78836
Consider \(A\) and \(B\) two square matrices of same order. Select the correct alternative.
1 \(|\mathrm{AB}|\) must be greater then \(|\mathrm{A}|\)
2 \(\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|\) is not unit matrix
3 \(|\mathrm{A}+\mathrm{B}|\) must be greater than \(|\mathrm{A}|\)
4 If \(A B=0\), either \(A\) or \(B\) must be zero matrix
Explanation:
(B) : According to question, A and B are square matrix. We know that, all the diagonal elements of the matrix is equal 1. And remaining all the elements are zero is called unit matrix. e.g. \(\quad A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\) Hence, \(\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right] \text { is not unit matrix or identity matrix. }\)
UPSEE-2016
Matrix and Determinant
78837
If \(A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0\end{array}\right]\) and \(B=(\operatorname{adj} A)\), and \(\mathbf{C}=\mathbf{5 A}\), then \(\frac{|\operatorname{adj} \mathbf{B}|}{|\mathrm{C}|}\)
78838
If \(A=\left[\begin{array}{ccc}1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1\end{array}\right]\), then \(\left.\operatorname{det}(\operatorname{adj})(\operatorname{adj} A)\right)\) is equal to
1 \(14^{4}\)
2 \(14^{3}\)
3 \(14^{2}\)
4 14
Explanation:
(A) : Given that, \(|A|=1(1+2)-2(-1-4)-1(1-2)\) \(|A|=3+10+1=14\) We know that, \(\operatorname{adj}(\operatorname{adj} A=|A|^{n-2} A \text { if only if }|A| \neq 0\) \(\operatorname{det}[\operatorname{adj}(\operatorname{adj} A)]=\left.|| A\right|^{n-2} A \mid\) \(=\left(|A|^{n-2}\right)^{n} \cdot|A|\) \(=|A|^{n^{2}-2 n+1}\) We know that, \(\mathrm{n}=3 \text { and }|\mathrm{A}|=14\) Hence, \(\operatorname{det}[\operatorname{adj}(\operatorname{adj} A)]=(14)^{3^{2}-2 \times 3+1}\) \(=(14)^{9-6+1}=(14)^{4}\)
(D) : According to question. A, B and C are nonsingular matrix. then, We know that, \((\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}\) Now, \(\left(\mathrm{AB} \mathrm{B}^{-1} \mathrm{C}\right)^{-1}\) \(= \mathrm{C}^{-1}\left(\mathrm{~B}^{-1}\right)^{-1} \mathrm{~A}^{-1}\) \(= \mathrm{C}^{-1} \mathrm{~B} \mathrm{~A}^{-1}\) \(\because\left(\mathrm{B}^{-1}\right)^{-1}=\mathrm{B}\).
78836
Consider \(A\) and \(B\) two square matrices of same order. Select the correct alternative.
1 \(|\mathrm{AB}|\) must be greater then \(|\mathrm{A}|\)
2 \(\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|\) is not unit matrix
3 \(|\mathrm{A}+\mathrm{B}|\) must be greater than \(|\mathrm{A}|\)
4 If \(A B=0\), either \(A\) or \(B\) must be zero matrix
Explanation:
(B) : According to question, A and B are square matrix. We know that, all the diagonal elements of the matrix is equal 1. And remaining all the elements are zero is called unit matrix. e.g. \(\quad A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\) Hence, \(\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right] \text { is not unit matrix or identity matrix. }\)
UPSEE-2016
Matrix and Determinant
78837
If \(A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0\end{array}\right]\) and \(B=(\operatorname{adj} A)\), and \(\mathbf{C}=\mathbf{5 A}\), then \(\frac{|\operatorname{adj} \mathbf{B}|}{|\mathrm{C}|}\)
78838
If \(A=\left[\begin{array}{ccc}1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1\end{array}\right]\), then \(\left.\operatorname{det}(\operatorname{adj})(\operatorname{adj} A)\right)\) is equal to
1 \(14^{4}\)
2 \(14^{3}\)
3 \(14^{2}\)
4 14
Explanation:
(A) : Given that, \(|A|=1(1+2)-2(-1-4)-1(1-2)\) \(|A|=3+10+1=14\) We know that, \(\operatorname{adj}(\operatorname{adj} A=|A|^{n-2} A \text { if only if }|A| \neq 0\) \(\operatorname{det}[\operatorname{adj}(\operatorname{adj} A)]=\left.|| A\right|^{n-2} A \mid\) \(=\left(|A|^{n-2}\right)^{n} \cdot|A|\) \(=|A|^{n^{2}-2 n+1}\) We know that, \(\mathrm{n}=3 \text { and }|\mathrm{A}|=14\) Hence, \(\operatorname{det}[\operatorname{adj}(\operatorname{adj} A)]=(14)^{3^{2}-2 \times 3+1}\) \(=(14)^{9-6+1}=(14)^{4}\)
(D) : According to question. A, B and C are nonsingular matrix. then, We know that, \((\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}\) Now, \(\left(\mathrm{AB} \mathrm{B}^{-1} \mathrm{C}\right)^{-1}\) \(= \mathrm{C}^{-1}\left(\mathrm{~B}^{-1}\right)^{-1} \mathrm{~A}^{-1}\) \(= \mathrm{C}^{-1} \mathrm{~B} \mathrm{~A}^{-1}\) \(\because\left(\mathrm{B}^{-1}\right)^{-1}=\mathrm{B}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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Matrix and Determinant
78836
Consider \(A\) and \(B\) two square matrices of same order. Select the correct alternative.
