A Method to Compute the Determinant of a $5\times 5$ Matrix

1. Introduction

The determinant of an $n\times n$ matrix $$A_n={\left[\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ a_{21}& a_{22}& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& \dots & a_{nn}\end{array}\right]}_{n\times n},$$ is denoted by $det(A_n)$ or $⏐A_n⏐$, and a basic formula to compute the determinant is $$det(A_n)=\left|\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ a_{21}& a_{22}& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& \dots & a_{nn}\end{array}\right|=\sum sgn(j_1,j_2,\dots ,j_n)a_{1j_1}{a}_{2j_2}\dots a_{n{j}_n},$$ where the summation is taken over all $n$ permutations $j_1,j_2,\dots ,j_n$ of the set of integers $1,2,\dots ,n$. Furthermore, the function $sgn(j_1,j_2,\dots ,j_n)$ is defined as: $$sgn(j_1,j_2,…,j_n)=\{\begin{array}{ll}+1& \text{if }{j}_1,j_2,\dots ,j_n\text{ is an even permutation}\\ -1& \text{if }{j}_1,j_2,\dots ,j_n\text{ is an odd permutation}\end{array}.$$ In this paper, we will present a new method to compute the determinant of a $5\times 5$ matrix.

2. Preliminaries: The main definitions and lemmas

In matrix theory, a square matrix is called \textit{nonsingular} if and only if its determinant is nonzero. We generalize the nonsingular matrices to doubly nonsingular:

Definition 2.1. An $n\times n$ matrix $A_n=[a_{ij}{]}_{n\times n}$ is doubly nonsingular if and only if $A_n$ is nonsingular and all $2\times 2$ matrices of adjacent terms within the $A_n$ are nonsingular.

Example 2.2. Let $A_3={\left[\begin{array}{ccc}1& 6& 3\\ 3& 2& 2\\ 6& 1& 3\end{array}\right]}_{3\times 3}$. We have $det(A_3)=-5$, consequently $A_3$ is nonsingular. Clearly, all $2\times 2$ determinants of adjacent terms are nonzero. Hence, the matrix $A_3$ is doubly nonsingular.

Now, we introduce a new function, which we call the \textit{star fraction}:

Definition 2.3. Given two $3\times 3$ matrices $A_3={\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]}_{3\times 3}$ and $B_3={\left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right]}_{3\times 3}$ such that $b_{ij}\ne 0(\mathrm{\forall }i,j=1,2,3)$ and $B_3$ is doubly nonsingular. The star fraction of $A_3$ on $B_3$ is defined as: $${\left(\frac{A_3}{B_3}\right)}^{\ast }={\left(\frac{{\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]}_{3\times 3}}{{\left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right]}_{3\times 3}}\right)}^{\ast }=\frac{\left|\begin{array}{cc}\frac{\left|\begin{array}{cc}\frac{a_{11}}{b_{11}}& \frac{a_{12}}{b_{12}}\\ \frac{a_{21}}{b_{21}}& \frac{a_{22}}{b_{22}}\end{array}\right|}{\left|\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right|}& \frac{\left|\begin{array}{cc}\frac{a_{12}}{b_{12}}& \frac{a_{13}}{b_{13}}\\ \frac{a_{22}}{b_{22}}& \frac{a_{23}}{b_{23}}\end{array}\right|}{\left|\begin{array}{cc}{b}_{12}& b_{13}\\ b_{22}& b_{23}\end{array}\right|}\\ \\ \frac{\left|\begin{array}{cc}\frac{a_{21}}{b_{21}}& \frac{a_{22}}{b_{22}}\\ \frac{a_{31}}{b_{31}}& \frac{a_{32}}{b_{32}}\end{array}\right|}{\left|\begin{array}{cc}{b}_{21}& b_{22}\\ b_{31}& b_{32}\end{array}\right|}& \frac{\left|\begin{array}{cc}\frac{a_{22}}{b_{22}}& \frac{a_{23}}{b_{23}}\\ \frac{a_{32}}{b_{32}}& \frac{a_{33}}{b_{33}}\end{array}\right|}{\left|\begin{array}{cc}{b}_{22}& b_{23}\\ b_{32}& b_{33}\end{array}\right|}\end{array}\right|}{\left|\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right|}.$$

