78449
If a matrix \(A\) is both symmetric and skew symmetric, then
1 A is diagonal matrix
2 A is a zero matrix
3 A is scalar matrix
4 A is square matrix
Explanation:
(B) : According to question, \(\mathrm{A}\) is both symmetric and skew symmetric matrix it means - \(\begin{array}{lll} & \mathrm{A}^{\mathrm{T}}=\mathrm{A} & \text { (Symmetric) } \\ \text { And } & \mathrm{A}^{\mathrm{T}}=-\mathrm{A} & \text { (Skew-symmetric) } \\ \therefore & \mathrm{A}=\mathrm{A}^{\mathrm{T}}=-\mathrm{A} & \end{array}\) It is only possible in case of zero matrix, \(A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=A^{T}=-A\)
Karnataka CET-2017
Matrix and Determinant
78450
If \(A\) and \(B\) are square matrices of order ' \(n\) ' such that \(A^{2}-B^{2}=(A-B)(A+B)\), then which of the following will be true?
1 Either of A or B is zero matrix.
2 \(\mathrm{A}=\mathrm{B}\)
3 \(\mathrm{AB}=\mathrm{BA}\)
4 Either of \(\mathrm{A}\) or \(\mathrm{B}\) is an identity matrix.
Explanation:
(C)Given that,\(\mathrm{A}^{2}-\mathrm{B}^{2}=(\mathrm{A}-\mathrm{B})(\mathrm{A}+\mathrm{B})\) Here, \(\quad A, B\) are order \(n \times n\) \(\therefore \quad \mathrm{A}^{2}-\mathrm{B}^{2}=\mathrm{A}^{2}-\mathrm{BA}+\mathrm{AB}-\mathrm{B}^{2}\) \(\mathrm{~A}^{2}-\mathrm{B}^{2}-\mathrm{A}^{2}+\mathrm{B}^{2}=\mathrm{AB}-\mathrm{BA}\) \(0=\mathrm{AB}-\mathrm{BA}\) Therefore, \(\mathrm{AB}=\mathrm{BA}\)
78449
If a matrix \(A\) is both symmetric and skew symmetric, then
1 A is diagonal matrix
2 A is a zero matrix
3 A is scalar matrix
4 A is square matrix
Explanation:
(B) : According to question, \(\mathrm{A}\) is both symmetric and skew symmetric matrix it means - \(\begin{array}{lll} & \mathrm{A}^{\mathrm{T}}=\mathrm{A} & \text { (Symmetric) } \\ \text { And } & \mathrm{A}^{\mathrm{T}}=-\mathrm{A} & \text { (Skew-symmetric) } \\ \therefore & \mathrm{A}=\mathrm{A}^{\mathrm{T}}=-\mathrm{A} & \end{array}\) It is only possible in case of zero matrix, \(A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=A^{T}=-A\)
Karnataka CET-2017
Matrix and Determinant
78450
If \(A\) and \(B\) are square matrices of order ' \(n\) ' such that \(A^{2}-B^{2}=(A-B)(A+B)\), then which of the following will be true?
1 Either of A or B is zero matrix.
2 \(\mathrm{A}=\mathrm{B}\)
3 \(\mathrm{AB}=\mathrm{BA}\)
