adj} A)=\left[\begin{array}{cc} 10 & 0 \tag{i}\\ 0 & 10 \end{array}\right]\) We know that, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\) \(\mathrm{AA}^{-1}|\mathrm{~A}| =\mathrm{A} \operatorname{adj} \mathrm{A}\) \(=\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}| \mathrm{I}\) From equation (i) and equation (ii), we get - \(|\mathrm{A}| \mathrm{I}=10\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) \(|\mathrm{A}| \mathrm{I}=10 \mathrm{I}\) \(|\mathrm{A}|=10\)
BCECE-2005
Matrix and Determinant
78989
If \(A=\left[\begin{array}{cc}1 & 3 \\ 3 & 10\end{array}\right]\) then adjoint of \(A\) is :