79140
If \(P=\left|\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right|\) is the adjoint of a \(3 \times 3\) matrix \(A\) and \(|A|=4\), then \(\alpha\) is equal to
1 11
2 5
3 0
4 4
Explanation:
(A) : It is given that, \(P=\left|\begin{array}{lll} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{array}\right|\) Let, \(\quad P=1(12-12)-\alpha(4-6)+3(4-6)=2 \alpha-6\) Also, \(\operatorname{det}(\operatorname{adj} A)=(\operatorname{det} A)^{2}\) \(2 \alpha-6=16\) \(\therefore \quad \alpha=11\) Note \(: \operatorname{det}(\operatorname{adj} A)=(\operatorname{det} A)^{\mathrm{n}-1}\), where \(\mathrm{A}\) is a matrix of order \(\mathrm{n}\),
COMEDK-2013
Matrix and Determinant
79141
If \(A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]\), then \(\operatorname{det}(\operatorname{adj} A)\) is
1 \(\mathrm{a}^{27}\)
2 \(a^{9}\)
3 \(a^{6}\)
4 \(a^{2}\)
Explanation:
(C) : We have, \(A =\left|\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{a} & 0 \\ 0 & 0 & \mathrm{a} \end{array}\right|\) \(|\mathrm{A}| =\mathrm{a}\left(\mathrm{a}^{2}-0\right)=\mathrm{a}^{3}\) Using formula, \(|\operatorname{adj} A|=|A|^{n-1}\), we get - \(\operatorname{det}(\operatorname{adj} A)=\left(a^{3}\right)^{3-1}=\left(a^{3}\right)^{2}=a^{6}\)
COMEDK-2015
Matrix and Determinant
79142
If \(a, b, c\) are cube roots of unity, then \(\left|\begin{array}{lll} \mathbf{e}^{a} & \mathbf{e}^{2 a} & e^{3 a}-1 \\ e^{b} & e^{2 b} & e^{3 b}-1 \\ e^{c} & e^{2 c} & e^{3 c}-1 \end{array}\right|=\)
79140
If \(P=\left|\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right|\) is the adjoint of a \(3 \times 3\) matrix \(A\) and \(|A|=4\), then \(\alpha\) is equal to
1 11
2 5
3 0
4 4
Explanation:
(A) : It is given that, \(P=\left|\begin{array}{lll} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{array}\right|\) Let, \(\quad P=1(12-12)-\alpha(4-6)+3(4-6)=2 \alpha-6\) Also, \(\operatorname{det}(\operatorname{adj} A)=(\operatorname{det} A)^{2}\) \(2 \alpha-6=16\) \(\therefore \quad \alpha=11\) Note \(: \operatorname{det}(\operatorname{adj} A)=(\operatorname{det} A)^{\mathrm{n}-1}\), where \(\mathrm{A}\) is a matrix of order \(\mathrm{n}\),
COMEDK-2013
Matrix and Determinant
79141
If \(A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]\), then \(\operatorname{det}(\operatorname{adj} A)\) is
1 \(\mathrm{a}^{27}\)
2 \(a^{9}\)
3 \(a^{6}\)
4 \(a^{2}\)
Explanation:
(C) : We have, \(A =\left|\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{a} & 0 \\ 0 & 0 & \mathrm{a} \end{array}\right|\) \(|\mathrm{A}| =\mathrm{a}\left(\mathrm{a}^{2}-0\right)=\mathrm{a}^{3}\) Using formula, \(|\operatorname{adj} A|=|A|^{n-1}\), we get - \(\operatorname{det}(\operatorname{adj} A)=\left(a^{3}\right)^{3-1}=\left(a^{3}\right)^{2}=a^{6}\)
COMEDK-2015
Matrix and Determinant
79142
If \(a, b, c\) are cube roots of unity, then \(\left|\begin{array}{lll} \mathbf{e}^{a} & \mathbf{e}^{2 a} & e^{3 a}-1 \\ e^{b} & e^{2 b} & e^{3 b}-1 \\ e^{c} & e^{2 c} & e^{3 c}-1 \end{array}\right|=\)
