1. Sep 2, 2005

### TonyC

How would I go about solving a systme of equations using an inverse matrix?

i.e. -10x - 6y = 6
7x + y = -7

2. Sep 2, 2005

### Galileo

As I`m sure this is stated in your book (it should be), this system of equations is equivalent to the single matrix equation:

$$\left( \begin{array}{cc} -10 & -6 \\ 7 & 1\end{array}\right)\left( \begin{array}{c} x \\ y \end{array}\right)=\left( \begin{array}{c} 6 \\ -7 \end{array}\right)$$

In general, you can write a linear system of n equations in n unknowns as a matrix equation Ax=b. If you can find the inverse of A (let's call it C, so that CA=AC=I) you have solved the problem, just put x=Cb, then A(Cb)=(AC)b=b.

3. Sep 2, 2005

### HallsofIvy

Staff Emeritus
Your question was "How would I go about solving a systme of equations using an inverse matrix?" Are you saying you don't have any problem with actually finding the inverse matrix?

4. Sep 2, 2005

### TonyC

Not exactly sure what I am saying. I am not having trouble with any other methods,
i.e. Cramer Method, Gaussian, elementary row operations. This one has me truly stumped. Maybe I'm too tied (or old) to grasp it today.

5. Sep 2, 2005

### TonyC

Still trying to figure this out

6. Sep 2, 2005

### VietDao29

So can you find the inverse of A?????
If:
$$A:= \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$$
Then:
$$A ^ {-1}= \frac{1}{ad - bc} \left[ \begin{array}{cc} d & -b \\ -c & a \end{array} \right]$$
After finding the inverse of A, note that:
$$Ax = b \Leftrightarrow A ^ {-1}Ax = A ^ {-1}b \Leftrightarrow x = A ^ {-1}b$$
Viet Dao,

7. Sep 2, 2005

### amcavoy

Although the above method will work (for finding the inverse), you might want to try it a different way. It will help when you get to larger matrices (at least until you learn the generalized form of the above equation).

$$\left[\begin{array}{cc|cc}a & b & 1 & 0 \\ c & d & 0 & 1\end{array}\right]$$

Use row operations to make the left side (your original matrix) the identity matrix, and what is left on the right will be the inverse. You are basically saying $A\mathbf{x}=I\implies\mathbf{x}=A^{-1}$.

8. Sep 2, 2005