## DETERMINANTS

We give here a treatment of the properties of determinants of square matrices. If A is a matrix we denote by A_{ij} the matrix obtained by crossing out the i th row and j th column of A.

Our definition of determinants is as follows. If A = [a] is one by one, then det(A) = a. If A is the 2 by 2 matrix

```a b
c d ```

then det(A) = ad - bc.

In the general case, we assume that one already knows how to compute determinants of size smaller than n by n. Let A be an n by n matrix. Then det(A) is defined as

```a_{11} det(A_{11}) - a_{12} det(A_{12})
+ ... +(-1)^(n-1) a_{1n} det(A_{1n})  ```

The typical term in the sum is

(-1)^(i-1) a_{1i} det(A_{1i}).

For example, if A =

```2 1 3
4 2 5
1 6 9
```

then A_{11}, A_{12}, A_{13} are

```2 5 ,  4 5 ,  4 2
6 9    1 9    1 6
```

and so det(A) = 2(18 - 30) - 1(36-5) + 3(24-2) = 11.

Fact 1. Switching any two rows of an n by n matrix A reverses the sign of its determinant.

Here is why:
Step 1. We first consider the case of the first and second rows. If n=2 the verification is an easy calculation. Assume that n > 2.

We carry out the expansion with respect to the first row, and then do that again for each of the size n-1 by n-1 smaller determinants. Let B be the matrix formed from A by omitting the first two rows and the i th and j th columns, where i is different from j, which is n-2 by n-2. The we get the sum of n(n-1) terms, each of which has the form

(+ or -)a_{1i} a_{2j} det(B)

where the sign is (-1)^(i-1) (-1) (j-2) if i < j and is (-1)^(i-1) (-1)(j-1) if i > j.

Thus, the sign is (-1)^(i+j-2) or (-1)^(i+j-3) depending on whether i > j or i < j.

Switching the first two rows gives the same terms when we expand, but all the signs are reversed.

Step 2. We now consider the case where two rows next to one another are switched. If the result is not true, pick n as small as possible for which it is false. If the two rows are first and second, we are already done by Step 1. If not, expand with respect to the first row. The two expansions are the same except that in each n-1 by n-1 matrix A_{1i}, two rows consecutive rows are switched. Since we know the result is true for this smaller size, it follows that all of the signs from the det(A_{1i}) are reversed, and the result follows.

Step 3. Now consider any two rows, and suppose there are k rows in between. Switch the upper of the two with each of these in turn, and then the lower. That is k+1 switches. (The lower is now just above the upper). Now switch the lower with each of the k rows originally in between. This involves k switches. The result is that the two rows have exchanged positions. This took 2k+1 switches of consecutive rows, an odd number. Hence, the sign has reversed.

We have now established the result in general.

Fact 2. It follows from Fact 1 that we can expand a determinant with respect to any row. For the i th row, if i is odd the formula is

```a_{i1} det(A_{i1}) - a_{i2} det(A_{i2})
+ ... + (-1)^(n-1) a_{in}det(A_{in})
```

while if i is even the formula is

```-a_{i1} det(A_{i1}) + a_{i2} det(A_{i2})
- ... + (-1)^n a_{in}det(A_{in})
```

(Moving the i th row to the top involves i-1 exchanges, and this is even when i is odd and odd when i is even.)

Fact 3. If two rows of a matrix are equal, its determinant is 0. (Interchanging the rows gives the same matrix, but reverses the sign of the determinant. Thus, det(A) = - det(A), and this implies that det(A) = 0.)

Fact 4. The determinant is a linear function of the i th row if the entries outside the i th row are held constant. This is immediate from our formula for the expansion with respect to the i th row.

We illustrate this more specifically if i = 1. Thus, if A is the matrix with rows

```v_1
v_2
.
.
.
v_n
```

and B has rows

```w_1
v_2
.
.
.
v_n
```

(only the first rows are different) while C has rows

```cv_1 + dw_1
v_2
.
.
.
v_n
```

then det(C) = c det(A) + d det(B). In particular, if we replace the first row v_1 of A by cv_1, the determinant of A is multiplied by c.

If we let the entries of the first row of A be x_1, ..., x_n and we let c_i = (-1)^(i-1) det(A_{1i}) (this is constant here) then det(A) = c_1 x_1 + ... + c_n x_n.

The argument for the i th row is similar (or switch it to the first position).

Fact 5. If A is an n by n matrix, adding a multiple of one row to a different row does not affect its determinant!!!

Here is why: For concreteness, we give the argument with the first and second row. The general case follows in exactly the same way. Let v be the first row of A and w second row. Think of det(A) as a function F(v) of v, which we allow to vary while keeping the rest of A fixed. Then F(v + cw) = F(v) + cF(w) by Fact 4. F(w) is the determinant of a matrix with its first and second rows equal: both are w. Thus, F(w) = 0, and we have that F(v+cw) = F(w), as required.

Fact 6. An n by n matrix with a row of zeros has determinant zero.

