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B Is this matrix invertible?

  1. Aug 13, 2016 #1
    I tried to find the inverse of below matrix and what I get is no inverse.

    ##
    \left(
    \begin{array}{rrr}
    1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9
    \end{array}
    \right)
    ##

    Can someone please check it whether I am correct or not?
     
  2. jcsd
  3. Aug 13, 2016 #2

    fresh_42

    Staff: Mentor

    What have you done so far?
     
  4. Aug 13, 2016 #3
    This is what I get

    ##
    \left(
    \begin{array}{rrr|rrr}
    1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & \frac{4}{3} & -\frac{1}{3} & 0 \\ 0 & 0 & 0 & -1 & -2 & 1
    \end{array}
    \right)
    ##
     
  5. Aug 13, 2016 #4

    fresh_42

    Staff: Mentor

    Difficult to check something you cannot see.

    What is the determinant? Maybe the matrix isn't invertible.
    (You could compute the determinant along the first row. If it's zero, it isn't invertible.)

    You could also try the computational method with the adjugate matrix
    https://en.wikipedia.org/wiki/Adjugate_matrix#Inverses
    but I would compute the determinant first.
     
  6. Aug 13, 2016 #5
    Basically, I rewrote the matrix in the below form:

    ##
    \left(
    \begin{array}{rrr|rrr}
    1 & 2 & 3 & 1 & 0 & 0 \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 7 & 8 & 9 & 0 & 0 & 1
    \end{array}
    \right)
    ##

    Next, I do row operation, and what I get is this:

    ##
    \left(
    \begin{array}{rrr|rrr}
    1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & \frac{4}{3} & -\frac{1}{3} & 0 \\ 0 & 0 & 0 & -1 & -2 & 1
    \end{array}
    \right)
    ##
     
  7. Aug 13, 2016 #6

    fresh_42

    Staff: Mentor

    Assuming you made no mistake, what does this tell you?
     
  8. Aug 13, 2016 #7
    Is this inconsistent matrix? Does this mean the matrix has no solutions/invers?
     
  9. Aug 13, 2016 #8

    fresh_42

    Staff: Mentor

    I don't know what an inconsistent matrix is or what you mean by a solution, since there is nowhere an equation to solve.
    But it has no inverse, yes. Do you know how to compute a determinant and what it says? If not you should read about it, because it is essential when dealing with matrices.
     
  10. Aug 13, 2016 #9
    OK, let me find the determinant of below matrix by row 3 expansion.

    ##
    \left(
    \begin{array}{rrr}
    1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9
    \end{array}
    \right)
    ##

    ##a_{31} = 7, a_{32} = 8, a_{33} = 9##

    ##C_{31} = (-1)^{3+1}
    \left|
    \begin{array}{rr}
    2 & 3 \\ 5 & 6 \\
    \end{array}
    \right|
    = (-1)^{4}(2.6 - 3.5) = (1)(12 - 15) = (1)(-3) = -3##

    ##C_{32} = (-1)^{3+2}
    \left|
    \begin{array}{rr}
    1 & 3 \\ 4 & 6 \\
    \end{array}
    \right|
    = (-1)^{5}(1.6 - 3.4) = (-1)(6 - 12) = (-1)(-6) = 6##

    ##C_{33} = (-1)^{3+3}
    \left|
    \begin{array}{rr}
    1 & 2 \\ 4 & 5 \\
    \end{array}
    \right|
    = (-1)^{6}(1.5 - 2.4) = (1)(5 - 8) = (1)(-3) = -3##

    ##\det = a_{31}.C_{31} + a_{32}.C_{32} + a_{33}.C_{33}##
    ##= 7(-3) + 8(6) + 9(-3)##
    ##= -21 + 48 -27##
    ##= 0##
     
  11. Aug 13, 2016 #10

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    0 is correct.

    There are matrices without an inverse, this is one example. And there is a nice relation between the determinant of a matrix and the existence of an inverse matrix.
     
  12. Aug 13, 2016 #11

    Stephen Tashi

    User Avatar
    Science Advisor

    If your text materials want you to find the inverse of the matrix by row operations, we should investigate what row operations you did.
     
  13. Aug 14, 2016 #12
    This the row operation:

    ##
    \left(
    \begin{array}{rrr|rrr}
    1 & 2 & 3 & 1 & 0 & 0 \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 7 & 8 & 9 & 0 & 0 & 1
    \end{array}
    \right)
    ##

    R2 - 4R1 and R3 - 7R1

    ##
    \left(
    \begin{array}{rrr|rrr}
    1 & 2 & 3 & 1 & 0 & 0 \\ 0 & -3 & -6 & -4 & 1 & 0 \\ 0 & -6 & -12 & -7 & 0 & 1
    \end{array}
    \right)
    ##

    R2 x ##(-\frac{1}{3})##

    ##
    \left(
    \begin{array}{rrr|rrr}
    1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & \frac{4}{3} & -\frac{1}{3} & 0 \\ 0 & -6 & -12 & -7 & 0 & 1
    \end{array}
    \right)
    ##

    R3 + 6R2

    ##
    \left(
    \begin{array}{rrr|rrr}
    1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & \frac{4}{3} & -\frac{1}{3} & 0 \\ 0 & 0 & 0 & 1 & -2 & 1
    \end{array}
    \right)
    ##

    As you can see in the last matrix, the 3rd row has 0 0 0. It means that this matrix is inconsistent.
     
    Last edited: Aug 14, 2016
  14. Aug 14, 2016 #13

    Stephen Tashi

    User Avatar
    Science Advisor

    Your steps are correct.
     
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