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### Determinant of an Invertible Matrix

```
Date: 08/04/99 at 03:37:14
From: Mike Smith
Subject: Determinants

I've been having problems understanding what determinants have to do
with matrix invertibility. Could you give me an explanation as well as
a proof of why a matrix is invertible if and only if the determinant
is non-zero? I would really like to know.

Mike Smith.
```

```
Date: 08/04/99 at 07:54:11
From: Doctor Floor
Subject: Re: Determinants

Dear Mike,

Let us consider the following general nxn matrix:

[a11 a12 ... a1n]
M = [a21 a22 ... a2n]
[ :   :       : ]
[an1 an2 ... ann]

We define the cofactor of a11, call it c11, by the determinant of the
matrix found when we leave row 1 and column 1 out of M, so:

|a22 ... a2n|
c11 = | :       : |
|an2 ... ann|

In almost the same way, the cofactor cpq of apq is found as the
determinant of the matrix when we leave row p and column q out of M.

Now we have to use the rule of signs:
If p+q is even then cpq = determinant,
if p+q is odd then cpq = -determinant.

Doing this we find that:

ap1*cp1 + ap2*cp2 + ap3*cp3 + ... + apn*cpn = det(M)   .....[1]

for any 1 <= p <= n.

Also, we find that:

aq1*cp1 + aq2*cp2 + aq3*cp3 + ... + aqn*cpn = 0   ..........[2]

when 1 <= p <= n and 1 <= q <= n, while p and q are unequal.

Expression [2] is the determinant of the matrix you get when in matrix
M you replace row p by row q. The matrix then has two equal rows, so
clearly its determinant must be 0.

We find the inverse of M as:

[c11/det(M) ... cn1/det(M)]
M^(-1) = [c12/det(M) ... cn2/det(M)]
[   :              :      ]
[c1n/det(M) ... cnn/det(M)]

because now M*M^(-1) = I (the nxn identity matrix) (check this).

This computation of the inverse of M clearly can't work when
det(M) = 0.

The fact that a matrix with determinant 0 really can't have an inverse
can be seen from the following:

We know that, when det(M) = 0, then row n is a linear combination of
rows 1, 2, 3, ..., n-1.

Now suppose that matrix N is the inverse of matrix M, so M*N = I (the
identity matrix). Note that the nth row of M*N is now a linear
combination of the rows 1, 2, 3, ..., n-1 of M*N too. The elements in
the nth column of rows 1, 2, 3, ..., n-1 of I are all zero, so that
means that the nth element of the nth row of M*N should be zero, too,
being a linear combination of zeroes. But the nth element of the nth
row of I is one, and we have a contradiction.

I hope this cleared things up. If you have more questions, don't
hesitate to send them to Dr. Math.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Linear Algebra

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