# Determinates in Terms of Cofactors

For square matrices, we can calculate a special value called the determinate. This value has a few uses, most notably, it can tell us if a system is consistent, as well as provide an alternative method for finding the inverse of a matrix.

First, consider a two-variable system of linear equations.

$a_{11}x_1 + a_{12}x_2 = b_1$

$a_{21}x_1 + a_{22}x_2 = b_2$

If we isolate the first equation for $x_1$ and the second equation for $x_2$, we get this:

$x_1 = \frac{a_{22}b_1 - a_{12}b_2}{a_{11}a_{22} - a_{12}a_{21}}$

$x_2 = \frac{a_{11}b_2 - a_{21}b_1}{a_{11}a_{22} - a_{12}a_{21}}$

Notice that with this definition, the only way we can have a solution is if $a_{11}a_{22} - a_{12}a_{21} \ne 0$. From here, let's look at the definition of the determinate for a two by two matrix.

$det(\left[\begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]) = a_{11}a_{22} - a_{12}a_{21}$

As you can see, we define our determinate as the condition for a solution to a linear system. As we discussed earlier, determinates were defined to tell us if a system has a consistent solution, which is why the determinate is defined in this way.

For three by three matrices, the determinate becomes more complicated. If you were to manipulate a three equation system for each variables, you would find that there is a solution if $a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} \ne 0$.

This is hard to remember if we write it in this format, however there is a trick to make this formula more approachable. We can apply the definition of a two by two matrix to take the determinate of parts of the three by three matrix. If we do this, the definition of the determinate will be as follows.

$D = a_{11} * det(\left[\begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right]) - a_12 * det(\left[\begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array}\right] + a_{13} * \left[\begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}\right]$

This definition is a little more approachable, as it requires less memorization to be able to use. We can even further simplify it using the concept of a cofactor. Let $A(i,j)$ be a two by two submatrix, obtained by removing row i and column j from the matrix A. The cofactor of a matrix is $C_{ij} = (-1)^{(i+j)} * det(A(i,j))$

Using this definition, we can redefine our determinate formula one last time to be: $det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$. This formula is much easier to remember and will be the one we use for the remainder of the section.

**Example:
**Find the determinate of $\left[\begin{array}{cc} 4 & -1
& 1 \\ 2 & 3 & 5 \\ 1 & 0 & 6 \end{array}\right]$

$det(A) = a_{11}C_{11} + a_{12}C_{12} +
a_{13}C_{13}$

$= 4* \left[\begin{array}{cc} 3 & 5 \\ 0 & 6 \end{array}\right] - (-1) *
\left[\begin{array}{cc} 2 & 5 \\ 1 & 6 \end{array}\right] + (1) \left[\begin{array}{cc} 2
& 3 \\ 1 & 0 \end{array}\right]$

$= 4 * (3)(6) - (0) (5) - (-1)(12) - 5 + (1) (0 - 3)$

$= 4 * 18 + 7 + 2 = 76$

We can generalize the definition of determinates for n by n matrices. Doing this ends up in a situation where you need to recursively apply the determinate definition until you reach the two by two matrices of the matrix you start with. For instance, a four by four matrix will involve you finding all the three by three matrix determinates which in turn requires you to find all the two by two matrix determinates. Once you reach two by two, you can apply the determinate and get an answer.

**Theorem:
**If one row or column of an n by n matrix contains
only zeros, then det(A) = 0

If we expand on cofactors, it is clear to see that each determinate or cofactor will result in 0, causing the determinate to be equal to zero as well.

**Theorem:
**If a matrix is lower triangular, or upper
triangular, then det(A) is equal to the product of the diagonal entries.

If you look at the form of triangular matrices, you will be able to see that cofactors will result in zero when they aren't a part of the diagonal elements. This is since everything below or above the diagonal is zero. Due to this, we can expect that zero will impact all the determinates, leaving us with just the product of the diagonal entries.

**Theorem:
**If A is an n by n matrix, then $det(A) = det(A^T)$

The determinate is an operation that is not impacted by transposes. You can think of the determinate of an n by n matrix as an operation that is just changing the order of multiplication within the determinates. The outcome that we find from a transpose yields the same multiplication in value, leading to the same determinate.