78912
Let \(A\) and \(B\) be two \(3 \times 3\) real matrices such that \(\left(A^{2}-B^{2}\right)\) is invertible matrix. If \(A^{5}=B^{5}\) and \(A^{3} B^{2}=A^{2} B^{3}\), then the value of the determinant of the matrix \(A^{3}+B^{3}\) is equal to
1 2
2 4
3 1
4 0
Explanation:
(D) : Given that, \(\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)\) is invertible matrix So, \(\quad\left|\mathrm{A}^{2}-\mathrm{B}_{5}^{2}\right| \neq 0\) And \(A^{5}=B^{5}, \quad A^{3} B^{2}=A^{2} B^{3}\) \(\therefore\left(\mathrm{A}^{3}+\mathrm{B}^{3}\right)\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)=\mathrm{A}^{5}-\mathrm{A}^{3} \mathrm{~B}^{2}+\mathrm{B}^{3} \mathrm{~A}^{2}-\mathrm{B}^{5}\) Taking determinant on both sides, we get - \(\left|\left(\mathrm{A}^{3}+\mathrm{B}^{3}\right)\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)\right|=0\) \(\left|\mathrm{~A}^{3}+\mathrm{B}^{3}\right|=0 \quad\left(\because\left|\mathrm{A}^{2}-\mathrm{B}^{2}\right| \neq 0\right)\)
JEE Main-2021-27.07.2021
Matrix and Determinant
78913
If \(A\) is a square matrix of order \(n \times n\) and \(K\) is a scalar, then adj (KA) is equal to
(C) : We know that, \(\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}\) So, \(\quad(\mathrm{KA})^{-1}=\frac{\operatorname{adj}(\mathrm{KA})}{|\mathrm{KA}|}\) \(\operatorname{adj}(\mathrm{KA}) =|\mathrm{KA}|(\mathrm{KA})^{-1}\) \(=\mathrm{K}^{\mathrm{n}}|\mathrm{A}| \mathrm{K}^{-1} \mathrm{~A}^{-1}\) \(=\frac{\mathrm{K}^{\mathrm{n}}}{\mathrm{K}}|\mathrm{A}| \mathrm{A}^{-1}=\mathrm{K}^{\mathrm{n}-1}|\mathrm{~A}| \frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}\) \(=\mathrm{K}^{\mathrm{n}-1} \operatorname{adj}(\mathrm{A})\)
CG PET-2021
Matrix and Determinant
78914
Let \(A\) be a matrix of order \(3 \times 3\) and \(\operatorname{det}(A)=\) 2. Then \(\operatorname{det}\left(\operatorname{det}(\mathrm{A}) \operatorname{adj}\left(5 \mathrm{adj}\left(\mathrm{A}^{3}\right)\right)\right.\) ) is equal to
78916
If \(A \cdot \operatorname{adj}(A)=0\), then \(|A|\) is
1 0
2 \(\frac{1}{|\operatorname{adj} \mathrm{A}|}\)
3 1
4 -1
Explanation:
(A) : Given that, A.adj (A) \(=0 \quad \ldots\) (i) We know that, \(\quad A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\) Or \(\operatorname{adj}(A)=A^{-1}|A|\) Putting these value in equation (i), we get - \(\mathrm{A} \cdot \mathrm{A}^{-1}|\mathrm{~A}|=0\) \(\mathrm{I}|\mathrm{A}|=0\) \(|\mathrm{~A}|=0\)
78912
Let \(A\) and \(B\) be two \(3 \times 3\) real matrices such that \(\left(A^{2}-B^{2}\right)\) is invertible matrix. If \(A^{5}=B^{5}\) and \(A^{3} B^{2}=A^{2} B^{3}\), then the value of the determinant of the matrix \(A^{3}+B^{3}\) is equal to
1 2
2 4
3 1
4 0
Explanation:
