The symmetric part of the matrix A = begin{bm | vedclass

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

Question

(B) The symmetric part of a square matrix $A$ is given by the formula $\frac{1}{2}(A + A^T)$.Given $A = \begin{bmatrix} 1 & 2 & 4 \ 6 & 8 & 2 \ 2 &

Vedclass · Accepted Answer

$ \begin{bmatrix} 1 & 4 & 3 \ 4 & 8 & 0 \ 3 & 0 & 7 \end{bmatrix} $

Vedclass · Answer

(B) The symmetric part of a square matrix $A$ is given by the formula $\frac{1}{2}(A + A^T)$.Given $A = \begin{bmatrix} 1 & 2 & 4 \ 6 & 8 & 2 \ 2 & -2 & 7 \end{bmatrix}$,its transpose $A^T$ is $\begin{bmatrix} 1 & 6 & 2 \ 2 & 8 & -2 \ 4 & 2 & 7 \end{bmatrix}$.Now,$A + A^T = \begin{bmatrix} 1+1 & 2+6 & 4+2 \ 6+2 & 8+8 & 2-2 \ 2+4 & -2+2 & 7+7 \end{bmatrix} = \begin{bmatrix} 2 & 8 & 6 \ 8 & 16 & 0 \ 6 & 0 & 14 \end{bmatrix}$.Finally,$\frac{1}{2}(A + A^T) = \frac{1}{2} \begin{bmatrix} 2 & 8 & 6 \ 8 & 16 & 0 \ 6 & 0 & 14 \end{bmatrix} = \begin{bmatrix} 1 & 4 & 3 \ 4 & 8 & 0 \ 3 & 0 & 7 \end{bmatrix}$.

The symmetric part of the matrix $A = \begin{bmatrix} 1 & 2 & 4 \\ 6 & 8 & 2 \\ 2 & -2 & 7 \end{bmatrix}$ is

Similar Questions

If $A$ and $B$ are square matrices of size $n \times n$ such that $A^2 - B^2 = (A - B)(A + B)$,then which of the following will be always true?

If $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$,then $(aI + bA)^n$ is (where $I$ is the identity matrix of order $2$)

If $A$ is a $3 \times 4$ matrix and $B$ is a matrix such that $A^{\prime}B$ and $BA^{\prime}$ are both defined,then $B$ is of the type:

If $A=\left[\begin{array}{cccc}2 & 1 & 3 & -1 \\ 1 & -2 & 2 & -3\end{array}\right]$,$B=\left[\begin{array}{cccc}2 & 1 & 0 & 3 \\ 1 & -1 & 2 & 3\end{array}\right]$ and $2A+3B-5C=0$,then $C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module