79140
If \(P=\left|\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right|\) is the adjoint of a \(3 \times 3\) matrix \(A\) and \(|A|=4\), then \(\alpha\) is equal to
1 11
2 5
3 0
4 4
Explanation:
(A) : It is given that, \(P=\left|\begin{array}{lll} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{array}\right|\) Let, \(\quad P=1(12-12)-\alpha(4-6)+3(4-6)=2 \alpha-6\) Also, \(\operatorname{det}(\operatorname{adj} A)=(\operatorname{det} A)^{2}\) \(2 \alpha-6=16\) \(\therefore \quad \alpha=11\) Note \(: \operatorname{det}(\operatorname{adj} A)=(\operatorname{det} A)^{\mathrm{n}-1}\), where \(\mathrm{A}\) is a matrix of order \(\mathrm{n}\),
COMEDK-2013
Matrix and Determinant
79141
If \(A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]\), then \(\operatorname{det}(\operatorname{adj} A)\) is
1 \(\mathrm{a}^{27}\)
2 \(a^{9}\)
3 \(a^{6}\)
4 \(a^{2}\)
Explanation:
(C) : We have, \(A =\left|\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{a} & 0 \\ 0 & 0 & \mathrm{a} \end{array}\right|\) \(|\mathrm{A}| =\mathrm{a}\left(\mathrm{a}^{2}-0\right)=\mathrm{a}^{3}\) Using formula, \(|\operatorname{adj} A|=|A|^{n-1}\), we get - \(\operatorname{det}(\operatorname{adj} A)=\left(a^{3}\right)^{3-1}=\left(a^{3}\right)^{2}=a^{6}\)
COMEDK-2015
Matrix and Determinant
79142
If \(a, b, c\) are cube roots of unity, then \(\left|\begin{array}{lll} \mathbf{e}^{a} & \mathbf{e}^{2 a} & e^{3 a}-1 \\ e^{b} & e^{2 b} & e^{3 b}-1 \\ e^{c} & e^{2 c} & e^{3 c}-1 \end{array}\right|=\)
79140
If \(P=\left|\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right|\) is the adjoint of a \(3 \times 3\) matrix \(A\) and \(|A|=4\), then \(\alpha\) is equal to
1 11
2 5
3 0
4 4
Explanation:
(A) : It is given that, \(P=\left|\begin{array}{lll} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{array}\right|\) Let, \(\quad P=1(12-12)-\alpha(4-6)+3(4-6)=2 \alpha-6\) Also, \(\operatorname{det}(\operatorname{adj} A)=(\operatorname{det} A)^{2}\) \(2 \alpha-6=16\) \(\therefore \quad \alpha=11\) Note \(: \operatorname{det}(\operatorname{adj} A)=(\operatorname{det} A)^{\mathrm{n}-1}\), where \(\mathrm{A}\) is a matrix of order \(\mathrm{n}\),
COMEDK-2013
Matrix and Determinant
79141
If \(A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]\), then \(\operatorname{det}(\operatorname{adj} A)\) is
1 \(\mathrm{a}^{27}\)
2 \(a^{9}\)
3 \(a^{6}\)
4 \(a^{2}\)
Explanation:
(C) : We have, \(A =\left|\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{a} & 0 \\ 0 & 0 & \mathrm{a} \end{array}\right|\) \(|\mathrm{A}| =\mathrm{a}\left(\mathrm{a}^{2}-0\right)=\mathrm{a}^{3}\) Using formula, \(|\operatorname{adj} A|=|A|^{n-1}\), we get - \(\operatorname{det}(\operatorname{adj} A)=\left(a^{3}\right)^{3-1}=\left(a^{3}\right)^{2}=a^{6}\)
COMEDK-2015
Matrix and Determinant
79142
If \(a, b, c\) are cube roots of unity, then \(\left|\begin{array}{lll} \mathbf{e}^{a} & \mathbf{e}^{2 a} & e^{3 a}-1 \\ e^{b} & e^{2 b} & e^{3 b}-1 \\ e^{c} & e^{2 c} & e^{3 c}-1 \end{array}\right|=\)
