Solve System of Equations with A^-1

Write the system in the vector matrix form Ax=b and solve by finding A^-1

Vector Matrix form?

[2,1;4,3]*[x;y] = [2,-4]

2x+y=2

4x+3x=-4

So A^-1 =

[1.5,-.5;-2,1]

So I don't understand what I am suppose to do with the A^-1. I understand how to use ref, rref and Cramer's rule to solve these equations but I am not understanding what method they are wanting me to use.

Reference: Answer is x=5, y=-8

Re: Solve System of Equations with A^-1

Quote:

Originally Posted by

**bustacap09** Write the system in the vector matrix form Ax=b and solve by finding A^-1

Vector Matrix form?

[2,1;4,3]*[x;y] = [2,-4]

2x+y=2

4x+3x=-4

So A^-1 =

[1.5,-.5;-2,1]

So I don't understand what I am suppose to do with the A^-1. I understand how to use ref, rref and Cramer's rule to solve these equations but I am not understanding what method they are wanting me to use.

Reference: Answer is x=5, y=-8

Well you have Au = v. You found A^{-1}, so then A^{-1}Au = A^{-1}v implies u = A^{-1}v.

Does that answer the question or am I not getting where you are confused?

-Dan

Re: Solve System of Equations with A^-1

"Well you have Au = v. You found A^{-1}, so then A^{-1}Au = A^{-1}v implies u = A^{-1}v."

I didn't understand how you used the A^-1 to find the values for [x,y]. But this makes perfect sense.

Thank you very much.

Re: Solve System of Equations with A^-1

If $\displaystyle A\overline{x}=\overline{b}$, then $\displaystyle \overline{x}=A^{-1}\overline{b}$. So, $\displaystyle \left(\begin{matrix}{\frac{3}{2} & -\frac{1}{2}\\-2 & 1\end{matrix}\right)\left(\begin{matrix}2\\-4\end{matrix}\right)=\left(\begin{matrix}5\\-8\end{matrix}\right)$. Straight up plug and chuggin'.