Calculating matrix determinant

Now that you are familiar with the determinant of matrices of size $2$, you will know its definition for square matrices of any size.
You'll even derive a general formula for the determinant of any matrix of size $3$. In addition, you'll identify some of the matrices whose determinant is very easy to calculate. Get ready to practice a lot!

The building blocks

You already know that the determinant of a $2 \times 2$ matrix is $\begin{vmatrix}a&b\\c&d \end{vmatrix} = ad -bc$. Maybe it's easy to grasp but not so easy to remember.

A good tip is to think about it in terms of the diagonals of the matrix. The determinant is just the product of its forward diagonal minus the product of its backward diagonal:

But what about the determinant of larger matrices? The focus here will be to define the determinant of $3 \times 3$ matrices through the determinant of several $2 \times 2$ matrices. And you'll use just the same strategy to define the determinant of any $n \times n$ matrix in terms of the determinant of several $(n-1) \times (n-1)$ matrices.

Before you go any further on this road, you need to know a bit of terminology.

The protagonists: minors and cofactors

In order to define the determinant of a larger matrix, you are going to start extracting smaller sub-matrices from it.

Let's take a square matrix $A$ of size $n$ and assume that you've already defined the determinant for matrices of size $n-1$.
The $(i,j)$-minor of $A$, denoted as $m_{ij}$, is the determinant of the $(n-1) \times(n-1)$ matrix that you obtain by deleting the $i$-th row and $j$-th column of $A$.

Let's illustrate this with a matrix of size $3$:

$$A=\begin{pmatrix}3&0&-1\\4&1&0 \\ 0&-5&1 \end{pmatrix}$$For example, the minor $m_{11}$ of $A$ is the determinant of the matrix that results from eliminating the first row and first column of $A$:

In turn, the minor $m_{23}$ of $A$ is the determinant of the matrix that results from eliminating the second row and the third column of A.

Let's see the last ingredient you need. The determinant has a certain "alternation" behavior. This is a very particular quality and is manifested in the fact that you need to slightly alter the minors by multiplying them by a simple factor. In some cases, this factor is simply $1$ and in the rest is $-1$. The $(i,j)$-cofactor of $A$ is:

$$c_{ij} = (-1)^{i+j}\, m_{ij}$$This means that the factor of the $m_{ij}$ is $1$ when $i +j$ is even, and $-1$ when it is odd. In the previous example, the factor of $m_{11}$ is then $1$ and the factor of $m_{23}$ is $-1$. Thus $c_{11} = m_{11}$ and $c_{23} = -m_{23}$

You can visualize the behavior of this factor as if it were a chessboard. White entries have a factor of $-1$ while dark entries have a factor of $1$

The determinant of any square matrix

Now that you're familiar with cofactors, it's time to take the next step. Let's combine them all. We define the cofactor expansion along the $i$-th row of $A$ as the linear combination of the cofactors of the entries of $i$-th row, using the entries of that row as the coefficients (notice that $a_{i1}, a_{i2}, \dots, a_{in}$ are all the entries of the $i$-th row of $A$):

$$a_{i1} \, c_{i1} + a_{i2} \, c_{i2} + \dots +a_{in} \, c_{in}$$Let's continue with the matrix from the previous section:

$$A=\begin{pmatrix}3&0&-1\\4&1&0 \\ 0&-5&1 \end{pmatrix}$$For example, the cofactor expansion along the first row from $A$ is:

$$a_{11} \, c_{11} + a_{12} \, c_{12} +a_{13} \, c_{13} = 3 \, c_{11} + 0 \, c_{12} - \, c_{13} = 3 \, m_{11} - \, m_{13} =3 \begin{vmatrix}1&0\\-5&1 \end{vmatrix} - \begin{vmatrix}4&1\\0&-5 \end{vmatrix} = 23$$

If you calculate the expansion of the other two rows, you will notice that the result is the same! You can even define the cofactor expansion along the $j$-th column in a similar way (here that $a_{1j}, a_{2j}, \dots, a_{nj}$ are all the entries of the $j$-th column of $A$):

$$a_{1j} \, c_{1j} + a_{2j} \, c_{2j} + \dots +a_{nj} \, c_{nj}$$Try to calculate the expansion along each column of the example matrix. Not only are they all the same, but their value is the same as the common value of the row expansions! This is more than a coincidence and is in fact true for any matrix. For this reason, we define the determinant of a matrix as any of these expansions!

The important thing about the definition is that after having defined the determinant for matrices of size $i$ you can extend its definition for those of size $2$, but then this definition is used for those of size $i$ and so on.

