• Support PF! Buy your school textbooks, materials and every day products Here!

Finding inverse of a matrix and solving given equations.

  • Thread starter cmcc3119
  • Start date
  • #1
16
1
[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:

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
8,107
72
So you've got this matrix [tex]\left(\begin{array}{ccc} -1 & 3 & 3 \\
-2&1&2 \\
0&1&1
\end{array}\right) [/tex] and you set of equations.

The first thing to note is that your equations are equal to [tex]\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) [/tex].

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
16
1
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
cristo
Staff Emeritus
Science Advisor
8,107
72
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 [itex]AA^{-1}=A^{-1}A=I[/itex], so you know that multiplying the equation by A inverse will leave you with [itex]x=A^{-1}b[/itex], from which you can read off values for x.
 
  • #5
16
1
Its so hard to put all these number and boxes back into simple ax = b and again rearrange. Thanks cristo for your help!
 
  • #6
cristo
Staff Emeritus
Science Advisor
8,107
72
No worries
 

Related Threads for: Finding inverse of a matrix and solving given equations.

Replies
6
Views
4K
  • Last Post
Replies
3
Views
477
Replies
4
Views
958
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
22
Views
3K
Replies
6
Views
1K
Replies
5
Views
6K
Replies
4
Views
20K
Top