The determinant of a matrix is a single numerical value which is used when calculating the inverse or when solving systems of linear equations.
The determinant of a matrix $\mathbf{A}$ is denoted $\lvert \mathbf{A} \rvert$, or sometimes $\det(\mathbf{A})$. The determinant is only defined for square matrices.
A matrix is said to be singular if its determinant is zero.
The general formula for the determinant of matrices of any size is very complicated. You will only be asked to compute the determinants of $2 \times 2$ or $3 \times 3$ matrices by hand.
Let $\mathbf{A} = \begin{pmatrix} a & b\\ c & d \end{pmatrix}$.
The determinant of $\mathbf{A}$ is \[\lvert \mathbf{A} \rvert = ad - bc\]
Find the determinant of the matrix $\mathbf{A} = \begin{pmatrix} 4 & 1\\ 0 & 2\end{pmatrix}$.
\begin{align} \lvert \mathbf{A} \rvert = \begin{vmatrix} 4 & 1 \\ 0 & 2 \end{vmatrix} &= 4 \times 2 - 1 \times 0\\ &=8 - 0\\ &=8 \end{align}
Find the determinant of the matrix $\mathbf{A} = \begin{pmatrix} -6 & 3\\ -1 & 1\end{pmatrix}$.
\begin{align} \lvert \mathbf{A} \rvert &= (-6) \times 1 - 3 \times (-1)\\ &=(-6) - (-3)\\ &=-6+3\\ &=-3 \end{align}
Find the determinant of the matrix $\mathbf{A} = \begin{pmatrix} 5 & -10\\ -2 & 4\end{pmatrix}$.
\begin{align} \lvert \mathbf{A} \rvert &= 5 \times 4 - (-10) \times (-2)\\ &=20 - 20\\ &=0 \end{align}
Note: Since its determinant is zero, this matrix has no inverse.
Hayley Bishop finds the determinant of the $2\times 2$ matrices $\mathbf{A} = \begin{pmatrix}3&-1\\0&4\end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix} -2&10\\-1&5\end{pmatrix}$.
The determinant of a $3 \times 3$ matrix can be calculated by breaking it down into smaller $2 \times 2$ matrices, as follows:
\[\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a \begin{vmatrix} e & f \\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix}\]
Or,
\[\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}\]
Here's one way of interpreting the formula: for each element $a_{1i}$ in the top row, block out the row and column it belongs to, and calculate the determinant of the remaining uncovered $2 \times 2$ matrix, then multiply that by $a_{1i}$. The determinant is the sum of those values, alternating addition and subtraction.
Find the determinant of the matrix $\begin{pmatrix} 1 & 2 & 1\\ 0 & 3 & 4\\ 3 & 1 & 4 \end{pmatrix}$.
\begin{align} \lvert \mathbf{A} \rvert &= 1\begin{vmatrix} 3 & 4\\ 1 & 4 \end{vmatrix} -2 \begin{vmatrix} 0 & 4\\ 3 & 4 \end{vmatrix} +1 \begin{vmatrix} 0 & 3\\ 3 & 1 \end{vmatrix}\\\\ &= 1(3\times 4 - 4\times 1) -2(0\times 4 - 4\times 3) +1(0 \times 1 - 3\times 3)\\ &= 1(12-4) -2(0-12) +1(0-9)\\ &= 8 +24-9\\ &=23 \end{align}
Find the determinant of the matrix $\begin{pmatrix} 1 & 0 & 3\\ -1 & -1 & -3\\ 0 & 0 & 6 \end{pmatrix}$.
\begin{align} \lvert \mathbf{A} \rvert &= 1\begin{vmatrix} -1 & -3\\ 0 & 6 \end{vmatrix} -0 \begin{vmatrix} -1 & -3\\ 0 & 6 \end{vmatrix} +3 \begin{vmatrix} -1 & -1\\ 0 & 0 \end{vmatrix}\\\\ &= 1\bigl((-1)\times 6 - (-3)\times 0\bigr) -0(\,\dotso\,) +3\bigl((-1) \times 0 - (-1)\times 0\bigr)\\ &= 1\bigr((-6)-0\bigl) -0 +3(0-0)\\ &=-6 +0\\ &=-6 \end{align}
Note: There was no need to work out the second $2 \times 2$ determinant as it was being multiplied by zero.
Hayley Bishop works out the determinant of the $3\times 3$ matrix $\mathbf{A} = \begin{pmatrix}1&0&2\\3&1&0\\-1&1&4\end{pmatrix}$.
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.
Test yourself: Numbas test on 2x2 determinants