यदि A^ = [ rr 3 & 4 -1 & 2 0 & 1 ] तथा B = [ rrr -1 & 2 & 1 1 & 2 & 3 ] हैं तो सत्यापित कीजिए कि (A + B)^ = A^ + B^

A = (A^ )^ = [ rr 3 & 4 -1 & 2 0 & 1 ]^ = [ ccc 3 & -1 & 0 4 & 2 & 1 ] यहाँ, A + B = [ ccc 3 & -1 & 0 4 & 2 & 1 ] + [ ccc -1 & 2 & 1 1 & 2 & 3 ] = [ lll 2 & 1 & 1 5 & 4 & 4 ] बायाँ पक्ष = (A + B)^ = [ lll 2 & 1 & 1 5 & 4 & 4 ] = [ ll 2 & 5 1 & 4 1 & 4 ] A^ [ cc 3 & 4 -1 & 2 0 & 1 ] तथा B^ = [ ccc -1 & 2 & 1 1 & 2 & 3 ]^ = [ cc -1 & 1 2 & 2 1 & 3 ], यहाँ, दायाँ पक्ष = A^ + B^ = [ cc 3 & 4 -1 & 2 0 & 1 ] + [ cc -1 & 1 2 & 2 1 & 3 ] = [ ll 2 & 5 1 & 4 1 & 4 ] अतः (A + B)^ = A^ + B^ सत्यापित हुआ।

यदि A^ = [ rr 3 & 4 -1 & 2 0 & 1 ] तथा B = [ rrr -1 & 2 & 1 1 & 2…

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यदि $\mathrm{A}^{\prime}$ = $ \left[\begin{array}{rr} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{array}\right]$ तथा B = $ \left[\begin{array}{rrr} -1 & 2 & 1 \\ 1 & 2 & 3 \end{array}\right] $ हैं तो सत्यापित कीजिए कि (A + B)$^{\prime}$ = A$^{\prime}$ + B$^{\prime}$

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Answer

$\because$ A = $\left(A^{\prime}\right)^{\prime}$ = $\left[\begin{array}{rr} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{array}\right]^{\prime}$ = $ \left[\begin{array}{ccc} 3 & -1 & 0 \\ 4 & 2 & 1 \end{array}\right]$
यहाँ, A + B = $ \left[\begin{array}{ccc}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]$ + $ \left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$ = $\left[\begin{array}{lll}2 & 1 & 1 \\ 5 & 4 & 4\end{array}\right]$
$ \therefore$ बायाँ पक्ष = (A + B)$^{\prime}$ = $ \left[\begin{array}{lll}2 & 1 & 1 \\ 5 & 4 & 4\end{array}\right]$ = $\left[\begin{array}{ll}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right]$
A$^{\prime}$ $\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]$ तथा B$^{\prime}$ = $\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]^{\prime}$= $ \left[\begin{array}{cc}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]$,
यहाँ, दायाँ पक्ष = A$^{\prime}$ + B$^{\prime}$ = $\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]$ +$ \left[\begin{array}{cc}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]$ = $ \left[\begin{array}{ll}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right]$
अतः (A + B)$^{\prime}$ = A$^{\prime}$ + B$^{\prime}$ सत्यापित हुआ।

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