Find All Possible Matrix Inverses

  • #1
RJLiberator
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Homework Statement



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.

Homework Equations


The insverse is defined as 1/determinat(matrix) * adj(matrix)

Otherwise shown as:
[itex]\frac{1}{ad-bc}\begin{bmatrix}
d & -b \\
-c & a
\end{bmatrix}[/itex]

The Attempt at a Solution


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

[itex]\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}[/itex]

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

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?
 

Answers and Replies

  • #2
pasmith
Homework Helper
1,882
528

Homework Statement



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.

Homework Equations


The insverse is defined as 1/determinat(matrix) * adj(matrix)

Otherwise shown as:
[itex]\frac{1}{ad-bc}\begin{bmatrix}
d & -b \\
-c & a
\end{bmatrix}[/itex]

The Attempt at a Solution


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

[itex]\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}[/itex]

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

I then realized an issue with my method:
  1. There are many possible situations that this occurs.#
That's fine; you only need to find one of them.

  1. How do I check that A^-1 and B^-1 exist in this scenario.
The determinants of A and B will be non-zero.

So it may seem like I am doing something wrong.

Any helpful words of advice here?
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].
 
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  • #3
SammyS
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Homework Statement



Find 2x2 matrices A and B, all of whose entries are [itex]\ \geq 0 \ [/itex], such that A^-1 and B^-1 exist, but (A+B)^-1 does not exist.

Homework Equations


The inverse is defined as 1/determinant(matrix) * adj(matrix)

Otherwise shown as:
[itex]\frac{1}{ad-bc}\begin{bmatrix}
d & -b \\
-c & a
\end{bmatrix}[/itex]

...

Any helpful words of advice here?
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?
 
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  • #4
RJLiberator
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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)≠0.
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 [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]
 
  • #5
pasmith
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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.
 
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  • #6
RJLiberator
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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
[itex]1+(a_2+d_2)+(a_2d_2-c_2b_2)=0[/itex]
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]

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
pasmith
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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
[itex]1+(a_2+d_2)+(a_2d_2-c_2b_2)=0[/itex]
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]
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.
 
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  • #8
RJLiberator
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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
RJLiberator
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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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