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)=(1)/(4\cdot6-3\cdot(-4))\left[\begin{array}{cc}6&-3\\4&4\end{array}\right]=(1)/(24+12)\left[\begin{array}{cc}6&-3\\4&4\end{array}\right]=`


`=(1)/(36)\left[\begin{array}{cc}6&-3\\4&4\end{array}\right]=\left[\begin{array}{cc}(1)/(6)&-(1)/(12)\\\\(1)/(9)&(1)/(9)\end{array}\right]`

so:


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


`=\left[\begin{array}{c}(24)/(12)\\\\(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)=(1)/(2\cdot6-6\cdot6)\left[\begin{array}{cc}6&-6\\-6&2\end{array}\right]=(1)/(12-36)\left[\begin{array}{cc}6&-6\\-6&2\end{array}\right]=`


`=(1)/(-24)\left[\begin{array}{cc}6&-6\\-6&2\end{array}\right]=\left[\begin{array}{cc}-(1)/(4)&(1)/(4)\\\\(1)/(4)&-(1)/(12)\end{array}\right]`

so:


`\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}-(1)/(4)&(1)/(4)\\\\(1)/(4)&-(1)/(12)\end{array}\right]\left[\begin{array}{c}-16\\-24\end{array}\right]=\left[\begin{array}{c}-(-16)/(4)+(-24)/(4)\\\\(-16)/(4)-(-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)=(1)/(-2\cdot(-4)-3\cdot2)\left[\begin{array}{cc}-4&-3\\-2&-2\end{array}\right]=(1)/(8-6)\left[\begin{array}{cc}-4&-3\\-2&-2\end{array}\right]=`


`=(1)/(2)\left[\begin{array}{cc}-4&-3\\-2&-2\end{array}\right]=\left[\begin{array}{cc}-2&-(3)/(2)\\\\-1&-1\end{array}\right]`

so:


`\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}-2&-(3)/(2)\\\\-1&-1\end{array}\right]\left[\begin{array}{c}9\\-16\end{array}\right]=\left[\begin{array}{c}-2\cdot9-(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)}`