# Finding the Determinant
### Created by: Orion Sinacori

In order to find the determinant of a matrix we must first understand what a determinant of a matrix is. Suppose we have a matrix $A$ with dimensions $n \times n$, then the determinant of $A$ is the volume (or area) of the parallelepiped formed by the columns of $A$ in absolute value. Now, to find a determinant we have a couple methods to choose from.

First, we will consider a $2 \times 2$ matrix. To find the determinant of one of these matrices we have the formula $ad - bc$ if we consider the matrix to be of the form $\begin{bmatrix}
a&b\\
c&d\\
\end{bmatrix}$. For example, let's have $A$ = $\begin{bmatrix}
1&4\\
-5&2\\
\end{bmatrix}$. To find the determinant we will let $a = 1, b = 4, c = -5$ and $d = 2$ so using our equation above we get det$A$$=1 \cdot 2 - 4 \cdot (-5)=22$. Thus our determinant is 22 and our area of the paralelepiped that is formed by the matrix $A$ is $22$.

Now that we know how to solve a $2 \times 2$ matrix we can start looking at how to solve $n \times n$ matrices. Unfortunately we do not have a simple equation for matrices that are larger than a $2\times 2$, but we have a couple of different methods. The first method we will learn about is the alternating sum. For $n \geq 2$ the determinant of $A = [a_{ij}]$ is the alternating sum det $A = a_{11} \cdot$ det $A_{11} - a_{12}\cdot$ det $A_{12} + a_{13}\cdot$ det $A_{13} - \cdots + (-1)^{n+1} a_{1n} \cdot$ det $A_{1n}$. To understand what is goning on here we will look at an example where matrix $A$ = $\begin{bmatrix}
1&5&0\\
2&4&-1\\
0&-2&0\\
\end{bmatrix}$. We will start with row one column one and work across that row to get $1 \cdot$ det $\begin{bmatrix}
4&-1\\
-2&0\\
\end{bmatrix}$ - $5 \cdot$ det $\begin{bmatrix}
2&-1\\
0&0\\
\end{bmatrix}$ + $0 \cdot$ det $\begin{bmatrix}
2&4\\
0&-2\\
\end{bmatrix}$. Now using what we learned about finding the determinant of $2 \times 2$ matrices we get $1(-2) - 5(0) + 0(-4) = -2$. So $-2$ is our determinant and 2 is the volume of the paralelepiped that is formed. Also, we could have used any row and we would have gotten the same answer but when using different rows you have to be mindful of the idea of alternating sums. For a $3 \times 3$ matrix, the signs for each value would look like $\begin{bmatrix}
+&-&+\\
-&+&-\\
+&-&+\\
\end{bmatrix}$ where the signs alternate. So if you started with row 2 column 1 the first term of your equation would be $-2 \cdot$ det $\begin{bmatrix}
5&0\\
-2&0\\
\end{bmatrix}$.

Now that we got a feel with alternating sums, we can take a look at another method of determining determinants. This method makes use of a triangle matrix or a reduced matrix. For example, in a $3 \times 3$ matrix it will look like $\begin{bmatrix}
a_{11}&a_{12}&a_{13}\\
0&a_{22}&a_{23}\\
0&0&a_{33}\\
\end{bmatrix}$ where each pivot has only zeroes below it. If our matrix is in this form then the determinant is $a_{11} \cdot a_{22} \cdot a_{33}$ or the product of the entries on the main diagnol. If our matrix is not in this form then we can perform row operations to get it to this for following some specific rules. Let $A$ be a square matrix:
1) If a multiple of one row of $A$ is added to another row to produce $B$, then the det $B =$ det $A$.
2) If two rows are interchanged to produce $B$, then det $B = -$det $A$.
3) If one row of $A$ is multiplied by $k$ to produce $B$, then det $B = k \cdot$ det $A$.

With these rules in mind lets try to find the determinant of $A$ = $\begin{bmatrix}
1&5&0\\
2&4&-1\\
0&-2&0\\
\end{bmatrix}$ by making it a triangular matrix, and if we do everything right we should end up with the same answer from above.
$\begin{bmatrix}
1&5&0\\
2&4&-1\\
0&-2&0\\
\end{bmatrix}$ $\rightarrow$ $\begin{bmatrix}
1&5&0\\
0&-6&-1\\
0&-2&0\\
\end{bmatrix}$ $\rightarrow$ $\begin{bmatrix}
1&5&0\\
0&-2&0\\
0&-6&-1\\
\end{bmatrix}$ $\rightarrow$ $\begin{bmatrix}
1&5&0\\
0&-2&0\\
0&0&-1\\
\end{bmatrix}$
Now that we have our triangular matrix we can solve for our determinant. since we had a row change our determinant will be: det $B = -$det$A$ $\implies$ $1 \cdot -2 \cdot -1 = -$det $A$ $\implies$ det $A = -1 \cdot 2 \cdot 1 = -2$. Thus our determinant is $-2$ which is the same as when we used the alternating sums.

Now that you know how to find determinants using two different methods it is time to try and solve one yourself. Using either method, find the determinant of the matrix $A = \begin{bmatrix}
2&3&1&5\\
2&4&-1&1\\
0&0&-3&4\\
3&2&7&0\\
\end{bmatrix}$.