4 Either of \(\mathrm{A}\) or \(\mathrm{B}\) is an identity matrix.
Explanation:
(C)Given that,\(\mathrm{A}^{2}-\mathrm{B}^{2}=(\mathrm{A}-\mathrm{B})(\mathrm{A}+\mathrm{B})\) Here, \(\quad A, B\) are order \(n \times n\) \(\therefore \quad \mathrm{A}^{2}-\mathrm{B}^{2}=\mathrm{A}^{2}-\mathrm{BA}+\mathrm{AB}-\mathrm{B}^{2}\) \(\mathrm{~A}^{2}-\mathrm{B}^{2}-\mathrm{A}^{2}+\mathrm{B}^{2}=\mathrm{AB}-\mathrm{BA}\) \(0=\mathrm{AB}-\mathrm{BA}\) Therefore, \(\mathrm{AB}=\mathrm{BA}\)
78449
If a matrix \(A\) is both symmetric and skew symmetric, then
1 A is diagonal matrix
2 A is a zero matrix
3 A is scalar matrix
4 A is square matrix
Explanation:
(B) : According to question, \(\mathrm{A}\) is both symmetric and skew symmetric matrix it means - \(\begin{array}{lll} & \mathrm{A}^{\mathrm{T}}=\mathrm{A} & \text { (Symmetric) } \\ \text { And } & \mathrm{A}^{\mathrm{T}}=-\mathrm{A} & \text { (Skew-symmetric) } \\ \therefore & \mathrm{A}=\mathrm{A}^{\mathrm{T}}=-\mathrm{A} & \end{array}\) It is only possible in case of zero matrix, \(A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=A^{T}=-A\)
Karnataka CET-2017
Matrix and Determinant
78450
If \(A\) and \(B\) are square matrices of order ' \(n\) ' such that \(A^{2}-B^{2}=(A-B)(A+B)\), then which of the following will be true?
1 Either of A or B is zero matrix.
2 \(\mathrm{A}=\mathrm{B}\)
3 \(\mathrm{AB}=\mathrm{BA}\)
4 Either of \(\mathrm{A}\) or \(\mathrm{B}\) is an identity matrix.
Explanation:
(C)Given that,\(\mathrm{A}^{2}-\mathrm{B}^{2}=(\mathrm{A}-\mathrm{B})(\mathrm{A}+\mathrm{B})\) Here, \(\quad A, B\) are order \(n \times n\) \(\therefore \quad \mathrm{A}^{2}-\mathrm{B}^{2}=\mathrm{A}^{2}-\mathrm{BA}+\mathrm{AB}-\mathrm{B}^{2}\) \(\mathrm{~A}^{2}-\mathrm{B}^{2}-\mathrm{A}^{2}+\mathrm{B}^{2}=\mathrm{AB}-\mathrm{BA}\) \(0=\mathrm{AB}-\mathrm{BA}\) Therefore, \(\mathrm{AB}=\mathrm{BA}\)
78449
If a matrix \(A\) is both symmetric and skew symmetric, then
1 A is diagonal matrix
2 A is a zero matrix
3 A is scalar matrix
4 A is square matrix
Explanation:
(B) : According to question, \(\mathrm{A}\) is both symmetric and skew symmetric matrix it means - \(\begin{array}{lll} & \mathrm{A}^{\mathrm{T}}=\mathrm{A} & \text { (Symmetric) } \\ \text { And } & \mathrm{A}^{\mathrm{T}}=-\mathrm{A} & \text { (Skew-symmetric) } \\ \therefore & \mathrm{A}=\mathrm{A}^{\mathrm{T}}=-\mathrm{A} & \end{array}\) It is only possible in case of zero matrix, \(A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=A^{T}=-A\)
Karnataka CET-2017
Matrix and Determinant
78450
If \(A\) and \(B\) are square matrices of order ' \(n\) ' such that \(A^{2}-B^{2}=(A-B)(A+B)\), then which of the following will be true?
1 Either of A or B is zero matrix.
2 \(\mathrm{A}=\mathrm{B}\)
3 \(\mathrm{AB}=\mathrm{BA}\)
4 Either of \(\mathrm{A}\) or \(\mathrm{B}\) is an identity matrix.
Explanation:
(C)Given that,\(\mathrm{A}^{2}-\mathrm{B}^{2}=(\mathrm{A}-\mathrm{B})(\mathrm{A}+\mathrm{B})\) Here, \(\quad A, B\) are order \(n \times n\) \(\therefore \quad \mathrm{A}^{2}-\mathrm{B}^{2}=\mathrm{A}^{2}-\mathrm{BA}+\mathrm{AB}-\mathrm{B}^{2}\) \(\mathrm{~A}^{2}-\mathrm{B}^{2}-\mathrm{A}^{2}+\mathrm{B}^{2}=\mathrm{AB}-\mathrm{BA}\) \(0=\mathrm{AB}-\mathrm{BA}\) Therefore, \(\mathrm{AB}=\mathrm{BA}\)
