# Finding inverse of a matrix and solving given equations.

1. May 14, 2008

### cmcc3119

[SOLVED] Finding inverse of a matrix and solving given equations.

Hi,

I have a quiz tomorrow and I am looking at the sample quiz right now but I don't understand how the answers work!

Here is the question:

Find the inverse of the matrix A=

-1 3 3
-2 1 2
0 1 1

Hence solve the following systems of equations:

-x1 + 3x2 + 3x3 = 5
-2x1 + x2 + 2x3 = 2
x2 + x3 = 1

and

-x + 3z = -5
2x - y - 4z = 2
-2x + y + 5z = -1

The answer says the inverse of A
-1 0 3
2 -1 -4
-2 1 5

Then it spells out how to solve the equations but the is the part that confuses me:

"The first system of equations corresponds to the matrix equation A
[x1 = [5
x2 = 2
x3 ] = 1]

Therefore, as A is invertible (i.e. A-1 exists),

[x1
x2
x3]

=A inverse of

[5
2
1]

which =

[-1 0 3 [5 [-2
2 -1 -4 2 = 4
-2 1 5 ] 1] -3 ]

So x1 = -2, x2 = 4, x3 = -3.

Can someone please explain how they just did that because I really can't understand where they derive -2, 4 , 3 from and what they do with [5, 2 ,1]

Last edited: May 14, 2008
2. May 14, 2008

### cristo

Staff Emeritus
So you've got this matrix $$\left(\begin{array}{ccc} -1 & 3 & 3 \\ -2&1&2 \\ 0&1&1 \end{array}\right)$$ and you set of equations.

The first thing to note is that your equations are equal to $$\left(\begin{array}{ccc} -1 & 3 & 3 \\-2&1&2 \\ 0&1&1 \end{array}\right) \left(\begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array}\right)=\left(\begin{array}{c}5\\2\\1\end{array}\right)$$.

So, in order to find (x1,x2,x3)T you need to multiply the equation on the left by the inverse of A. This will give (x1,x2,x3)T= something.. where you can calculate the "something" by matrix algebra.

3. May 14, 2008

### cmcc3119

So how would I know that I have to multiply the left by the inverse of A to find the answer? Is there some basic principle or law that I am unaware of?

4. May 14, 2008

### cristo

Staff Emeritus
Well, yes. The easiest way to solve the equation Ax=b (for a matrix A and vectors x and b) is to try and get x on it's own, no? Now, you know that $AA^{-1}=A^{-1}A=I$, so you know that multiplying the equation by A inverse will leave you with $x=A^{-1}b$, from which you can read off values for x.

5. May 14, 2008

### cmcc3119

Its so hard to put all these number and boxes back into simple ax = b and again rearrange. Thanks cristo for your help!

6. May 14, 2008

### cristo

Staff Emeritus
No worries