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Homework Help: 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

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    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

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    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

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    Staff Emeritus
    Science Advisor

    No worries
     
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