(B) Given that $A$ is a symmetric matrix,so $A^T = A$. Given that $B$ is a skew-symmetric matrix,so $B^T = -B$. We need to check the nature of the mat

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(B) Given that $A$ is a symmetric matrix,so $A^T = A$. Given that $B$ is a skew-symmetric matrix,so $B^T = -B$. We need to check the nature of the matrix $M = AB - BA$. Taking the transpose of $M$: $M^T = (AB - BA)^T = (AB)^T - (BA)^T$ Using the property $(XY)^T = Y^T X^T$: $M^T = B^T A^T - A^T B^T$ Substituting $A^T = A$ and $B^T = -B$: $M^T = (-B)(A) - (A)(-B)$ $M^T = -BA + AB = AB - BA = M$ Since $M^T = M$,the matrix $AB - BA$ is a symmetric matrix.

Class 12 Mathematics 3 and 4 .Determinants and Matrices Special types of matrices, Transpose of matrices and Trace of matrices

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Let $A$ and $B$ be any two $3 \times 3$ matrices. If $A$ is symmetric and $B$ is skew-symmetric,then the matrix $AB - BA$ is

JEE MAIN 2014 All JEE MAIN

A
skew-symmetric
B
symmetric
C
neither symmetric nor skew-symmetric
D
$I$ or $-I$,where $I$ is an identity matrix

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3 and 4 .Determinants and Matrices

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Special types of matrices, Transpose of matrices and Trace of matrices

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Let $A$ and $B$ be any two $3 \times 3$ matrices. If $A$ is symmetric and $B$ is skew-symmetric,then the matrix $AB - BA$ is

Similar Questions

An orthogonal matrix is

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$, then $(AA')' = $

If $A$ and $B$ are symmetric matrices of the same order,then show that $AB$ is symmetric if and only if $A$ and $B$ commute,that is $AB = BA$.

If $A^{\prime}=\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}$,then verify that $(A+B)^{\prime}=A^{\prime}+B^{\prime}$.

If $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 3 & 2 \\ 3 & 4 & 5\end{array}\right]$, then $(A+A^T)(A-A^T)=$

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