The determinant calculator calculates the determinant of a square matrix of order $\hspace{0.2em} 10 \hspace{0.2em}$ and below.

The size of the square matrix can be any positive integer from $\hspace{0.2em} 1 \hspace{0.2em}$ to $\hspace{0.2em} 10 \hspace{0.2em}$.

Each element in the matrix can be a real number in any format — integers, decimals, fractions, or even mixed numbers. Here are a few examples.

- Whole numbers or decimals → $\hspace{0.2em} 2 \hspace{0.2em}$, $\hspace{0.2em} -4.25 \hspace{0.2em}$, $\hspace{0.2em} 0 \hspace{0.2em}$, $\hspace{0.2em} 0.33 \hspace{0.2em}$
- Fractions → $\hspace{0.2em} 2/3 \hspace{0.2em}$, $\hspace{0.2em} -1/5 \hspace{0.2em}$
- Mixed numbers → $\hspace{0.2em} 5 \hspace{0.5em} 1/4 \hspace{0.2em}$

If you would like to see an example of the calculator's working, just click the "example" button.

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Here's a quick overview of the concept of determinants.

A determinant is a unique scalar value associated with a square matrix and is a function of the elements of the matrix. It is common denoted as $\hspace{0.2em} \text{det} (A) \hspace{0.2em}$ or $\hspace{0.2em} |A| \hspace{0.2em}$, $\hspace{0.2em} A \hspace{0.2em}$ being the relevant matrix.

By definition, the determinant of a $\hspace{0.2em} 2 \times 2 \hspace{0.2em}$ matrix is the difference between the products of the elements along the main diagonal ↘ and those along the antidiagonal ↙.

Let me explain. Consider the matrix below.

$A \hspace{0.25em} = \hspace{0.25em} \begin{bmatrix}
a & b\\
c & d
\end{bmatrix}$

The determinant for the matrix above would be

$|A| = ad - bc$

Before extending the idea of determinants to matrices of a higher order, it's helpful to understand what we mean by the terms "minor" and "cofactor" mean.

Consider the determinant below. $\hspace{0.2em} a_{ij} \hspace{0.2em}$ represents the element in the $\hspace{0.2em} i^{\text{th}} \hspace{0.2em}$ row and $\hspace{0.2em} j^{\text{th}} \hspace{0.2em}$ column.

$\begin{vmatrix}
a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\
a_{21} & a_{22} & a_{23} & \cdots & a_{2n} \\
a_{31} & a_{32} & a_{33} & \cdots & a_{3n} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & a_{n3} & \cdots & a_{nn} \\
\end{vmatrix}$

Now the minor corresponding to the element $\hspace{0.2em} a_{ij} \hspace{0.2em}$ is the smaller determinant obtained by omitting the the $\hspace{0.2em} i^{\text{th}} \hspace{0.2em}$ row and $\hspace{0.2em} j^{\text{th}} \hspace{0.2em}$ column of the original matrix.

So the minor corresponding to the element $\hspace{0.2em} a_{22} \hspace{0.2em}$ (denoted by $\hspace{0.2em} M_{22} \hspace{0.2em}$) would be —

$M_{22} \hspace{0.25em} = \hspace{0.25em}
\begin{vmatrix}
a_{11} & a_{13} & \cdots & a_{1n} \\
a_{31} & a_{33} & \cdots & a_{3n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n3} & \cdots & a_{nn} \\
\end{vmatrix}$

Cofactor of an element $\hspace{0.2em} a_{ij} \hspace{0.2em}$ is defined as —

$C_{ij} \hspace{0.25em} = \hspace{0.25em} (-1)^{(i + j)} \cdot M_{ij}$

So the cofactor of $\hspace{0.2em} a_{22} \hspace{0.2em}$ would be

$\begin{align*} C_{22} \hspace{0.25em} &= \hspace{0.25em} (-1)^{(2 + 2)} \cdot M_{22} \\[1em] &= \hspace{0.25em} M_{22} \end{align*}$

To calculate the value of a determinant larger than $\hspace{0.2em} 2 \times 2 \hspace{0.2em}$, we pick one row (or column) and add the products of each element in that row (or column) with its cofactor.

So for a square matrix of order $\hspace{0.2em} n \hspace{0.2em}$, the determinant is given by the formula

$|A| \hspace{0.25em} = \hspace{0.25em} \sum_{j = 1}^{n} a_{ij} \hspace{0.15em} C_{ij}$

Here, $\hspace{0.2em} i \hspace{0.2em}$ can take integral value from $\hspace{0.2em} 1 \hspace{0.2em}$ to $\hspace{0.2em} n \hspace{0.2em}$.

This method is known as Laplace expansion along one particular row or column.

Example

Calculate the determinant of the matrix below.

$A \hspace{0.25em} = \hspace{0.25em} \begin{bmatrix}
7 & 2 & 4 \\
-1 & 5 & 0\\
3 & 9 & -2
\end{bmatrix}$

Solution

To calculate the determinant of this $\hspace{0.2em} 3 \times 3 \hspace{0.2em}$ matrix, let's go with Laplace expansion along the second row. Why we choose the second row will make sense in a moment.

So let's calculate the cofactors for each of the elements in the second row. We start with the cofactor of $\hspace{0.2em} a_{21} \hspace{0.2em}$.

$\begin{align*} C_{21} \hspace{0.25em} &= \hspace{0.25em} (-1)^{(2 + 1)} \cdot \begin{vmatrix}
2 & 4 \\
9 & -2
\end{vmatrix} \\[1.5em] &= \hspace{0.25em} -1 \cdot (2 \times -2 + 4 \times 9) \\[1.5em] &= \hspace{0.25em} -32 \end{align*}$

Similarly, the cofactor of $\hspace{0.2em} a_{22} \hspace{0.2em}$

$\begin{align*} C_{22} \hspace{0.25em} &= \hspace{0.25em} (-1)^{(2 + 2)} \cdot \begin{vmatrix}
7 & 4 \\
3 & -2
\end{vmatrix} \\[1.5em] &= \hspace{0.25em} 1 \cdot (7 \times -2 + 4 \times 3) \\[1.5em] &= \hspace{0.25em} -2 \end{align*}$

We don't need to calculate $\hspace{0.2em} C_{23} \hspace{0.2em}$ because as you'll see, it will get multiplied with $\hspace{0.2em} 0 \hspace{0.2em}$ and hence won't matter. and that's precisely why we selected the second row for expansion — less calculation.

Now, as we learned earlier to calculate the determinant, we add the products of each element in the row (or column) with its cofactor. So,

$\begin{align*} |A| \hspace{0.25em} &= \hspace{0.25em} a_{21} \cdot C_{21} + a_{22} \cdot C_{22} + a_{23} \cdot C_{23} \\[1em] &= \hspace{0.25em} -1 \cdot (-32) + 5 \cdot (-2) + 0 \cdot C_{23} \\[1em] &= \hspace{0.25em} 32 + -10 + 0 \\[1em] &= \hspace{0.25em} 22 \end{align*}$

And that's it.

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