यदि A = [ rrr 0 & a & b -a & 0 & c -b & -c & 0 ] तो 1 2 (A + A^ ) तथा 1 2 (A - A^ ) ज्ञात कीजिए।

अब, 1 2 (A + A^ ) = 1 2 ( [ ccc 0 & a & b -a & 0 & c -b & -c & 0 ]+ [ ccc 0 & a & b -a & 0 & c -b & -c & 0 ]^ )= 1 2 ( [ ccc 0 & a & b -a & 0 & c -b & -c & 0 ]+ [ ccc 0 & -a & -b a & 0 & -c b & c & 0 ] ) = 1 2 [ lll 0 & 0 & 0 0 & 0 & 0 0 & 0 & 0 ] = [ lll 0 & 0 & 0 0 & 0 & 0 0 & 0 & 0 ] तथा 1 2 (A - A^ ) = 1 2 ( [ rrr 0 & a & b -a & 0 & c -b & -c & 0 ]- [ rrr 0 & a & b -a & 0 & c -b & -c & 0 ]^ ) = 1 2 ( [ ccc 0 & a & b -a & 0 & c -b & -c & 0 ]- [ rrr 0 & -a & -b a & 0 & -c b & c & 0 ] ) = 1 2 [ ccc 0 & 2 a & 2 b -2 a & 0 & 2 c -2 b & -2 c & 0 ] = [ ccc 0 & a & b -a & 0 & c -b & -c & 0 ]

यदि A = [ rrr 0 & a & b -a & 0 & c -b & -c & 0 ] तो 1 2 (A + A^ )…

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Question

यदि A = $ \left[\begin{array}{rrr} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]$ तो $ \frac{1}{2}$(A + A$^{\prime}$) तथा $ \frac{1}{2}$(A - A$^{\prime}$) ज्ञात कीजिए।

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Answer

अब, $ \frac{1}{2}$ (A + A$^{\prime}$)
= $\frac{1}{2}\left(\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]+\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]^{\prime}\right)$= $\frac{1}{2}\left(\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]+\left[\begin{array}{ccc} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right]\right)$
= $\frac{1}{2}$ $\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$ = $\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$
तथा $\frac{1}{2}$ (A - A$^{\prime}$) = $\frac{1}{2}\left(\left[\begin{array}{rrr}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]-\left[\begin{array}{rrr}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]^{\prime}\right)$
= $\frac{1}{2}$$ \left(\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]-\left[\begin{array}{rrr} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array}\right]\right)$ = $ \frac{1}{2}\left[\begin{array}{ccc} 0 & 2 a & 2 b \\ -2 a & 0 & 2 c \\ -2 b & -2 c & 0 \end{array}\right]$ = $\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right] $