(D) : Given that, \(\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)\) is invertible matrix So, \(\quad\left|\mathrm{A}^{2}-\mathrm{B}_{5}^{2}\right| \neq 0\) And \(A^{5}=B^{5}, \quad A^{3} B^{2}=A^{2} B^{3}\) \(\therefore\left(\mathrm{A}^{3}+\mathrm{B}^{3}\right)\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)=\mathrm{A}^{5}-\mathrm{A}^{3} \mathrm{~B}^{2}+\mathrm{B}^{3} \mathrm{~A}^{2}-\mathrm{B}^{5}\) Taking determinant on both sides, we get - \(\left|\left(\mathrm{A}^{3}+\mathrm{B}^{3}\right)\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)\right|=0\) \(\left|\mathrm{~A}^{3}+\mathrm{B}^{3}\right|=0 \quad\left(\because\left|\mathrm{A}^{2}-\mathrm{B}^{2}\right| \neq 0\right)\)
JEE Main-2021-27.07.2021
Matrix and Determinant
78913
If \(A\) is a square matrix of order \(n \times n\) and \(K\) is a scalar, then adj (KA) is equal to
(C) : We know that, \(\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}\) So, \(\quad(\mathrm{KA})^{-1}=\frac{\operatorname{adj}(\mathrm{KA})}{|\mathrm{KA}|}\) \(\operatorname{adj}(\mathrm{KA}) =|\mathrm{KA}|(\mathrm{KA})^{-1}\) \(=\mathrm{K}^{\mathrm{n}}|\mathrm{A}| \mathrm{K}^{-1} \mathrm{~A}^{-1}\) \(=\frac{\mathrm{K}^{\mathrm{n}}}{\mathrm{K}}|\mathrm{A}| \mathrm{A}^{-1}=\mathrm{K}^{\mathrm{n}-1}|\mathrm{~A}| \frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}\) \(=\mathrm{K}^{\mathrm{n}-1} \operatorname{adj}(\mathrm{A})\)
CG PET-2021
Matrix and Determinant
78914
Let \(A\) be a matrix of order \(3 \times 3\) and \(\operatorname{det}(A)=\) 2. Then \(\operatorname{det}\left(\operatorname{det}(\mathrm{A}) \operatorname{adj}\left(5 \mathrm{adj}\left(\mathrm{A}^{3}\right)\right)\right.\) ) is equal to
78916
If \(A \cdot \operatorname{adj}(A)=0\), then \(|A|\) is
1 0
2 \(\frac{1}{|\operatorname{adj} \mathrm{A}|}\)
3 1
4 -1
Explanation:
(A) : Given that, A.adj (A) \(=0 \quad \ldots\) (i) We know that, \(\quad A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\) Or \(\operatorname{adj}(A)=A^{-1}|A|\) Putting these value in equation (i), we get - \(\mathrm{A} \cdot \mathrm{A}^{-1}|\mathrm{~A}|=0\) \(\mathrm{I}|\mathrm{A}|=0\) \(|\mathrm{~A}|=0\)
78912
Let \(A\) and \(B\) be two \(3 \times 3\) real matrices such that \(\left(A^{2}-B^{2}\right)\) is invertible matrix. If \(A^{5}=B^{5}\) and \(A^{3} B^{2}=A^{2} B^{3}\), then the value of the determinant of the matrix \(A^{3}+B^{3}\) is equal to
1 2
2 4
3 1
4 0
Explanation:
(D) : Given that, \(\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)\) is invertible matrix So, \(\quad\left|\mathrm{A}^{2}-\mathrm{B}_{5}^{2}\right| \neq 0\) And \(A^{5}=B^{5}, \quad A^{3} B^{2}=A^{2} B^{3}\) \(\therefore\left(\mathrm{A}^{3}+\mathrm{B}^{3}\right)\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)=\mathrm{A}^{5}-\mathrm{A}^{3} \mathrm{~B}^{2}+\mathrm{B}^{3} \mathrm{~A}^{2}-\mathrm{B}^{5}\) Taking determinant on both sides, we get - \(\left|\left(\mathrm{A}^{3}+\mathrm{B}^{3}\right)\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)\right|=0\) \(\left|\mathrm{~A}^{3}+\mathrm{B}^{3}\right|=0 \quad\left(\because\left|\mathrm{A}^{2}-\mathrm{B}^{2}\right| \neq 0\right)\)
JEE Main-2021-27.07.2021
Matrix and Determinant
78913
If \(A\) is a square matrix of order \(n \times n\) and \(K\) is a scalar, then adj (KA) is equal to
