Question

Find X and Y,if \((i)X+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}\) and \(X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}\)\((ii)2X+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}\)and \(3X+2Y=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}\)

Answer 1

\((i)X+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}-(1)\)
\(X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}-(2)\)
Adding equations (1) and (2), we get:
\(2X=\begin{bmatrix}7&0\\2&5\end{bmatrix}+\begin{bmatrix}3&0\\0&3\end{bmatrix}\)\(=\begin{bmatrix}7+3& 0+0\\ 2+0& 5+3\end{bmatrix}\)\(=\begin{bmatrix}10& 0\\ 2& 8\end{bmatrix}\)
\(X=\frac{1}{2}\begin{bmatrix}10& 0\\ 2& 8\end{bmatrix}=\begin{bmatrix}5&0\\1&4\end{bmatrix}\)
Now \(X+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}\)
\(\implies\begin{bmatrix}5&0\\1&4\end{bmatrix}+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}\)
\(\implies Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}-\begin{bmatrix}5&0\\1&4\end{bmatrix}\)
\(\implies Y=\begin{bmatrix}7-5& 0-0\\ 2-1& 5-4\end{bmatrix}\)
\(\therefore Y=\begin{bmatrix}2&0\\1&1\end{bmatrix}\)
\((ii)2X+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}-(3)\)
\(3X+2Y=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}-(4)\)
Multiplying equation (3) with (2), we get:
\(2(2X+3Y)=2\begin{bmatrix}2&3\\4&0\end{bmatrix}\)
\(\implies 4X+6Y=\begin{bmatrix}4&6\\8&0\end{bmatrix}....(5)\)
Multiplying equation (4) with (3), we get:
\(3(3X+2Y)=3\begin{bmatrix}2&-2\\-1&5\end{bmatrix}\)
\(\implies 9X+6Y=\begin{bmatrix}6&-6\\-3&15\end{bmatrix}....(6)\)
From (5) and (6), we have: 
\((4X+6Y)-(9X+6Y)\)
\(=\begin{bmatrix}4&6\\8&0\end{bmatrix}-\begin{bmatrix}6&-6\\-3&15\end{bmatrix}\)
\(\implies-5X=\begin{bmatrix}4-6& 6-(-6)\\ 8-(-3)& 0-15\end{bmatrix}=\begin{bmatrix}-2& 12\\ 11& -15\end{bmatrix}\)
\(\therefore X=\frac{-1}{5}\begin{bmatrix}-2& 12\\ 11& -15\end{bmatrix}=\begin{bmatrix}\frac{2}{5}& \frac{-12}{5}\\ \frac{-11}{5}& 3\end{bmatrix}\)
Now,\(2X+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}\)
\(\implies 2\begin{bmatrix}\frac{2}{5}& \frac{-12}{5}\\ \frac{-11}{5}& 3\end{bmatrix}+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}\)
\(\implies \begin{bmatrix}\frac{4}{5}& \frac{-24}{5}\\ \frac{-22}{5}& 6\end{bmatrix}+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}\)
\(\implies 3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}-\begin{bmatrix}\frac{4}{5}& \frac{-24}{5}\\ \frac{-22}{5}& 6\end{bmatrix}\)

\(\implies 3Y=\begin{bmatrix}2-\frac{4}{5}& 3+\frac{24}{5}\\ 4+\frac{22}{5}& 0-6\end{bmatrix}=\begin{bmatrix}\frac{6}{5}& \frac{39}{5}\\ \frac{42}{5}& -6\end{bmatrix}\)

\(\therefore Y=\frac{1}{3}\begin{bmatrix}\frac{6}{5}&\frac{ 39}{5}\\ \frac{42}{5}&-6\end{bmatrix}==\begin{bmatrix}\frac{2}{5}& \frac{13}{5}\\ \frac{14}{5}& -2\end{bmatrix}\)