In the next section, we will show that the star fraction is a useful function for calculating the determinant of a $5\times 5$ matrix. Now, we shall know about the Dodgson’s condensation of a matrix that was introduced by Charles Lutwidge Dodgson in 1866 [1]:

Definition 2.4. The Dodgson’s condensation of an $n\times n$ matrix $A_n=[a_{ij}{]}_{n\times n}$ is an $(n-1)\times (n-1)$ matrix such as $[m_{ij}{]}_{(n-1)\times (n-1)}$ such that $$m_{ij}=\left|\begin{array}{cc}{a}_{ij}& a_{i(j+1)}\\ a_{(i+1)j}& a_{(i+1)(j+1)}\end{array}\right|.$$

Henceforth the notation $DC(A_n)$ is denote the first Dodgson’s condensation of a matrix $A_n$, and the second condensation is $DC(DC(A_n))$ and so on. Clearly a square matrix $A_n$ is doubly nonsingular if and only if all elements of $DC(A_n)$ are nonzero.
To prove the main theorem we need the following lemmas.

Lemma 2.5.[3] We have $$\left|\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& a_{14}& a_{15}\\ a_{21}& a_{22}& a_{23}& a_{24}& a_{25}\\ a_{31}& a_{32}& a_{33}& a_{34}& a_{35}\\ a_{41}& a_{42}& a_{43}& a_{44}& a_{45}\\ a_{51}& a_{52}& a_{53}& a_{54}& a_{55}\end{array}\right|=\frac{\left|\begin{array}{cc}\left|\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right|& \left|\begin{array}{cccc}{a}_{12}& a_{13}& a_{14}& a_{15}\\ a_{22}& a_{23}& a_{24}& a_{25}\\ a_{32}& a_{33}& a_{34}& a_{35}\\ a_{42}& a_{43}& a_{44}& a_{45}\end{array}\right|\\ \\ \left|\begin{array}{cccc}{a}_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\\ a_{51}& a_{52}& a_{53}& a_{54}\end{array}\right|& \left|\begin{array}{cccc}{a}_{22}& a_{23}& a_{24}& a_{25}\\ a_{32}& a_{33}& a_{34}& a_{35}\\ a_{42}& a_{43}& a_{44}& a_{45}\\ a_{52}& a_{53}& a_{54}& a_{55}\end{array}\right|\end{array}\right|}{\left|\begin{array}{ccc}{a}_{22}& a_{23}& a_{24}\\ a_{32}& a_{33}& a_{34}\\ a_{42}& a_{43}& a_{44}\end{array}\right|},$$ where $\left|\begin{array}{ccc}{a}_{22}& a_{23}& a_{24}\\ a_{32}& a_{33}& a_{34}\\ a_{42}& a_{43}& a_{44}\end{array}\right|\ne 0$.

Lemma 2.6. [2, Theorem 1] We have $$\left|\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right|=\frac{\left|\begin{array}{cc}\frac{\left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|& \left|\begin{array}{cc}{a}_{12}& a_{13}\\ a_{22}& a_{23}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|& \left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|\end{array}\right|}{a_{22}}& \frac{\left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{12}& a_{13}\\ a_{22}& a_{23}\end{array}\right|& \left|\begin{array}{cc}{a}_{13}& a_{14}\\ a_{23}& a_{24}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|& \left|\begin{array}{cc}{a}_{23}& a_{24}\\ a_{33}& a_{34}\end{array}\right|\end{array}\right|}{a_{23}}\\ \\ \frac{\left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|& \left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{31}& a_{32}\\ a_{41}& a_{42}\end{array}\right|& \left|\begin{array}{cc}{a}_{32}& a_{33}\\ a_{42}& a_{43}\end{array}\right|\end{array}\right|}{a_{32}}& \frac{\left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|& \left|\begin{array}{cc}{a}_{23}& a_{24}\\ a_{33}& a_{34}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{32}& a_{33}\\ a_{42}& a_{43}\end{array}\right|& \left|\begin{array}{cc}{a}_{33}& a_{34}\\ a_{43}& a_{44}\end{array}\right|\end{array}\right|}{a_{33}}\end{array}\right|}{\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|},$$ where $a_{22},a_{23},a_{32},a_{33}$ are nonzero numbers and $\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|\ne 0$.

