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  • Find 1 / 2 (A + A') and 1 / 2 (A - A'), when A = [ 0 a b - a 0 c]

Q9.    Find \frac{1}{2}(A+A') and \frac{1}{2}(A-A'), when A = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}

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A = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}

the transpose of the matrix is obtained by interchanging rows and columns

A' = \begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix}

\frac{1}{2}(A+A') = \frac{1}{2}(\begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}  +\begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix})

\frac{1}{2}(A+A') = \frac{1}{2}(\begin{bmatrix} 0+0 & a+(-a) & b+(-b)\\ -a+a & 0+0 & c+(-c)\\ -b+b & -c+c & 0+0 \end{bmatrix})

\frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}

\frac{1}{2}(A+A') = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}

\frac{1}{2}(A+A') = 0

 

\frac{1}{2}(A-A') = \frac{1}{2}(\begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}- \begin{bmatrix} 0 & -a & -b\\ a & 0 & -c\\ b & c & 0 \end{bmatrix})

\frac{1}{2}(A-A') = \frac{1}{2}(\begin{bmatrix} 0-0 & a-(-a) & b-(-b)\\ -a-a & 0-0 & c-(-c)\\ -b-b & -c-c & 0-0 \end{bmatrix})

\frac{1}{2}(A-A') = \frac{1}{2}\begin{bmatrix} 0 & 2a &2 b\\ -2a & 0 & 2c\\ -2b & -2c & 0 \end{bmatrix}

\frac{1}{2}(A-A') = \begin{bmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}

 

 

 

 

Posted by

seema garhwal

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