Question

I need help answering these problems. Im dealing with Systems of Linear Equations using an inverse. I posted a pic on how to do it, plz show the work

Answer 1

1.

$\begin{cases}4x+3y=11\\-4x+6y=-2\end{cases}\\\\\\\\ \left[\begin{array}{cc}4&3\\-4&6\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}11\\-2\end{array}\right]\\\\\\\\A=\left[\begin{array}{cc}4&3\\-4&6\end{array}\right]\\\\\\\\A^{-1}=\dfrac{1}{4\cdot6-3\cdot(-4)}\left[\begin{array}{cc}6&-3\\4&4\end{array}\right]=\dfrac{1}{24+12}\left[\begin{array}{cc}6&-3\\4&4\end{array}\right]=$

$=\dfrac{1}{36}\left[\begin{array}{cc}6&-3\\4&4\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6}&-\frac{1}{12}\\\\\frac{1}{9}&\frac{1}{9}\end{array}\right]$

so:

$\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6}&-\frac{1}{12}\\\\\frac{1}{9}&\frac{1}{9}\end{array}\right]\left[\begin{array}{c}11\\-2\end{array}\right]=\left[\begin{array}{c}\frac{11}{6}-\frac{-2}{12}\\\\\frac{11}{9}+\frac{-2}{9}\end{array}\right]=\left[\begin{array}{c}\frac{22}{12}+\frac{2}{12}\\\\\frac{11}{9}-\frac{2}{9}\end{array}\right]=$

$=\left[\begin{array}{c}\frac{24}{12}\\\\\frac{9}{9}\end{array}\right]=\left[\begin{array}{c}2\\1\end{array}\right]$

Answer is $\boxed{(2,1)}$


2.

$\begin{cases}2x+6y=-16\\6x+6y=-24\end{cases}\\\\\\\\ \left[\begin{array}{cc}2&6\\6&6\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}-16\\-24\end{array}\right]\\\\\\\\A=\left[\begin{array}{cc}2&6\\6&6\end{array}\right]\\\\\\\\A^{-1}=\dfrac{1}{2\cdot6-6\cdot6}\left[\begin{array}{cc}6&-6\\-6&2\end{array}\right]=\dfrac{1}{12-36}\left[\begin{array}{cc}6&-6\\-6&2\end{array}\right]=$

$=\dfrac{1}{-24}\left[\begin{array}{cc}6&-6\\-6&2\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{4}&\frac{1}{4}\\\\\frac{1}{4}&-\frac{1}{12}\end{array}\right]$

so:

$\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{4}&\frac{1}{4}\\\\\frac{1}{4}&-\frac{1}{12}\end{array}\right]\left[\begin{array}{c}-16\\-24\end{array}\right]=\left[\begin{array}{c}-\frac{-16}{4}+\frac{-24}{4}\\\\\frac{-16}{4}-\frac{-24}{12}\end{array}\right]=\left[\begin{array}{c}4-6\\-4+2\end{array}\right]=\\\\\\=\left[\begin{array}{c}-2\\-2\end{array}\right]$

Answer is $\boxed{(-2,-2)}$


3.

$\begin{cases}-2x+3y=9\\2x-4y=-16\end{cases}\\\\\\\\ \left[\begin{array}{cc}-2&3\\2&-4\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}9\\-16\end{array}\right]\\\\\\\\A=\left[\begin{array}{cc}-2&3\\2&-4\end{array}\right]\\\\\\\\A^{-1}=\dfrac{1}{-2\cdot(-4)-3\cdot2}\left[\begin{array}{cc}-4&-3\\-2&-2\end{array}\right]=\dfrac{1}{8-6}\left[\begin{array}{cc}-4&-3\\-2&-2\end{array}\right]=$

$=\dfrac{1}{2}\left[\begin{array}{cc}-4&-3\\-2&-2\end{array}\right]=\left[\begin{array}{cc}-2&-\frac{3}{2}\\\\-1&-1\end{array}\right]$

so:

$\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}-2&-\frac{3}{2}\\\\-1&-1\end{array}\right]\left[\begin{array}{c}9\\-16\end{array}\right]=\left[\begin{array}{c}-2\cdot9-\frac{3\cdot(-16)}{2}\\\\-9-(-16)\end{array}\right]=\\\\\\=\left[\begin{array}{c}-18+24\\-9+16\end{array}\right]=\left[\begin{array}{c}6\\7\end{array}\right]$

Answer is $\boxed{(6,7)}$