(C) : We know that, \(\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}\) So, \(\quad(\mathrm{KA})^{-1}=\frac{\operatorname{adj}(\mathrm{KA})}{|\mathrm{KA}|}\) \(\operatorname{adj}(\mathrm{KA}) =|\mathrm{KA}|(\mathrm{KA})^{-1}\) \(=\mathrm{K}^{\mathrm{n}}|\mathrm{A}| \mathrm{K}^{-1} \mathrm{~A}^{-1}\) \(=\frac{\mathrm{K}^{\mathrm{n}}}{\mathrm{K}}|\mathrm{A}| \mathrm{A}^{-1}=\mathrm{K}^{\mathrm{n}-1}|\mathrm{~A}| \frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}\) \(=\mathrm{K}^{\mathrm{n}-1} \operatorname{adj}(\mathrm{A})\)
CG PET-2021
Matrix and Determinant
78914
Let \(A\) be a matrix of order \(3 \times 3\) and \(\operatorname{det}(A)=\) 2. Then \(\operatorname{det}\left(\operatorname{det}(\mathrm{A}) \operatorname{adj}\left(5 \mathrm{adj}\left(\mathrm{A}^{3}\right)\right)\right.\) ) is equal to
78916
If \(A \cdot \operatorname{adj}(A)=0\), then \(|A|\) is
1 0
2 \(\frac{1}{|\operatorname{adj} \mathrm{A}|}\)
3 1
4 -1
Explanation:
(A) : Given that, A.adj (A) \(=0 \quad \ldots\) (i) We know that, \(\quad A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\) Or \(\operatorname{adj}(A)=A^{-1}|A|\) Putting these value in equation (i), we get - \(\mathrm{A} \cdot \mathrm{A}^{-1}|\mathrm{~A}|=0\) \(\mathrm{I}|\mathrm{A}|=0\) \(|\mathrm{~A}|=0\)
78912
Let \(A\) and \(B\) be two \(3 \times 3\) real matrices such that \(\left(A^{2}-B^{2}\right)\) is invertible matrix. If \(A^{5}=B^{5}\) and \(A^{3} B^{2}=A^{2} B^{3}\), then the value of the determinant of the matrix \(A^{3}+B^{3}\) is equal to
1 2
2 4
3 1
4 0
Explanation:
(D) : Given that, \(\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)\) is invertible matrix So, \(\quad\left|\mathrm{A}^{2}-\mathrm{B}_{5}^{2}\right| \neq 0\) And \(A^{5}=B^{5}, \quad A^{3} B^{2}=A^{2} B^{3}\) \(\therefore\left(\mathrm{A}^{3}+\mathrm{B}^{3}\right)\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)=\mathrm{A}^{5}-\mathrm{A}^{3} \mathrm{~B}^{2}+\mathrm{B}^{3} \mathrm{~A}^{2}-\mathrm{B}^{5}\) Taking determinant on both sides, we get - \(\left|\left(\mathrm{A}^{3}+\mathrm{B}^{3}\right)\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)\right|=0\) \(\left|\mathrm{~A}^{3}+\mathrm{B}^{3}\right|=0 \quad\left(\because\left|\mathrm{A}^{2}-\mathrm{B}^{2}\right| \neq 0\right)\)
JEE Main-2021-27.07.2021
Matrix and Determinant
78913
If \(A\) is a square matrix of order \(n \times n\) and \(K\) is a scalar, then adj (KA) is equal to
(C) : We know that, \(\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}\) So, \(\quad(\mathrm{KA})^{-1}=\frac{\operatorname{adj}(\mathrm{KA})}{|\mathrm{KA}|}\) \(\operatorname{adj}(\mathrm{KA}) =|\mathrm{KA}|(\mathrm{KA})^{-1}\) \(=\mathrm{K}^{\mathrm{n}}|\mathrm{A}| \mathrm{K}^{-1} \mathrm{~A}^{-1}\) \(=\frac{\mathrm{K}^{\mathrm{n}}}{\mathrm{K}}|\mathrm{A}| \mathrm{A}^{-1}=\mathrm{K}^{\mathrm{n}-1}|\mathrm{~A}| \frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}\) \(=\mathrm{K}^{\mathrm{n}-1} \operatorname{adj}(\mathrm{A})\)
CG PET-2021
Matrix and Determinant
78914
