Find X and Y,if X+Y=left[begin{array}{ll}7 & | vedclass

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(A) Given equations are:$X+Y=\left[\begin{array}{ll}7 & 0 \ 2 & 5\end{array}ight]$  ............ $(1)$$X-Y=\left[\begin{array}{ll}3 & 0 \ 0 & 3\en

Vedclass · Accepted Answer

$X = \left[\begin{array}{ll}5 & 0 \ 1 & 4\end{array}ight]$,$Y=\left[\begin{array}{ll}2 & 0 \ 1 & 1\end{array}ight]$

Vedclass · Answer

(A) Given equations are:$X+Y=\left[\begin{array}{ll}7 & 0 \ 2 & 5\end{array}ight]$  ............ $(1)$$X-Y=\left[\begin{array}{ll}3 & 0 \ 0 & 3\end{array}ight]$  ............ $(2)$Adding equations $(1)$ and $(2)$,we get:$2X = \left[\begin{array}{ll}7 & 0 \ 2 & 5\end{array}ight] + \left[\begin{array}{ll}3 & 0 \ 0 & 3\end{array}ight] = \left[\begin{array}{ll}7+3 & 0+0 \ 2+0 & 5+3\end{array}ight] = \left[\begin{array}{ll}10 & 0 \ 2 & 8\end{array}ight]$$	herefore X = \frac{1}{2}\left[\begin{array}{ll}10 & 0 \ 2 & 8\end{array}ight] = \left[\begin{array}{ll}5 & 0 \ 1 & 4\end{array}ight]$Now,substituting $X$ in equation $(1)$:$\left[\begin{array}{ll}5 & 0 \ 1 & 4\end{array}ight] + Y = \left[\begin{array}{ll}7 & 0 \ 2 & 5\end{array}ight]$$Y = \left[\begin{array}{ll}7 & 0 \ 2 & 5\end{array}ight] - \left[\begin{array}{ll}5 & 0 \ 1 & 4\end{array}ight]$$Y = \left[\begin{array}{ll}7-5 & 0-0 \ 2-1 & 5-4\end{array}ight] = \left[\begin{array}{ll}2 & 0 \ 1 & 1\end{array}ight]$Thus,$X = \left[\begin{array}{ll}5 & 0 \ 1 & 4\end{array}ight]$ and $Y = \left[\begin{array}{ll}2 & 0 \ 1 & 1\end{array}ight]$.

Find $X$ and $Y$,if $X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$ and $X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$.

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