Determinant

The determinant of a square matrix is a scalar (a number) that tells you something about how that matrix behaves. You can calculate the determinant 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 $\left|{\begin{bmatrix}a&b\\c&d\end{bmatrix}}\right|$, we just write $\det {\begin{bmatrix}a&b\\c&d\end{bmatrix}}$ and $\left|{\begin{matrix}a&b\\c&d\end{matrix}}\right|$.

Interpretation

There are a few ways to understand what the determinant says about the 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 this case, the determinant tells you 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. 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 the volume was mirrored over an odd number of axes.

"System of equations" interpretation

You can see 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 means 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, you 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, and 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, you can remember the formulas:

  $\det {\begin{bmatrix}a\end{bmatrix}}=a,\qquad \det {\begin{bmatrix}a&b\\c&d\end{bmatrix}}=ad-bc.$

• For $3\times 3$ matrices, the formula is:

  ${\det {\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}}={\color {blue}{aei}+{dhc}+{gbf}}{\color {OrangeRed}{}-{gec}-{ahf}-{dbi}}}$

  You 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.

Let's say 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 you can see here, we can save work by choosing a row or column that has many zeros. If $a_{ij}$ is 0, we don't need to calculate $C_{ij}$.

See also

• Volume