# Matrices are not my strong suit! Please help

#### TonyC

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

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

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#### Galileo

Science Advisor
Homework Helper
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.

#### HallsofIvy

Science Advisor
Homework Helper
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?

#### 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.

#### TonyC

Still trying to figure this out

#### VietDao29

Homework Helper
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,

#### 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 $A\mathbf{x}=I\implies\mathbf{x}=A^{-1}$.

#### neurocomp2003

SHOW YOUR WORK first.

#### TonyC

I posted the work I have been doing with Matrices previously, thanks for your concern

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