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Homework Help: Find All Possible Matrix Inverses

  1. Sep 5, 2015 #1


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    1. The problem statement, all variables and given/known data

    Find 2x2 matrices A and B, all of whose entries are \begin{align} &\geq 0 \end{align}, such that A^-1 and B^-1 exist, but (A+B)^-1 does not exist.

    2. Relevant equations
    The insverse is defined as 1/determinat(matrix) * adj(matrix)

    Otherwise shown as:
    d & -b \\
    -c & a

    3. The attempt at a solution

    My idea was to write it all out in unknown variable form. But I came to a problem.

    (d_1+d_2) & -(b_1+b_2) \\
    -(c_1+c_2) & (a_1+a_2)

    I then noted that the matrix inverse does not exist if:

    I then realized an issue with my method:
    1. There are many possible situations that this occurs.
    2. How do I check that A^-1 and B^-1 exist in this scenario.
    So it may seem like I am doing something wrong.

    Any helpful words of advice here?
  2. jcsd
  3. Sep 5, 2015 #2


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    That's fine; you only need to find one of them.

    The determinants of A and B will be non-zero.

    Fix A to be the identity, which we know to be invertible. That leaves you with four variables to play with in order to achieve [itex]\det(A + B) = 0[/itex] and [itex]\det(B) \neq 0[/itex].
  4. Sep 5, 2015 #3


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    You're making this more difficult than it needs to be.

    Look at the general form for the inverse of a 2×2 matrix.

    ##\displaystyle \ \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} =

    \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \ ##

    Under what conditions does this inverse not exist?
  5. Sep 5, 2015 #4


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    I like the ideas that you presented and ran with them.


    This allowed me to get det [itex](A+B) = (1+a_2+d_2)+(a_2d_2-c_2b_2)=0[/itex]

    My other concern is det(B) cannot equal 0. So [itex]a_2d_2-c_2b_2 =/= 0[/itex]

    I see the connection between these two, but I'm trying to put the final touches on this.

    So [itex](1+a_2+d_2) =/= -(a_2d_2-c_2b_2)[/itex]
  6. Sep 5, 2015 #5


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    It's better to deal with equations than inequalities. Stick with [tex]1 + (a_2 + d_2) + (a_2d_2 - c_2b_2) = 0.[/tex] Start setting variables to zero until only two unknowns remain and see what happens. You do have the constraints that [itex]a_2d_2[/itex] and [itex]b_2c_2[/itex] are not both zero and none of the variables can be negative.
  7. Sep 7, 2015 #6


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    Okay, I'm back at this problem set after a bit of a break.
    Let's see:

    So I can manageably get to the equation
    by setting A as the identity Matrix.

    I see that there is a connection in the sense that:
    [itex](a_2d_2-c_2b_2) =/= 0[/itex]

    So we should be able to represent it as an unknown variable, say, x.

    [itex] 1+(a_2+d_2)+x=0.[/itex]


    Since our initial condition is all the components must be positive numbers, we see that x has to be a negative value for this to work.
    so c_2*b_2 > a_2d_2.
  8. Sep 7, 2015 #7


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    I would write [tex]1 + (a_2 + d_2 + a_2d_2) = (b_2c_2).[/tex] The left hand side is at least 1. You can, by appropriate choice of [itex]a_2\geq 0[/itex] and [itex]d_2 \geq 0[/itex], set it to be exactly 1.
  9. Sep 10, 2015 #8


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    Okay, so, by setting a_2 and d_2 equal to 0, we get the following:


    I'm sitting here thinking why this helps me, Hmmm....
    How does this help me determine possible 2x2 matrices in such a scenario?

    Do we say:
    As 2x2 matrix B

    and the identity matrix as matrix A?
  10. Sep 10, 2015 #9


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    Oh, Perhaps I've had an understanding:

    The question is simply asking to find 2x2 Matrices A and B such that the properties hold. I can find any 2x2 matrices I want to finish this question, I do not need some general formula.

    Essentially, by setting matrix A to the easy to work with Identity Matrix, I can find matrix B by the condition 1=c_2*b_2

    If we let b_2=c_2, we see that 1=c_2 and b_2 is a viable solution.

    So if we set A = Identity Matrix, than B = [0, 1, 1, 0] matrix, this should be a solution.
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