3. Main results

In the following theorem we present a new method, just to compute the determinant of a $5\times 5$ matrix.

Theorem 3.1. Given a $5\times 5$ matrix $$A_5={\left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& a_{14}& a_{15}\\ a_{21}& a_{22}& a_{23}& a_{24}& a_{25}\\ a_{31}& a_{32}& a_{33}& a_{34}& a_{35}\\ a_{41}& a_{42}& a_{43}& a_{44}& a_{45}\\ a_{51}& a_{52}& a_{53}& a_{54}& a_{55}\end{array}\right]}_{5\times 5},$$ where $\left[\begin{array}{ccc}{a}_{22}& a_{23}& a_{24}\\ a_{32}& a_{33}& a_{34}\\ a_{42}& a_{43}& a_{44}\end{array}\right]$ is a doubly nonsingular matrix with all nonzero elements. Then $$det(A_5)={\left(\begin{array}{c}\frac{DC(DC(A_5))}{{\left[\begin{array}{ccc}{a}_{22}& a_{23}& a_{24}\\ a_{32}& a_{33}& a_{34}\\ a_{42}& a_{43}& a_{44}\end{array}\right]}_{3\times 3}}\end{array}\right)}^{\ast }.$$

Proof. Since $\left|\begin{array}{ccc}{a}_{22}& a_{23}& a_{24}\\ a_{32}& a_{33}& a_{34}\\ a_{42}& a_{43}& a_{44}\end{array}\right|\ne 0$, using Lemma 2.5 we have $$det(A_5)=\frac{\left|\begin{array}{cc}\left|\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right|& \left|\begin{array}{cccc}{a}_{12}& a_{13}& a_{14}& a_{15}\\ a_{22}& a_{23}& a_{24}& a_{25}\\ a_{32}& a_{33}& a_{34}& a_{35}\\ a_{42}& a_{43}& a_{44}& a_{45}\end{array}\right|\\ \\ \left|\begin{array}{cccc}{a}_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\\ a_{51}& a_{52}& a_{53}& a_{54}\end{array}\right|& \left|\begin{array}{cccc}{a}_{22}& a_{23}& a_{24}& a_{25}\\ a_{32}& a_{33}& a_{34}& a_{35}\\ a_{42}& a_{43}& a_{44}& a_{45}\\ a_{52}& a_{53}& a_{54}& a_{55}\end{array}\right|\end{array}\right|}{\left|\begin{array}{ccc}{a}_{22}& a_{23}& a_{24}\\ a_{32}& a_{33}& a_{34}\\ a_{42}& a_{43}& a_{44}\end{array}\right|}.$$ Besides, we know that all $a_{22},a_{23},a_{24},a_{32},a_{33},a_{34},a_{42},a_{43},a_{44}$ are nonzero numbers. So, using Lemma 2.6 for all $4\times 4$ within (1), we have $$det(A_5)=\frac{\left|\begin{array}{cc}{S}_{11}& S_{12}\\ S_{21}& S_{22}\end{array}\right|}{\left|\begin{array}{ccc}{a}_{22}& a_{23}& a_{24}\\ a_{32}& a_{33}& a_{34}\\ a_{42}& a_{43}& a_{44}\end{array}\right|},$$ where $S_{ij}(\mathrm{\forall }i,j=1,2)$ is equal to $$\frac{\left|\begin{array}{cc}\frac{\left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{ij}& a_{i(j+1)}\\ a_{(i+1)j}& a_{(i+1)(j+1)}\end{array}\right|& \left|\begin{array}{cc}{a}_{i(j+1)}& a_{i(j+2)}\\ a_{(i+1)(j+1)}& a_{(i+1)(j+2)}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{(i+1)j}& a_{(i+1)(j+1)}\\ a_{(i+2)j}& a_{(i+2)(j+1)}\end{array}\right|& \left|\begin{array}{cc}{a}_{(i+1)(j+1)}& a_{(i+1)(j+2)}\\ a_{(i+2)(j+1)}& a_{(i+2)(j+2)}\end{array}\right|\end{array}\right|}{a_{(i+1)(j+1)}}& \frac{\left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{i(j+1)}& a_{i(j+2)}\\ a_{(i+1)(j+1)}& a_{(i+1)(j+2)}\end{array}\right|& \left|\begin{array}{cc}{a}_{i(j+2)}& a_{i(j+3)}\\ a_{(i+1)(j+2)}& a_{(i+1)(j+3)}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{(i+1)(j+1)}& a_{(i+1)(j+2)}\\ a_{(i+2)(j+1)}& a_{(i+2)(j+2)}\end{array}\right|& \left|\begin{array}{cc}{a}_{(i+1)(j+2)}& a_{(i+1)(j+3)}\\ a_{(i+2)(j+2)}& a_{(i+2)(j+3)}\end{array}\right|\end{array}\right|}{a_{(i+1)(j+2)}}\\ \\ \frac{\left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{(i+1)j}& a_{(i+1)(j+1)}\\ a_{(i+2)j}& a_{(i+2)(j+1)}\end{array}\right|& \left|\begin{array}{cc}{a}_{(i+1)(j+1)}& a_{(i+1)(j+2)}\\ a_{(i+2)(j+1)}& a_{(2+1)(j+2)}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{(i+2)j}& a_{(i+2)(j+1)}\\ a_{(i+3)j}& a_{(i+3)(j+1)}\end{array}\right|& \left|\begin{array}{cc}{a}_{(i+2)(j+1)}& a_{(i+2)(j+2)}\\ a_{(i+3)(j+1)}& a_{(i+3)(j+2)}\end{array}\right|\end{array}\right|}{a_{(i+2)(j+1)}}& \frac{\left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{(i+1)(j+1)}& a_{(i+1)(j+2)}\\ a_{(i+2)(j+1)}& a_{(i+2)(j+2)}\end{array}\right|& \left|\begin{array}{cc}{a}_{(i+1)(j+2)}& a_{(i+1)(j+3)}\\ a_{(i+2)(j+2)}& a_{(2+1)(j+3)}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{(i+2)(j+1)}& a_{(i+2)(j+2)}\\ a_{(i+3)(j+1)}& a_{(i+3)(j+2)}\end{array}\right|& \left|\begin{array}{cc}{a}_{(i+2)(j+2)}& a_{(i+2)(j+3)}\\ a_{(i+3)(j+2)}& a_{(i+3)(j+3)}\end{array}\right|\end{array}\right|}{a_{(i+2)(j+2)}}\end{array}\right|}{\left|\begin{array}{cc}{a}_{(i+1)(j+1)}& a_{(i+1)(j+2)}\\ a_{(i+2)(j+1)}& a_{(i+2)(j+2)}\end{array}\right|}.$$ Using Definition 2.3, we obtain $$\frac{\left|\begin{array}{cc}{S}_{11}& S_{12}\\ S_{21}& S_{22}\end{array}\right|}{\left|\begin{array}{ccc}{a}_{22}& a_{23}& a_{24}\\ a_{32}& a_{33}& a_{34}\\ a_{42}& a_{43}& a_{44}\end{array}\right|}$$ $$={\left(\begin{array}{c}\frac{{\left[\begin{array}{ccc}\left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|& \left|\begin{array}{cc}{a}_{12}& a_{13}\\ a_{22}& a_{23}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|& \left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|\end{array}\right|& \left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{12}& a_{13}\\ a_{22}& a_{23}\end{array}\right|& \left|\begin{array}{cc}{a}_{13}& a_{14}\\ a_{23}& a_{24}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|& \left|\begin{array}{cc}{a}_{23}& a_{24}\\ a_{33}& a_{34}\end{array}\right|\end{array}\right|& \left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{13}& a_{14}\\ a_{23}& a_{24}\end{array}\right|& \left|\begin{array}{cc}{a}_{14}& a_{15}\\ a_{24}& a_{25}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{23}& a_{24}\\ a_{33}& a_{34}\end{array}\right|& \left|\begin{array}{cc}{a}_{24}& a_{25}\\ a_{34}& a_{35}\end{array}\right|\end{array}\right|\\ \\ \left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|& \left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{31}& a_{32}\\ a_{41}& a_{42}\end{array}\right|& \left|\begin{array}{cc}{a}_{32}& a_{33}\\ a_{42}& a_{43}\end{array}\right|\end{array}\right|& \left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|& \left|\begin{array}{cc}{a}_{23}& a_{24}\\ a_{33}& a_{34}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{32}& a_{33}\\ a_{42}& a_{43}\end{array}\right|& \left|\begin{array}{cc}{a}_{33}& a_{34}\\ a_{43}& a_{44}\end{array}\right|\end{array}\right|& \left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{23}& a_{24}\\ a_{33}& a_{34}\end{array}\right|& \left|\begin{array}{cc}{a}_{24}& a_{25}\\ a_{34}& a_{35}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{33}& a_{34}\\ a_{43}& a_{44}\end{array}\right|& \left|\begin{array}{cc}{a}_{34}& a_{35}\\ a_{44}& a_{45}\end{array}\right|\end{array}\right|\\ \\ \left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{31}& a_{32}\\ a_{41}& a_{42}\end{array}\right|& \left|\begin{array}{cc}{a}_{32}& a_{33}\\ a_{42}& a_{43}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{41}& a_{42}\\ a_{51}& a_{52}\end{array}\right|& \left|\begin{array}{cc}{a}_{42}& a_{43}\\ a_{52}& a_{53}\end{array}\right|\end{array}\right|& \left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{32}& a_{33}\\ a_{42}& a_{43}\end{array}\right|& \left|\begin{array}{cc}{a}_{33}& a_{34}\\ a_{43}& a_{44}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{42}& a_{43}\\ a_{52}& a_{53}\end{array}\right|& \left|\begin{array}{cc}{a}_{43}& a_{44}\\ a_{53}& a_{54}\end{array}\right|\end{array}\right|& \left|\begin{array}{cc}\left|\begin{array}{cc}{a}_{33}& a_{34}\\ a_{43}& a_{44}\end{array}\right|& \left|\begin{array}{cc}{a}_{34}& a_{35}\\ a_{44}& a_{45}\end{array}\right|\\ \\ \left|\begin{array}{cc}{a}_{43}& a_{44}\\ a_{53}& a_{54}\end{array}\right|& \left|\begin{array}{cc}{a}_{44}& a_{45}\\ a_{54}& a_{55}\end{array}\right|\end{array}\right|\end{array}\right]}_{3\times 3}}{{\left[\begin{array}{ccc}{a}_{22}& a_{23}& a_{24}\\ a_{32}& a_{33}& a_{34}\\ a_{42}& a_{43}& a_{44}\end{array}\right]}_{3\times 3}}\end{array}\right)}^{\ast }.$$ Clearly using Definition 2.4, the top part of fraction (3) is equal to $DC(DC(A_5))$, consequently (2) and (3) give $$det(A_5)={\left(\begin{array}{c}\frac{DC(DC(A_5))}{{\left[\begin{array}{ccc}{a}_{22}& a_{23}& a_{24}\\ a_{32}& a_{33}& a_{34}\\ a_{42}& a_{43}& a_{44}\end{array}\right]}_{3\times 3}}\end{array}\right)}^{\ast }.$$ The theorem is proved.

