If A and B are square matrices of order 3,the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $S_A = A - A^T$. Since $S_A^T = (A - A^T)^T = A^T - A = -(A - A^T) = -S_A$,$S_A$ is a skew-symmetric matrix of order $3$.Similarly,let $S_B =

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $S_A = A - A^T$. Since $S_A^T = (A - A^T)^T = A^T - A = -(A - A^T) = -S_A$,$S_A$ is a skew-symmetric matrix of order $3$.Similarly,let $S_B = B - B^T$. Since $S_B^T = (B - B^T)^T = B^T - B = -(B - B^T) = -S_B$,$S_B$ is a skew-symmetric matrix of order $3$.Let $M = S_A + S_B$. Then $M^T = (S_A + S_B)^T = S_A^T + S_B^T = -S_A - S_B = -(S_A + S_B) = -M$.Thus,$M$ is a skew-symmetric matrix of order $3$.The determinant of a skew-symmetric matrix of odd order $n$ is always $0$,because $|M| = |M^T| = |-M| = (-1)^n |M| = -|M|$ for odd $n$,which implies $2|M| = 0$,so $|M| = 0$.Since $n = 3$ is odd,$|M| = |(A-A^T)+(B-B^T)| = 0$.

If $A$ and $B$ are square matrices of order $3$,then $|(A-A^T)+(B-B^T)|=$

Similar Questions

Let $A$ and $B$ be $3 \times 3$ real matrices such that $A$ is a symmetric matrix and $B$ is a skew-symmetric matrix. Then the system of linear equations $(A^2 B^2 - B^2 A^2) X = 0$,where $X$ is a $3 \times 1$ column matrix of unknown variables and $0$ is a $3 \times 1$ null matrix,has:

Let $B$ and $C$ be $n \times n$ matrices such that $A=B+C$, $BC=CB$, and $C^2=0$ (where $0$ is the null matrix). Then, $B^{2020}[B+(2021)C]=$

If $A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$,prove that $A^{n}=\begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix}$ for all $n \in N$.

Let $A$ and $B$ be two $3 \times 3$ real matrices such that $(A^{2}-B^{2})$ is an 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:

Let $A = \begin{bmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{bmatrix}$ and $B = A - I$. If $\omega = \frac{\sqrt{3}i - 1}{2}$, then the number of elements in the set $\{n \in \{1, 2, \ldots, 100\} : A^n + (\omega B)^n = A + B\}$ is equal to $..........$

Vedclass Test Series

Exam Paper Generator

Online Exam Module