# Elementary Row Operators and Determinates

From the previous section, we saw that the process of calculating a determinate can be long and complex. Due to this, we often may want to manipulate a matrix in such a way that the determinate is easier to calculate, or the matrix resembles one we already calculate the determinate for. When we do these manipulations, there are several rules we can follow to ensure the determinate remains the same for the new matrix.

**Theorem:
**Let $A = \left[\begin{array}{cc} a_{11} & a_{12} &
a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}
\end{array}\right]$, and let B be the matrix resulting from multiplying a row of A
by a real number b. Then, det(B) = r * det(A)

**Proof:
**Suppose we were to take the determinate of the
matrix $B = \left[\begin{array}{cc} a_{11} & a_{12} & a_{13} \\ a_{21} &
a_{22} & a_{23} \\ r*a_{31} & r*a_{32} & r*a_{33} \end{array}\right]$. Observe
that the determinate would impact all values that contain the third row by a
factor of r. This would give us the result: $det(B) = r*det(A)$.

More generally, if we multiply a whole matrix by a scalar r, we get the result $det(rA) = r^n * det(A)$, where n is the number of rows in A. This result is very useful because it allows us to find the determinate of a matrix, and as a result, know the determinate of all the scalar multiples of that same matrix.

We want to next look at how the manipulations putting matrices into reduced row echelon form impact the determinant of a matrix. Recall that upper triangular and lower triangular matrices have a property of their determinate being the multiple of their diagonals. If we can row reduce a matrix to one of these formats, we can easily solve the determinate. So, understanding how these manipulations impact the matrices will help us be able to create these scenarios.

**Theorem:
**Suppose that A is some n x n matrix, and B is a
matrix obtained by swapping the position of two rows in A. Then, det(B) =
-det(A).

We can show that this is true using induction. First, the base case of a 2 x 2 matrix is easy to show.

$det(\left[\begin{array}{cc} a & b \\ c & d
\end{array}\right]) = a*d - b*c$

$det(\left[\begin{array}{cc} c & d \\ a & b \end{array}\right]) = c*b - d*a = - a*d +
b*c = - (a*d - b*c)$

From here, we assume that the theorem holds for an (n - 1) x (n - 1) matrix. Now, we just need to show that it holds for the n x n case. The (n - 1) x (n - 1) case could only be true if $C_{ij} = -C_{ij}$. Knowing this, we can manipulate the n x n as follows.

$det(b) = a_{i1}C_{i1} + \dots + a_{in}C_{in} = a_{i1} (-C_{i1}) _ \dots + a_{in}(-C_{in}) = -(a_{i1}C_{i1} + \dots + a_{in}C_{in}) = -det(A)$

**Theorem:
**If two rows in A are equal, then det(A) = 0.

Using the result above, we can show that this is true. If we let B represent the matrix obtained by swapping the two equal rows of A, we would have det(B) = - det(A). Since the rows are equal, the swap yields the same determinate which implies that det(A) = -det(A). This can only be true if det(A) = 0.

**Theorem:
**Suppose that A is some n x n matrix, and B is
obtained by adding r times the i-th row of A to the k-th row of A. Then det(B)
= det(A).

This can be shown easily using induction like the proof of interchanging rows. Using these theorems, we can now easily solve any determinate by row reducing it to the closest triangular form and multiplying the diagonal elements. This turns determinates into a much easier problem compared to the recursive definition presented with cofactors.

**Example:
**Find the determinate of $A = \left[\begin{array}{cc} 1
& 3 & 1 & 5 \\ 1 & 3 & -3 & -3 \\ 0 & 3 & 1
& 0 \\ 1 & 6 & 2 & 11 \end{array}\right]$

You could use the recursive cofactor definition to solve this problem, but it would take far too long compared to finding the upper triangular form of the matrix. Let's look at how we can reduce this matrix.

First, we subtract row 2 by row 1, and row 4 by row 1. This will not change the determinate at all based on the theorems we proved. Doing so would give us:

$\left[\begin{array}{cc} 1 & 3 & 1 & 5 \\ 0 & 0 & -4 & -8 \\ 0 & 3 & 1 & 0 \\ 0 & 3 & 1 & 6 \end{array}\right]$

From here, we can swap rows 2 and 3. This means we need to negate the determinate when we take it.

$\left[\begin{array}{cc} 1 & 3 & 1 & 5 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & -4 & -8 \\ 0 & 3 & 1 & 6 \end{array}\right]$

Next, we can factor the -4 out of the third row, meaning we need to multiply our determinate by -4 as well.

$\left[\begin{array}{cc} 1 & 3 & 1 & 5 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 3 & 1 & 6 \end{array}\right]$

Finally, we need to subtract row 4 from row 2, which doesn't impact the determinate. This gives us:

$\left[\begin{array}{cc} 1 & 3 & 1 & 5 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 6 \end{array}\right]$

This is now in upper triangular form, meaning we can calculate the determinate. Here is a review of the operations we did.

- Subtracted row 2 by row 1: No impact on the determinate
- Subtracted row 4 by row 1: No impact on the determinate
- Swapped rows 2 and 3: Multiply the determinate by -1
- Factored out -4 from row 3: Multiply the determinate by -4
- Subtracted row 4 by row 2: No impact on the determinate

This means that our final determinate is calculate as: det(A) = (-1) * (-4) * (1) * (3) * (1) * (6) = 72.

There are a few more interesting theories we can discuss before moving on to applications of determinates.

**Theorem:
**If A is a n x n matrix, the following statements
are equivalent:

- $det(A) \ne 0$
- Rank(A) = n
- A is invertible

This allows us to easily conclude information about our matrix if we know the determinate, rank, or can invert the matrix. In general, this just allows us to tie these properties together so that they are related in some way.

As a final note, we can also conclude information about multiply matrices as well.

**Theorem:
**If A and B are n x n matrices, then det(AB) =
det(A) * det(B)