Note that in the Theorem 3.1, if the matrix $\left[\begin{array}{ccc}{a}_{22}& a_{23}& a_{24}\\ a_{32}& a_{33}& a_{34}\\ a_{42}& a_{43}& a_{44}\end{array}\right]$ is not doublynonsingular or if some elements of it are zero, then by adding a multiple of one row to another row, or a multiple of one column to another column of the main matrix $A_5$, these problems can be resolved (since the determinant of the main matrix does not change). \newpage

Example 3.2. Let $A_5={\left[\begin{array}{ccccc}3& 5& 1& 0& 4\\ 2& 1& 6& 3& 2\\ 4& 3& 2& 2& 5\\ 1& 6& 1& 3& 4\\ 7& 5& 4& 4& 3\end{array}\right]}_{5\times 5}$. To obtain the $det(A_5)$, we have $$\stackrel{DC(A_5)}{\to }{\left[\begin{array}{cccc}-7& 29& 3& -12\\ 2& -16& 6& 11\\ 21& -9& 4& -7\\ -37& 19& -8& -7\end{array}\right]}_{4\times 4}\stackrel{DC(DC(A_5))}{\to }{\left[\begin{array}{ccc}54& 222& 105\\ 318& -10& -86\\ 66& -4& -84\end{array}\right]}_{3\times 3}.$$ $$det(A_5)={\left(\begin{array}{c}\frac{{\left[\begin{array}{ccc}54& 222& 105\\ 318& -10& -86\\ 66& -4& -84\end{array}\right]}_{3\times 3}}{{\left[\begin{array}{ccc}1& 6& 3\\ 3& 2& 2\\ 6& 1& 3\end{array}\right]}_{3\times 3}}\end{array}\right)}^{\ast }=\frac{\left|\begin{array}{cc}\frac{\left|\begin{array}{cc}\frac{54}{1}& \frac{222}{6}\\ \frac{318}{3}& \frac{-10}{2}\end{array}\right|}{\left|\begin{array}{cc}1& 6\\ 3& 2\end{array}\right|}& \frac{\left|\begin{array}{cc}\frac{222}{6}& \frac{105}{3}\\ \frac{-10}{2}& \frac{-86}{2}\end{array}\right|}{\left|\begin{array}{cc}6& 3\\ 2& 2\end{array}\right|}\\ \\ \frac{\left|\begin{array}{cc}\frac{318}{3}& \frac{-10}{2}\\ \frac{66}{6}& \frac{-4}{1}\end{array}\right|}{\left|\begin{array}{cc}3& 2\\ 6& 1\end{array}\right|}& \frac{\left|\begin{array}{cc}\frac{-10}{2}& \frac{-86}{2}\\ \frac{-4}{1}& \frac{-84}{3}\end{array}\right|}{\left|\begin{array}{cc}2& 2\\ 1& 3\end{array}\right|}\end{array}\right|}{\left|\begin{array}{ccc}1& 6& 3\\ 3& 2& 2\\ 6& 1& 3\end{array}\right|}$$ $$=\frac{\left|\begin{array}{cc}262& -236\\ 41& -8\end{array}\right|}{\left|\begin{array}{ccc}1& 6& 3\\ 3& 2& 2\\ 6& 1& 3\end{array}\right|}=\frac{7580}{-5}=-1516.$$

4. Conclusion

We presented a new method to compute the determinant of a $5\times 5$ matrix. In fact, this is a generalization of a simpler method for $4\times 4$ matrices which was previously provided in [2]. It seems to be possible to generalize this method for matrices of order $n$. Of course, for more generalizations, more calculations are required.

Competing Interests

The author do not have any competing interests in the manuscript.

Acknowledgments

The author would like to thank the editor and the anonymous referee for their helpful comments.