1 \(|\mathrm{AB}|\) must be greater then \(|\mathrm{A}|\)
2 \(\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|\) is not unit matrix
3 \(|\mathrm{A}+\mathrm{B}|\) must be greater than \(|\mathrm{A}|\)
4 If \(A B=0\), either \(A\) or \(B\) must be zero matrix
Explanation:
(B) : According to question, A and B are square matrix. We know that, all the diagonal elements of the matrix is equal 1. And remaining all the elements are zero is called unit matrix. e.g. \(\quad A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\) Hence, \(\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right] \text { is not unit matrix or identity matrix. }\)
UPSEE-2016
Matrix and Determinant
78837
If \(A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0\end{array}\right]\) and \(B=(\operatorname{adj} A)\), and \(\mathbf{C}=\mathbf{5 A}\), then \(\frac{|\operatorname{adj} \mathbf{B}|}{|\mathrm{C}|}\)
78838
If \(A=\left[\begin{array}{ccc}1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1\end{array}\right]\), then \(\left.\operatorname{det}(\operatorname{adj})(\operatorname{adj} A)\right)\) is equal to
1 \(14^{4}\)
2 \(14^{3}\)
3 \(14^{2}\)
4 14
Explanation:
(A) : Given that, \(|A|=1(1+2)-2(-1-4)-1(1-2)\) \(|A|=3+10+1=14\) We know that, \(\operatorname{adj}(\operatorname{adj} A=|A|^{n-2} A \text { if only if }|A| \neq 0\) \(\operatorname{det}[\operatorname{adj}(\operatorname{adj} A)]=\left.|| A\right|^{n-2} A \mid\) \(=\left(|A|^{n-2}\right)^{n} \cdot|A|\) \(=|A|^{n^{2}-2 n+1}\) We know that, \(\mathrm{n}=3 \text { and }|\mathrm{A}|=14\) Hence, \(\operatorname{det}[\operatorname{adj}(\operatorname{adj} A)]=(14)^{3^{2}-2 \times 3+1}\) \(=(14)^{9-6+1}=(14)^{4}\)
(D) : According to question. A, B and C are nonsingular matrix. then, We know that, \((\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}\) Now, \(\left(\mathrm{AB} \mathrm{B}^{-1} \mathrm{C}\right)^{-1}\) \(= \mathrm{C}^{-1}\left(\mathrm{~B}^{-1}\right)^{-1} \mathrm{~A}^{-1}\) \(= \mathrm{C}^{-1} \mathrm{~B} \mathrm{~A}^{-1}\) \(\because\left(\mathrm{B}^{-1}\right)^{-1}=\mathrm{B}\).
78836
Consider \(A\) and \(B\) two square matrices of same order. Select the correct alternative.
1 \(|\mathrm{AB}|\) must be greater then \(|\mathrm{A}|\)
2 \(\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|\) is not unit matrix
3 \(|\mathrm{A}+\mathrm{B}|\) must be greater than \(|\mathrm{A}|\)
4 If \(A B=0\), either \(A\) or \(B\) must be zero matrix
Explanation:
(B) : According to question, A and B are square matrix. We know that, all the diagonal elements of the matrix is equal 1. And remaining all the elements are zero is called unit matrix. e.g. \(\quad A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\) Hence, \(\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right] \text { is not unit matrix or identity matrix. }\)
UPSEE-2016
Matrix and Determinant
78837
If \(A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0\end{array}\right]\) and \(B=(\operatorname{adj} A)\), and \(\mathbf{C}=\mathbf{5 A}\), then \(\frac{|\operatorname{adj} \mathbf{B}|}{|\mathrm{C}|}\)
78838
If \(A=\left[\begin{array}{ccc}1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1\end{array}\right]\), then \(\left.\operatorname{det}(\operatorname{adj})(\operatorname{adj} A)\right)\) is equal to
1 \(14^{4}\)
2 \(14^{3}\)
3 \(14^{2}\)
4 14
Explanation:
(A) : Given that, \(|A|=1(1+2)-2(-1-4)-1(1-2)\) \(|A|=3+10+1=14\) We know that, \(\operatorname{adj}(\operatorname{adj} A=|A|^{n-2} A \text { if only if }|A| \neq 0\) \(\operatorname{det}[\operatorname{adj}(\operatorname{adj} A)]=\left.|| A\right|^{n-2} A \mid\) \(=\left(|A|^{n-2}\right)^{n} \cdot|A|\) \(=|A|^{n^{2}-2 n+1}\) We know that, \(\mathrm{n}=3 \text { and }|\mathrm{A}|=14\) Hence, \(\operatorname{det}[\operatorname{adj}(\operatorname{adj} A)]=(14)^{3^{2}-2 \times 3+1}\) \(=(14)^{9-6+1}=(14)^{4}\)
(D) : According to question. A, B and C are nonsingular matrix. then, We know that, \((\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}\) Now, \(\left(\mathrm{AB} \mathrm{B}^{-1} \mathrm{C}\right)^{-1}\) \(= \mathrm{C}^{-1}\left(\mathrm{~B}^{-1}\right)^{-1} \mathrm{~A}^{-1}\) \(= \mathrm{C}^{-1} \mathrm{~B} \mathrm{~A}^{-1}\) \(\because\left(\mathrm{B}^{-1}\right)^{-1}=\mathrm{B}\).