1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding inverse of a matrix and solving given equations.

  1. May 14, 2008 #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: May 14, 2008
  2. jcsd
  3. May 14, 2008 #2

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  4. May 14, 2008 #3
    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?
     
  5. May 14, 2008 #4

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  6. May 14, 2008 #5
    Its so hard to put all these number and boxes back into simple ax = b and again rearrange. Thanks cristo for your help!
     
  7. May 14, 2008 #6

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    No worries
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Finding inverse of a matrix and solving given equations.
Loading...