# Invertible Matrices (Ring)

1. Aug 4, 2014

### Justabeginner

1. The problem statement, all variables and given/known data
Let R be the ring of all 2*2 matrices over Zp, p a prime,. Show that if det(a b c d) = ad - bc ≠ 0, then (a b c d) is invertible in R.

2. Relevant equations

3. The attempt at a solution
I don't know how to start if Zp, with p a prime, is the clause. I know that since ad- bc ≠ 0, it automatically makes the matrix (a b c d) invertible in R.. but does the determinant have to be one?

Last edited: Aug 4, 2014
2. Aug 4, 2014

### PieceOfPi

The most straight-forward way of showing something is invertible is simply find an inverse element and check that it is indeed the inverse. I think there is a very good candidate for the inverse in this case.

3. Aug 4, 2014

### Justabeginner

A matrix A= (a b c d) has an inverse B = [1/(ad-bc)] (d -b -c a).
Does Zp, p a prime, give any information about the elements of Matrix A, specifically? Does that mean the elements must be prime themselves, or is the determinant prime?...I'm not sure how to focus on what's relevant here.
If ad-bc = 1, then can I prove something about the prime nature of a, b, c, d?

4. Aug 4, 2014

### jbunniii

For one thing, it ensures that $1/(ad - bc)$ exists as long as $ad - bc \neq 0$. This would not be true in general for $Z_n$ where $n$ is not prime. (For example, in $Z_4$, the element $2$ has no inverse.) What is special about $Z_p$ for prime $p$?

5. Aug 4, 2014

### PieceOfPi

One thing to note is that in the set of real numbers, 1/(ad-bc) is an inverse of ad-bc provided that ad-bc is not equal to 0.

6. Aug 4, 2014

### Justabeginner

Zp is a field if and only if p is prime.

Suppose p, prime, where P>=2. and Zp = {0, 1,...p-1}
Zp satisfies field axioms of closure, commutativity, and associativity by addition through integer properties. Additive identity also exists as in integers. Additive inverse also exists: For x in Zp, take -x(mod p). Then x + (-x) = 0 mod p.

Closure under multiplication also occurs since for x and y in Zp, xy mod p in Zp exists.
Commutativity, associativity, and identity of multiplication also exists by property of integers.
For x in Zp, the multiplicative inverse 1/x cannot occur in Zp. So ax + bp = 1
Then ax= 1 mod p
So, a is the multiplicative inverse of x.

If p is not prime, then Zp is not a field.

Since p is not prime, p= mn where m,n ≠0 and m,n≠1. If Zp is a field, then m would have a multiplicative inverse such that mm-1= 1.
Then n*1 = n *(mm-1) = (mn)m-1 = pm-1 = 0.
But n≠0, so proof by contradiction holds.
Zp cannot be a field if p is not prime.

7. Aug 4, 2014

### jbunniii

Yes, that's correct. Every nonzero element of $Z_p$ has an inverse if and only if $p$ is prime. This is true of matrices, too: if we were working in $Z_4$, for example, then the matrix
$$\begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}$$
would have no inverse.

8. Aug 4, 2014

### Justabeginner

How may I 'show' that the matrix has an inverse then? Using a property of a field?

9. Aug 4, 2014

### jbunniii

You could show that there is no matrix $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ satisfying
$$\begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}\begin{pmatrix}a & b \\ c & d\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$$
This is clear if we simply carry out the matrix multiplication on the left hand side. Indeed, multiplying the first row of $\begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}$ by the first column of $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$, and setting equal to the $(1,1)$'th element of the right hand side, we get the equation $2a = 1$, which has no solution in $Z_4$.

10. Aug 4, 2014

### jbunniii

 Sorry, I misread your question. I thought you were asking how to show that $\begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}$ has no inverse in $Z_4$.

To show that a matrix $M$ DOES have an inverse, it suffices to find another matrix $X$ such that $MX = I$.

As you said above, if you start with an arbitrary matrix $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$, then
$$\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}d & -b \\ -c & a\end{pmatrix} = \begin{pmatrix}ad-bc & 0 \\ 0 & ad-bc \end{pmatrix}$$
Now if $ad-bc$ is nonzero, then it has an inverse since $Z_p$ is a field. Call this inverse $1/(ad-bc)$. Now multiply both sides by $1/(ad-bc)$ to obtain the result you need.