Question

2X+3Y=matrix [2 3,4 0] and 3X+ 2Y =[2 -2, -1 5]

Answer 1

Question:

Find matrices 'X' and 'Y' such that,

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

Solution:

First we add the given two matrices.

$\begin{aligned}&(2X+3Y)+(3X+2Y)&=&\ \ \left[\begin{array}{cc}2&3\\ 4&0\end{array}\right]+\left[\begin{array}{cc}2&-2\\ -1&5\end{array}\right]\\ \\ &2X+3Y+3X+2Y&=&\ \ \left[\begin{array}{cc}2+2&3+(-2)\\ 4+(-1)&0+5\end{array}\right]\\ \\ &5X+5Y&=&\ \ \left[\begin{array}{cc}4&1\\ 3&5\end{array}\right]\\ \\ &5(X+Y)&=&\ \ \left[\begin{array}{cc}4&1\\ 3&5\end{array}\right]\end{aligned}$

Now we divide both sides by 5. In RHS we multiply the obtained matrix by 1/5.

$\begin{aligned}&X+Y&=&\ \ \dfrac{1}{5}\left[\begin{array}{cc}4&1\\ 3&5\end{array}\right]\\ \\ &X+Y&=&\ \ \left[\begin{array}{cc}0.8&0.2\\ 0.6&1\end{array}\right]\end{aligned}$

Now we double both sides.

$\begin{aligned}&2(X+Y)&=&\ \ 2\left[\begin{array}{cc}0.8&0.2\\ 0.6&1\end{array}\right]\\ \\ &2X+2Y&=&\ \ \left[\begin{array}{cc}1.6&0.4\\ 1.2&2\end{array}\right]\end{aligned}$

Now this matrix is subtracted from each given matrix in the question,

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

So,

$\begin{aligned}&(3X+2Y)-(2X+2Y)&=&\ \ \left[\begin{array}{cc}2&-2\\ -1&5\end{array}\right]-\left[\begin{array}{cc}1.6&0.4\\ 1.2&2\end{array}\right]\\ \\ &3X+2Y-2X-2Y&=&\ \ \left[\begin{array}{cc}2-1.6&-2-0.4\\ -1-1.2&5-2\end{array}\right]\\ \\ &\textbf{X}&=&\ \ \mathbf{\left[\begin{array}{cc}0.4&-2.4\\ -2.2&3\end{array}\right]}\end{aligned}$

and,

$\begin{aligned}&(2X+3Y)-(2X+2Y)&=&\ \ \left[\begin{array}{cc}2&3\\ 4&0\end{array}\right]-\left[\begin{array}{cc}1.6&0.4\\ 1.2&2\end{array}\right]\\ \\ &2X+3Y-2X-2Y&=&\ \ \left[\begin{array}{cc}2-1.6&3-0.4\\ 4-1.2&0-2\end{array}\right]\\ \\ &\textbf{Y}&=&\ \ \mathbf{\left[\begin{array}{cc}0.4&2.6\\ 2.8&-2\end{array}\right]}\end{aligned}$

Hence Found!