Contents

**Determinant|Definition & Meaning**

**Definition**

The determinant of a matrix is a function that operates on the entries of a square matrix, producing a scalar value as the output.

**Square matrices** have an **equal** number of rows and columns. Thus, if the number of columns is n, then the number of rows must also be n, as illustrated below:

**Finding the Determinant**

There are a few methods to calculate the determinant. By definition, the determinant is found using the **Laplace Expansion **method. Without getting too technical right away, we illustrate the process for 1×1 square matrices and higher.

**1×1 Matrix**

A 1×1 matrix is essentially a **single scalar value**. The determinant is the **value itself**. Very simple!

\[ A = \begin{bmatrix} \,a\, \end{bmatrix} \Rightarrow |A| = a \]

**2×2 Matrix**

We illustrate the process and formula for the determinant of a 2×2 matrix in the figure below.

**3×3 Matrix**

For this, we need to **expand** along the first row. The steps are mentioned and illustrated below:

**Write out**the determinant form (optional).- Take the
**first element**along the row (blue and circled in figure) and find the**2×2 sub-matrix**that**does not contain**the elements in**its****row and column**(blue background in figure).**Repeat**for the**second**and**third**elements. -
**Multiply**the**element**with the**determinant**of its corresponding**sub-matrix**. You now have three terms:**add**the first,**subtract**the second, and**add**the third. This order is important! - Write the
**determinant**of the 2×2 sub-matrices in**expanded form**and**evaluate**the result.

Figure 3 – Calculating the determinant of a 3×3 matrix by expanding along the first row.

If you apply the same process to the 2×2 matrix, you will still arrive at the correct result. In that case, the sub-matrix will be a 1×1 matrix (a scalar value).

**4×4 and Higher-order Square Matrices**

By now, we have established a **pattern** that we can use to find the determinant of **higher-order** square matrices. When expanding over the first row, the determinant has as many terms as the elements in that row. Therefore, we need two things:

- The
**sign**of each term. - The
**sub-matrix**(or “**minor**”) of each element in the row.

The second one is easy enough. However, we must dive deeper and formally define the determinant for the first part.

**Formal Definition of Determinant**

The **Laplace Expansion definition** for the determinant (for expansion along the **first row**) is **formulated** as:

\[\text{det}(A) = |A| =\sum_{k\,=\,1}^{n} \left(-1\right)^{k+1} \times \left| A_{1,\, k} \right| \times a_{1, \,k}\]

Here, $a_{1,\,k}$ denotes the** k**^{th}** entry** in the first row (you can think of **k** as the **column number**), and $A_{1,\,k}$ represents the **minor (sub-matrix)** with the first row and k^{th} column **removed**. The term $(-1)^{k+1} \times \left| A_{1, \,k} \right|$ is called the “**cofactor**.”

**Determining the Sign**

The $(-1)^{k+1}$ term from the formula **assigns the sign** to each expanded term in the determinant’s expression. It is simple: even-numbered entries have the minus, while odds have the plus.

However, this is true **only** for expanding the first row. The **truly general pattern** is that if **(i, k)** represents the **row** and **column number** of the expanding element, then if:

- i + k = even,
**sign = positive** - i + k = odd,
**sign = negative**

You can see this is true by checking the expansion of the first row. The first entry has (i, k) = (1, 1) and 1 + 1 = 2, which is even, so this term has a positive sign. There is a useful **visual mnemonic** to remember this **alternating plus-minus pattern**, which we illustrate below:

Figure 4 – A visual representation of the alternating plus-minus patterns created by the $(-1)^{i+k}$ term in the generalized Laplace expansion of a matrix.

