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### Matrix Inverses

```
Date: 04/07/98 at 21:26:12
From: Jonathan Hornbuckle
Subject: Linear algebra, matrix inverses

Can you find a 2 x 2 matrix inverse to itself where every entry is
non-zero and an integer? Can you find a 2 x 2 matrix with all distinct
and non-zero entries which has no inverse?

Thanks!
```

```
Date: 04/13/98 at 09:30:25
From: Doctor Lester
Subject: Re: Linear algebra, matrix inverses

For your first question, we want to find a matrix A such that AA = I,
the identity matrix.

Let A be the matrix:

[a b]
[c d]

Now we'll try to find values for a, b, c and d that satisfy your
requirements.

We need:

[a b][a b] = [1 0]
[c d][c d]   [0 1]

Multiplying out the lefthand side we get:

[a^2+bc   ab+bd] = [1 0]
[ac+dc   bc+d^2]   [0 1]

Comparing each element in the lefthand matrix with the corresponding
element in the righthand matrix, we get the four equations:

i) a^2 + bc = 1
ii) ab  + db = 0
iii) ac  + dc = 0
iv) d^2 + bc = 1

We now need to find out what restrictions these put on our possible
choices for a, b, c and d.

Subtracting (iv) from (i), you find a simple relation between a and d.
Try this to see exactly how they are related. It tells you that once a
value for a has been chosen, there are only two possible choices for d
(set in terms of a). Let's call these choices 1 and 2.

Let's make an arbitrary choice for the value of a, and call it k.
Obviously, k cannot be 0, or we have introduced a zero element (which
you don't want). It turns out that k cannot be +1 or -1 either (but we
don't know that yet!).

Find out now what following choice 1 or 2 implies. It turns out that
only one of these can lead to a matrix satisfying your conditions.

For choice 1 (or 2 depending on how you've labelled them!), equation
(ii) gives us kb = 0 (try it). This requires that either k or b is 0,
again introducing zero elements.

For this choice of d in terms of k, now (ii) and (iii) are
automatically satisfied, whatever our choices of b and c. (Please
check this!)

So can we have ANY non-zero values for b and c? Well, (i) and (iv)
both give us an equation listing bc in terms of k. Please find out
what this is. You will find that, as long as k is not +1 or -1, then
bc = <an integer>, so you can always choose b and c to fit this
equation.

You should have found that, to build an example matrix of this form,
you must set a to any arbitrary number. We have seen that this cannot
be 0, -1 or +1. Now you need to set d equal to -a and then choose b
and c as you wish, such that bc = 1 - a^2. But make sure you
understand the steps above that lead us to this recipe.

Once you've made such a matrix, try working out the inverse just to
check.

Obviously, you could have just guessed a solution, but that wouldn't
have shown you why that solution worked. I have tried to leave much of
the working for you to try yourself, so please write back if this has
confused things for you!

For your second question, to find a matrix with non-zero integer
entries that is not invertible, you just need to know that the
determinant of a matrix is zero if and only if the matrix has no
inverse. Do you know how to evaluate determinants? The determinant of
a matrix is just a special number associated with a matrix. For 2x2
matrices of the form:

[a b]
[c d]

the determinant is just the value ad - bc.

So we just have to choose elements so that ad - bc is zero.

For example you can easily make a matrix such that ad = 12 and bc = 12
but where a and d factorise 12 in a different way from b and c
(because 3x4 = 12 = 2x6). Then ad - bc = 12 - 12 = 0.

-Doctor Lester,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
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Associated Topics:
High School Linear Algebra

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