Here is why: expand with respect to that row.

Fact 7. The determinant of a lower triangular matrix (or an upper triangular matrix) is the product of the diagonal entries. In particular, the determinant of a diagonal matrix is the product of the diagonal entries.

Here is why: expand with respect to the first row, which gives only one nonzero term, and then continue in the same way (for the upper triangular case expand with respect to the last row).

Fact 8. The determinant of an n by n matrix A is 0 if and only if the rows are linearly dependent (and not zero if and only if they are linearly independent). That is, the determinant of A is not zero if and only if A is invertible.

Here is why: The issue is not affected by switching rows, adding a multiple of one row to another, or multiplying a row by a nonzero scalar. Thus, we may assume that A is a square matrix in RREF. If the rows are independent, it will then be the identity, while otherwise it has a row of zeros.

HOW TO EVALUATE DETERMINANTS: Do row operations until the matrix is upper triangular. If you factor out a scalar you need to keep track of it. If you switch two rows, you need to keep track of the sign change. You need to clear the entries in a column below the diagonal, but not the ones above, so this is partial row reduction. When you add or subtract a multiple of one row to or from another, the determinant does not change!

Example: To find the determinant of

```1 1 0 1
2 5 4 2
3 6 1 4
4 7 2 9
```

subtract 2, 3 or 4 times the first row from the second, third and fourth rows to get

```1 1 0 1
0 3 4 0
0 3 1 1
0 3 2 5
```

Subtract the second row from the third and fourth rows to get

```1 1  0 1
0 3  4 0
0 0 -3 1
0 0 -2 5
```

Subtract 2/3 the third row from the fourth to get

```1 1  0    1
0 3  4    0
0 0 -3    1
0 0  0  13/3
```

The determinant is then 1(3)(-3)(13/3) = -39.

Fact 9. When a determinant of an n by n matrix A is expanded out it is the sum of n! terms, all of which are products of n elements, one from each row, no two from the same column, with certain signs attached to the products.

Here is why: assume it for smaller sizes. Then (+ or -)a_{1i} A_{1i} will give all such products involving a_{1i}, with various signs on them.

Fact 10. If one column of the n by n matrix is allowed to vary and the other entries are fixed, the determinant is a linear function of that column. (This corresponds to Fact 4 for rows.)

Here is why: This follows immediately from the kind of formula implied by Fact 9.

Fact 11. If two columns of an n by n matrix A are equal, the determinant is zero.

Here is why: this implies that the rank is less than n, which means that the rows are dependent, and therefore det(A) = 0.

Fact 12. Adding a multiple of one column of A to a different column does not change the determinant.

Here is why: The reasoning is exactly the same as for rows (see Fact 5) but using Facts 10 and 11 in place of Facts 4 and 3.

Fact 13. If two columns of an n by n matrix are switched, the sign is reversed.

Here is why: exactly as in the case of rows, it suffices to check this when the columns are next to each other. Look at a supposed counterexample of smallest size. When one expands with respect to the first row, the two terms coming from those two columns are the same but with signs switched. The other terms involve smaller size determinants with two columns switched. Thus, all terms have their signs switched.

Fact 14. If A is an n by n matrix, det(A) = det(A^T). Thus, whatever one knows for rows, one knows for columns, and conversely.

Here is why: do elementary row operations on A (and then one is doing elementary column operations on A^T) until A is upper triangular). When rows (columns of A^T) are switched, the sign changes in the same way. All of these operations have the same affect on det(A) as on det(A^T) (either none, a sign switch, or multiplication by the same nonzero constant). If A is not invertible the same is true of A^T and so both determinants are 0. If A is invertible we eventually reach an upper triangular matrix (A^T is lower triangular) and we already know these two have the same determinant.

Fact 15. det(AB) = det(A)det(B). Perform successive elementary row operations on A. Each of these has the same effect on A as on AB. (E.g., if one switches two rows of A, the same two rows are switched in AB. If one adds c times the i th row of A to the j th for j different from i, the same happens to AB. If one multiplies a row of A by c, the same row of AB gets multiplied by c.) This means that we can assume that A is in RREF. If A has a row of zeros then so does AB, and both determinants sides are 0. Otherwise, A has become the identity matrix, so that det(A) = 1, and the result is clear, since AB = B.

Fact 16. Let A be an n by n matrix. Let B be the matrix whose i,j entry is (-1)^(i+j) det(A_{ji}) (called the transpose of the cofactor matrix, or the classical adjoint of A). Then AB = BA = det(A) 1_n (a diagonal matrix with det(A) everywhere on the diagonal).

Here is why: each off diagonal entry of the product is the expansion of a matrix with two rows or columns equal with respect to a row or column, while each diagonal entry is the expansion of det(A) with respect to a row or column, and therefore is equal to det(A).

Fact 17. With notation as in Fact 16, if A is invertible then A^(-1) = (1/det A)B.

Here is why: this is immediate from Fact 16.