Find X and Y, if 2X+3Y=left[begin{array}{cc}2 | vedclass

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(A) Given equations are:$2X+3Y=\left[\begin{array}{cc}2 & 3 \ 4 & 0\end{array}ight]$  ... $(1)$$3X+2Y=\left[\begin{array}{cc}2 & -2 \ -1 & 5\end{a

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$X=\left[\begin{array}{cc}\frac{2}{5} & -\frac{12}{5} \ -\frac{11}{5} & 3\end{array}ight], Y=\left[\begin{array}{cc}\frac{2}{5} & \frac{13}{5} \ \frac{14}{5} & -2\end{array}ight]$

Vedclass · Answer

(A) Given equations are:$2X+3Y=\left[\begin{array}{cc}2 & 3 \ 4 & 0\end{array}ight]$  ... $(1)$$3X+2Y=\left[\begin{array}{cc}2 & -2 \ -1 & 5\end{array}ight]$ ... $(2)$Multiply equation $(1)$ by $2$ and equation $(2)$ by $3$:$4X+6Y=\left[\begin{array}{cc}4 & 6 \ 8 & 0\end{array}ight]$ ... $(3)$$9X+6Y=\left[\begin{array}{cc}6 & -6 \ -3 & 15\end{array}ight]$ ... $(4)$Subtracting $(3)$ from $(4)$:$(9X+6Y)-(4X+6Y)=\left[\begin{array}{cc}6 & -6 \ -3 & 15\end{array}ight]-\left[\begin{array}{cc}4 & 6 \ 8 & 0\end{array}ight]$$5X=\left[\begin{array}{cc}2 & -12 \ -11 & 15\end{array}ight] \Rightarrow X=\left[\begin{array}{cc}\frac{2}{5} & -\frac{12}{5} \ -\frac{11}{5} & 3\end{array}ight]$Now,substitute $X$ in $(1)$:$3Y=\left[\begin{array}{cc}2 & 3 \ 4 & 0\end{array}ight]-2\left[\begin{array}{cc}\frac{2}{5} & -\frac{12}{5} \ -\frac{11}{5} & 3\end{array}ight]$$3Y=\left[\begin{array}{cc}2 & 3 \ 4 & 0\end{array}ight]-\left[\begin{array}{cc}\frac{4}{5} & -\frac{24}{5} \ -\frac{22}{5} & 6\end{array}ight]=\left[\begin{array}{cc}\frac{6}{5} & \frac{39}{5} \ \frac{42}{5} & -6\end{array}ight]$$Y=\left[\begin{array}{cc}\frac{2}{5} & \frac{13}{5} \ \frac{14}{5} & -2\end{array}ight]$

Find $X$ and $Y,$ if $2X+3Y=\left[\begin{array}{cc}2 & 3 \\ 4 & 0\end{array}\right]$ and $3X+2Y=\left[\begin{array}{cc}2 & -2 \\ -1 & 5\end{array}\right].$

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