# Find All Possible Matrix Inverses

1. Sep 5, 2015

### RJLiberator

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

3. The attempt at a solution

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

$\frac{1}{(a_1+a_2)(d_1+d_2)-(b_1+b_2)(c_1+c_2)}\begin{bmatrix} (d_1+d_2) & -(b_1+b_2) \\ -(c_1+c_2) & (a_1+a_2) \end{bmatrix}$

I then noted that the matrix inverse does not exist if:
$(a_1+a_2)(d_1+d_2)-(b_1+b_2)(c_1+c_2)=0$

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.

2. Sep 5, 2015

### pasmith

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 $\det(A + B) = 0$ and $\det(B) \neq 0$.

3. Sep 5, 2015

### SammyS

Staff Emeritus
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?

4. Sep 5, 2015

### RJLiberator

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

a_1=1
d_1=1
b_1=0
c_1=0

This allowed me to get det $(A+B) = (1+a_2+d_2)+(a_2d_2-c_2b_2)=0$

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

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

So $(1+a_2+d_2) =/= -(a_2d_2-c_2b_2)$

5. Sep 5, 2015

### pasmith

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

6. Sep 7, 2015

### RJLiberator

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
$1+(a_2+d_2)+(a_2d_2-c_2b_2)=0$
by setting A as the identity Matrix.

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

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

$1+(a_2+d_2)+x=0.$

1+a_2+d_2=-x

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.

7. Sep 7, 2015

### pasmith

I would write $$1 + (a_2 + d_2 + a_2d_2) = (b_2c_2).$$ The left hand side is at least 1. You can, by appropriate choice of $a_2\geq 0$ and $d_2 \geq 0$, set it to be exactly 1.

8. Sep 10, 2015

### RJLiberator

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

1=b_2c_2.

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:
a_2=0
b_2=b_2
c_2=c_2
d_2=0
As 2x2 matrix B

and the identity matrix as matrix A?

9. Sep 10, 2015

### RJLiberator

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.