Let \(A\) be a matrix of order \(3 \times 3\) and \(\operatorname{det}(A)=\) 2. Then \(\operatorname{det}\left(\operatorname{det}(\mathrm{A}) \operatorname{adj}\left(5 \mathrm{adj}\left(\mathrm{A}^{3}\right)\right)\right.\) ) is equal to
78916
If \(A \cdot \operatorname{adj}(A)=0\), then \(|A|\) is
1 0
2 \(\frac{1}{|\operatorname{adj} \mathrm{A}|}\)
3 1
4 -1
Explanation:
(A) : Given that, A.adj (A) \(=0 \quad \ldots\) (i) We know that, \(\quad A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\) Or \(\operatorname{adj}(A)=A^{-1}|A|\) Putting these value in equation (i), we get - \(\mathrm{A} \cdot \mathrm{A}^{-1}|\mathrm{~A}|=0\) \(\mathrm{I}|\mathrm{A}|=0\) \(|\mathrm{~A}|=0\)
78912
Let \(A\) and \(B\) be two \(3 \times 3\) real matrices such that \(\left(A^{2}-B^{2}\right)\) is invertible matrix. If \(A^{5}=B^{5}\) and \(A^{3} B^{2}=A^{2} B^{3}\), then the value of the determinant of the matrix \(A^{3}+B^{3}\) is equal to
1 2
2 4
3 1
4 0
Explanation:
(D) : Given that, \(\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)\) is invertible matrix So, \(\quad\left|\mathrm{A}^{2}-\mathrm{B}_{5}^{2}\right| \neq 0\) And \(A^{5}=B^{5}, \quad A^{3} B^{2}=A^{2} B^{3}\) \(\therefore\left(\mathrm{A}^{3}+\mathrm{B}^{3}\right)\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)=\mathrm{A}^{5}-\mathrm{A}^{3} \mathrm{~B}^{2}+\mathrm{B}^{3} \mathrm{~A}^{2}-\mathrm{B}^{5}\) Taking determinant on both sides, we get - \(\left|\left(\mathrm{A}^{3}+\mathrm{B}^{3}\right)\left(\mathrm{A}^{2}-\mathrm{B}^{2}\right)\right|=0\) \(\left|\mathrm{~A}^{3}+\mathrm{B}^{3}\right|=0 \quad\left(\because\left|\mathrm{A}^{2}-\mathrm{B}^{2}\right| \neq 0\right)\)
JEE Main-2021-27.07.2021
Matrix and Determinant
78913
If \(A\) is a square matrix of order \(n \times n\) and \(K\) is a scalar, then adj (KA) is equal to
(C) : We know that, \(\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}\) So, \(\quad(\mathrm{KA})^{-1}=\frac{\operatorname{adj}(\mathrm{KA})}{|\mathrm{KA}|}\) \(\operatorname{adj}(\mathrm{KA}) =|\mathrm{KA}|(\mathrm{KA})^{-1}\) \(=\mathrm{K}^{\mathrm{n}}|\mathrm{A}| \mathrm{K}^{-1} \mathrm{~A}^{-1}\) \(=\frac{\mathrm{K}^{\mathrm{n}}}{\mathrm{K}}|\mathrm{A}| \mathrm{A}^{-1}=\mathrm{K}^{\mathrm{n}-1}|\mathrm{~A}| \frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}\) \(=\mathrm{K}^{\mathrm{n}-1} \operatorname{adj}(\mathrm{A})\)
CG PET-2021
Matrix and Determinant
78914
Let \(A\) be a matrix of order \(3 \times 3\) and \(\operatorname{det}(A)=\) 2. Then \(\operatorname{det}\left(\operatorname{det}(\mathrm{A}) \operatorname{adj}\left(5 \mathrm{adj}\left(\mathrm{A}^{3}\right)\right)\right.\) ) is equal to
78916
If \(A \cdot \operatorname{adj}(A)=0\), then \(|A|\) is
1 0
2 \(\frac{1}{|\operatorname{adj} \mathrm{A}|}\)
3 1
4 -1
Explanation:
(A) : Given that, A.adj (A) \(=0 \quad \ldots\) (i) We know that, \(\quad A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\) Or \(\operatorname{adj}(A)=A^{-1}|A|\) Putting these value in equation (i), we get - \(\mathrm{A} \cdot \mathrm{A}^{-1}|\mathrm{~A}|=0\) \(\mathrm{I}|\mathrm{A}|=0\) \(|\mathrm{~A}|=0\)
