1. Limited time only! Sign up for a free 30min personal 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!

How to find the inverse?

  1. Oct 15, 2015 #1
    1. The problem statement, all variables and given/known data
    The problem is posted below.

    2. Relevant equations
    In part b,I think I can use E1E2....A=I this property.
    But I cannot find the answer.

    3. The attempt at a solution
    By using the property,I multiply the inverse giving with σ2and σ1.
    However,the answer is not correct!
    So how to get the correct answer?

    Attached Files:

  2. jcsd
  3. Oct 15, 2015 #2


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    My neck hurts. Doesn't it say something in the guidelines about posting pictures ?
  4. Oct 15, 2015 #3


    User Avatar
    Science Advisor

    It's not that hard to type in the matrix itself. Also, if you know the basic definitions, you should not have much difficulty deciding what elementary matrices are used. For example, we are given the matrix,
    [tex]A_1= \begin{bmatrix}1 & 2 & 1 \\ 2 & 5 & 4 \\ 0 & -1 & -3 \end{bmatrix}[/tex]
    and are asked what elementary matrix, [itex]\sigma_1[/itex] converts that to
    [tex]A_2= \begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & -1 & -3 \end{bmatrix}[/tex]

    There are three kinds of row operations:
    1) multiply every number in a single row by a number
    2) swap two rows
    3) add a multiple of one row to another
    Comparing [itex]A_1[/itex] and [itex]A_2[/itex], we see that rows 1 and 2 have not changed so we have not swapped two rows. The second row has changed from [2 5 4] to [0 1 2]. If we were multiplying by a number, to go from "2" to "0" we would have to multiply by 0 but then we would have "0" in all positions so the row operation was not multiplying by a number. And since the first number in the last row is 0, adding any multiple of that row to the second row would not have changed that number. So the new second row must be [2 5 4] plus a multiple of [1 2 1]. To go from 2 to 0, we must have subtracted 2 which is the same as adding -2 times 1. Adding -2 times 2 to 5 gives 1 and adding -2 times 1 to 4 gives 2. That is exactly what we wanted. So the row operation must be "add -2 times each number in the first row to the corresponding number in the second row".

    To get the elementary matrix that gives that, do that row operation to the identity matrix. Adding -2 times each member in the first row to the corresponding number in the second row of the identity matrix gives
    [tex]\begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}[/tex]
    That is the elementary matrix [itex]\sigma_1[/itex].

    Now, continue like that
  5. Oct 17, 2015 #4
    ...I know how to find the corresponding elementary matrix,but I don't know how to find the inverse using the A-1:frown:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted