# Determinant

The determinant of a square matrix is a scalar (a number) that indicates how that matrix behaves. It can be calculated from the numbers in the matrix.

The determinant of the matrix $A$ is written as $\det(A)$ or $|A|$ in a formula. Sometimes, instead of $\det \left({\begin{bmatrix}a&b\\c&d\end{bmatrix}}\right)$ and $\left|{\begin{bmatrix}a&b\\c&d\end{bmatrix}}\right|$ , one simply writes $\det {\begin{bmatrix}a&b\\c&d\end{bmatrix}}$ and $\left|{\begin{matrix}a&b\\c&d\end{matrix}}\right|$ .

## Interpretation

There are a few ways to understand what the determinant says about a matrix.

### Geometric interpretation For a $2\times 2$ matrix ${\begin{bmatrix}a&c\\b&d\end{bmatrix}}$ , the determinant is the area of a parallellogram. (The area is equal to $ad-bc$ .)

An $n\times n$ matrix can be seen as describing a linear map in $n$ dimensions. In which case, the determinant indicates the factor by which this matrix scales (grows or shrinks) a region of $n$ -dimensional space.

For example, a $2\times 2$ matrix $A$ , seen as a linear map, will turn a square in 2-dimensional space into a parallelogram. That parallellogram's area will be $\det(A)$ times as big as the square's area.

In the same way, a $3\times 3$ matrix $B$ , seen as a linear map, will turn a cube in 3-dimensional space into a parallelepiped. That parallelepiped's volume will be $\det(B)$ times as big as the cube's volume.

The determinant can be negative or zero. A linear map can stretch and scale a volume, but it can also reflect it over an axis. Whenever this happens, the sign of the determinant changes from positive to negative, or from negative to positive. A negative determinant means that the volume was mirrored over an odd number of axes.

### "System of equations" interpretation

One can think of a matrix as describing a system of linear equations. That system has a unique non-trivial solution exactly when the determinant is not 0 (non-trivial meaning that the solution is not just all zeros).

If the determinant is zero, then there is either no unique non-trivial solution, or there are infinitely many.

## Singular matrices

A matrix has an inverse matrix exactly when the determinant is not 0. For this reason, a matrix with a non-zero determinant is called invertible. If the determinant is 0, then the matrix is called non-invertible or singular.

Geometrically, one can think of a singular matrix as "flattening" the parallelepiped into a parallelogram, or a parallelogram into a line. Then the volume or area is 0, which means that there is no linear map that will bring the old shape back.

## Calculating a determinant

There are a few ways to calculate a determinant.

### Formulas for small matrices The $3\times 3$ determinant formula is a sum of products. Those products go along diagonals that "wrap around" to the top of the matrix. This trick is called the Rule of Sarrus.
• For $1\times 1$ and $2\times 2$ matrices, the following simple formulas hold:

$\det {\begin{bmatrix}a\end{bmatrix}}=a,\qquad \det {\begin{bmatrix}a&b\\c&d\end{bmatrix}}=ad-bc.$ • For $3\times 3$ matrices, the formula is:

${\det {\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}}={\color {blue}{aei}+{dhc}+{gbf}}{\color {OrangeRed}{}-{gec}-{ahf}-{dbi}}}$ One can use the Rule of Sarrus (see image) to remember this formula.

### Cofactor expansion

For larger matrices, the determinant is harder to calculate. One way to do it is called cofactor expansion.

Suppose that we have an $n\times n$ matrix $A$ . First, we choose any row or column of the matrix. For each number $a_{ij}$ in that row or column, we calculate something called its cofactor $C_{ij}$ . Then $\det(A)=\sum a_{ij}C_{ij}$ .

To compute such a cofactor $C_{ij}$ , we erase row $i$ and column $j$ from the matrix $A$ . This gives us a smaller $(n-1)\times (n-1)$ matrix. We call it $M$ . The cofactor $C_{ij}$ then equals $(-1)^{i+j}\det(M)$ .

Here is an example of a cofactor expansion of the left column of a $3\times 3$ matrix:

{\begin{aligned}\det {\begin{bmatrix}{\color {red}1}&3&2\\{\color {red}2}&1&1\\{\color {red}0}&3&4\end{bmatrix}}&={\color {red}1}\cdot C_{11}+{\color {red}2}\cdot C_{21}+{\color {red}0}\cdot C_{31}\\&=\left({\color {red}1}\cdot (-1)^{1+1}\det {\begin{bmatrix}1&1\\3&4\end{bmatrix}}\right)+\left({\color {red}2}\cdot (-1)^{2+1}\det {\begin{bmatrix}3&2\\3&4\end{bmatrix}}\right)+\left({\color {red}0}\cdot (-1)^{3+1}\det {\begin{bmatrix}3&2\\1&1\end{bmatrix}}\right)\\&=({\color {red}1}\cdot 1\cdot 1)+({\color {red}2}\cdot (-1)\cdot 6)+{\color {red}0}\\&=-11.\end{aligned}} As illustrated above, one can simplify the computation of determinant by choosing a row or column that has many zeros; if $a_{ij}$ is 0, then one can skip calculating $C_{ij}$ altogether.