# 27. Further properties of determinants

## 27. Further properties of determinants

Recall that the 2-by-2 matrix a, b; c, d is invertible if and only if its determinant a d minus b c is nonzero. We'll prove the analogue of this for n-by-n matrices.

Theorem:

Let A be an n-by-n matrix. Then A is invertible if and only if det A is nonzero.

We'll also prove the following:

Theorem:

det of the product (A B) equals det A times det B.

### Proof of the invertibility criterion

Recall that A is invertible if and only if its reduced echelon form (say A_{red}) is the identity matrix. To get from A to A_{red}, we used some row operations. The effect of these on the determinant is:

• (type I) determinant unchanged,

• (type II) determinant rescaled by a nonzero factor,

• (type III) determinant switches sign.

Therefore det A differs from det A_{red} by a nonzero factor. In particular, det A is nonzero if and only if det A_{red} is nonzero.

If A is invertible, det of A_{red} equals det of the identity, which equals 1, so det A is nonzero.

If A is not invertible, A_{red} has a row of zeros (A_{red} is a square matrix in reduced echelon form, so if every row is nonzero then the leading entries have to go along the diagonal, and you get A_{red} = I; but we're assuming A is not invertible). Since A_{red} has a row of zeros, det of A_{red}, so det A = 0 too.

### Proof of multiplicativity of the determinant

We first prove det of the product (A B) equals det A times det B in the case where A is an elementary matrix.

• If A = E_{i j}(lambda) then A B is obtained from B by R_i maps to R_i + lambda R_j, so det A B equals det B (such a row operation doesn't change the determinant). We also have det A = 1, so we can verify the formula det (A B) = det A det B in this case just by calculating both sides.

• If A = E_i(lambda) then A B is obtained from B by R_i maps to lambda R_i, so det (A B) = lambda det B. We also have det A = lambda, so again we can verify the formula just by computing both sides.

This shows that det (A B) = det A times det B whenever A is an elementary matrix.

By induction, we can now show that det (A B) = det A times det B whenever A is a product of elementary matrices, in other words whenever A is an invertible matrix.

If A is not invertible, then det A = 0 (by the previous theorem) so to verify the formula in this case, we need to prove that det A B = 0. Suppose for a contradiction that det A B is nonzero 0. Then A B is invertible. But if A B is invertible then B times the inverse of the product A B is an inverse for A: A times B times the inverse of A B equals A B times the inverse of A B, which equals the identity. This gives a contradiction, so we deduce that det A B = 0.

We have now checked the formula for all possible matrices A, which proves the theorem.