So, you can find the determinant of any square matrix of size $3$ by expanding along any row or column, and you will always have the same result. For example, if you choose the first row, then:

$$\begin{align*} \begin{vmatrix}a&b&c\\d&e&f \\ g&h&i \end{vmatrix} &= a \begin{vmatrix}e&f \\ h&i \end{vmatrix} -b \begin{vmatrix}d&f \\ g&i \end{vmatrix} + c\begin{vmatrix}d&e \\ g&h \end{vmatrix} \\ &= a (ei-fh) -b(di-fg)+c(dh-eg) \\ &= aei+bfg+cdh−ceg−bdi−afh \end{align*}$$That's a pretty hard number to memorize. But you can remember it with a trick similar to the one you used for matrices of size $2$. The positive terms are the products of the $3$ possible diagonals of the matrix, while the negative terms are the products of the $3$ possible backward diagonals:

Putting theory to work

Let's see some examples. What better than starting with the simplest matrices, the upper triangular matrices? They only have zeros below their diagonal. For instance:

$$\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \qquad \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{pmatrix} \qquad \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 5 & 6 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$Their determinant is extremely easy to calculate, it's simply the product of its main diagonal! Actually, the same result holds for lower triangular matrices (those that only have zeros below their diagonal), no matter how big they are.

Here's a little shortcut: if the matrix is upper triangular and one of its diagonal entries is $0$, then its determinant is $0$.

Let's end up with a slightly larger matrix, say $5 \times 5$:

$$A=\begin{pmatrix} 2&0&-3&2&3\\ -3&0&4&-5&1\\ 1&0&5&1&0\\ 1&0&-3&0&0\\ 4&-3&4&2&-2\\ \end{pmatrix}$$
Remember that to calculate the determinant, you can expand along any row or column you want. The best strategy is to choose one that has a lot of $0$'s, as this will reduce the number of cofactors you have to calculate. If you look closely, the second column has a lot of zeros, so it's a good candidate:

$$\begin{align*} \det(A) &= a_{12} \, c_{12} + a_{22} \, c_{22} + a_{32} \, c_{32} + a_{42} \, c_{42} + a_{52} \, c_{52} \\ &= 0 + 0 + 0 + 0 -3 \, c_{52} \\ &= -3 \, (-1)^{5+2} m_{52}\\ &= -3 \,( -1) \, m_{52}\\ &= 3 \, m_{52}\\ &= 3 \begin{vmatrix} 2&-3&2&3\\ -3&4&-5&1\\ 1&5&1&0\\ 1&-3&0&0\\\end{vmatrix} \end{align*}$$Great, now there is only one determinant to solve for a matrix of size $4$. If you define:

$$B= \begin{pmatrix} 2&-3&2&3\\ -3&4&-5&1\\ 1&5&1&0\\ 1&-3&0&0\\\end{pmatrix}$$Then $\det(A) = 3 \det(B)$. But now, which is the best row or column to expand along? Undoubtedly the fourth row, since it has $2$ zeros.

$$\begin{align*} \det(B) &= b_{41} c_{41} + b_{42} c_{42} + b_{43} c_{43} +b_{44} c_{44} \\ &= b_{41} c_{41} + b_{42} c_{42} + 0 c_{43} +0 c_{44} \\ &= c_{41} -3 c_{42} \\ &= -m_{41} - 3 m_{42} \\ &= - \begin{vmatrix} -3&2&3\\ 4&-5&1\\ 5&1&0\\ \end{vmatrix} -3 \begin{vmatrix} 2&2&3\\ -3&-5&1\\ 1&1&0 \end{vmatrix} \\ \end{align*}$$Now there are $2$ determinants of size $3$. You can either use the explicit formula or expand across its third line (because it has a $0$).

For the first one:

$$\begin{vmatrix} -3&2&3\\ 4&-5&1\\ 5&1&0\\ \end{vmatrix} = 5 c_{31} + c_{32} + 0 c_{33} = 5 \begin{vmatrix} 2&3\\ -5&1\ \end{vmatrix} - \begin{vmatrix} -3&3\\ 4&1\ \end{vmatrix} = 5(2+15) -(-3-12) = 100$$And for the second one:

$$\begin{vmatrix} 2&2&3\\ -3&-5&1\\ 1&1&0 \end{vmatrix} = \begin{vmatrix} 2&3\\ -5&1\\ \end{vmatrix} - \begin{vmatrix} 2&3\\ -3&1\\ \end{vmatrix} = (2+15)-(2+9) = 6$$Thus, $\det(B) = -100 -3(6) = -118$. Finally, $\det(A) = 3 \det(B) = 3(-118) = -354$

Conclusion

Let's take a square matrix $A$ of size $n$

• The $(i,j)$-minor of $A$, denoted as $m_{ij}$, is the determinant of the $(n-1) \times(n-1)$ matrix that you obtain by deleting the $i$-th row and $j$-th column of $A$.

• The $(i,j)$-cofactor of $A$ is $c_{ij} = (-1)^{i+j}\, m_{ij}$.

• The cofactor expansion along the $i$-th row of $A$ is $a_{i1} \, c_{i1} + a_{i2} \, c_{i2} + \dots +a_{in} \, c_{in}$.The cofactor expansion along the $j$-th column of $A$ is $a_{1j} \, c_{1j} + a_{2j} \, c_{2j} + \dots +a_{nj} \, c_{nj}$.

• The determinant of $A$ is the value of its expansion along any row or column.

• The determinant of a matrix of size $3$ is $\begin{vmatrix}a&b&c\\d&e&f \\ g&h&i \end{vmatrix} = aei+bfg+cdh −ceg−bdi−afh$.

• The determinant of a triangular matrix is the product of its diagonal.