Thus, if A is a** 4×4 matrix**:

\[ A = \begin{bmatrix} \,a & b & c & d\, \\ \,e & f & g & h\, \\ \,i & j & k & l\, \\ \,m & n & o & p\, \end{bmatrix} \]

The row and column numbers for the first row’s elements are a = (1, 1), b = (1, 2), c = (1, 3), and d = (1, 4). Thus, the signs for a, b, c, and d are respectively +, -, +, and -. The **determinant of A** would be:

\[ |A| = a \times \begin{vmatrix} \,f & g & h\, \\ \,j & k & l\, \\ \,n & o & p\, \end{vmatrix}-b \times \begin{vmatrix} \,e & g & h\, \\ \,i & k & l\, \\ \,m & o & p\, \end{vmatrix}+c \times \begin{vmatrix} \,e & f & h\, \\ \,i & j & l\, \\ \,m & n & p\, \end{vmatrix}-d \times \begin{vmatrix} \,e & f & g\, \\ \,i & j & k\, \\ \,m & n & o\, \end{vmatrix} \]

We can extend this procedure to any **nxn square matrix**.

**Calculating the Determinant by Expanding Other Rows and Columns**

We do not need to always expand along the first row. It is often **more convenient** to expand along **another** row or column in the matrix (e.g., the second row has more zeros, which would simplify the calculation). For that, we need to keep a few things in mind, as discussed below.

**Determinant by Expanding Other Rows**

We can **extend** the original Laplace Expansion for the determinant to expand along any row ‘i’ as:

\[ \text{det}(A) = |A| = \!\!\!\!\!\!\!\!\!\! \sum_{\substack{k\,=\,1 \\ i\,\in\,\{1,\,2,\,\dots,\,n\}}}^{n} \!\!\!\!\!\!\!\!\! \left(-1\right)^{k+i} \times a_{i,\,k} \times \left| A_{i,\,k} \right| \]

Since the method to find the minor does not change, we can similarly **expand from any row** other than the first as long as we remember the **plus-minus** **patterns**. We illustrate the process for expanding a 3×3 matrix using the second row below:

Figure 5 – Finding the determinant using Laplace expansion over the second row. Notice the sign changes compared to Figure 3 and verify this from the mnemonic in Figure 4.

**Determinant by Expanding the Columns**

Formally, we can extend the definition of the Laplace Expansion to the **first column**:

\[\text{det}(A) = |A| =\sum_{i\,=\,1}^{n} \left(-1\right)^{i+1} \times \left| A_{i,\,1} \right| \times a_{i,\,1}\]

Here, ‘i’ represents the** i ^{th} row**. Similarly, we can expand along

**any column**‘k’ as:

\[ \text{det}(A) = |A| = \!\!\!\!\!\!\!\!\!\! \sum_{\substack{i\,=\,1 \\ k\,\in\,\{1,\,2,\,\dots,\,n\}}}^{n} \!\!\!\!\!\!\!\!\! \left(-1\right)^{i+k} \times a_{i,\,k} \times \left| A_{i,\,k} \right| \]

All that changed was the **notation order** (reflecting that the row number changes instead of the column number now).

Again, we must remember the **plus-minus patterns**. The minor-finding process remains the same.

## An Example of Calculating the Determinant

Find the determinant of the following matrix:

\[A = \begin{bmatrix} \,3 & -1 & 0\, \\ \,2 & -2 & 3\, \\ \,9 & -3 & 0\, \end{bmatrix}\]

**Solution**

Expanding from the third column (it has two 0s, so the calculation will be easy):

\[ |A| = 0-3[3(-3)-9(-1)]+0 = 3(-9+9) = \mathbf{0} \]

A matrix with a **zero determinant** is called a **singular matrix**. The inverse of such a matrix **does not exist**.

Additionally, we can verify our result using the fact that when **any two** rows or columns of a matrix are **multiples** of each other (the first and third rows in our case), the determinant is zero.

**Reason:** Row-order operations simplify such a matrix to have one row or column completely 0. Since the determinant is constant for a given matrix and can be found by expanding any row or column, having a **row full of zeros** means the determinant must be zero.

*All images/mathematical drawings were created with GeoGebra.*