Computation of Matrix Determinants by Cross-Multiplication: A Rethinking of Dodgson’s Condensation and Reduction by Elementary Row Operations Methods

Abstract

We formulate a more straightforward, symmetry-based technique for manually computing the determinant of any $n\times n$ matrix by revisiting Dodgson’s condensation method, as well as strategically applying elementary row (column) operations and the definition and properties of determinants. The result yields a more streamlined algorithm that is generalized through formulas and employs a smaller number of operations and succeeding matrices than the existing methods.

1. Introduction

The study of determinants continues as a relevant mathematical area pursued for its utility in various fields of mathematics, engineering, and sciences. The invention of the determinant as a unique number associated with a square array of numbers by Seki Kowa, in the late 17th century, predated the introduction of matrices in the 19th century by James Sylvester and Charles Dodgson (who preferred to use the term blocks) and, consequently, the formal study of linear algebra [1,2]. With the development of linear algebra, determinants and matrices have become fundamental and significant mathematical tools due to their efficiency in representing quantities in a compact form. The solution of systems of linear equations, matrix inversion, computations of areas, and representations of complicated expressions in applied sciences, mechanical, and electrical systems can be efficiently modeled by determinants and matrices [3,4,5]. Consequently, explorations of various methods of computing determinants whether of explicit numbers or symbolic representations have become significant mathematical pursuits [4,6].

The methods of computing determinants usually presented in elementary linear algebra texts for instructional exposition purposes include the application of Leibniz’s definition of determinants, Laplace’s co-factor expansion, LU decomposition, and reduction by elementary row or column operations [5,7,8]. Research literature also offers a plethora of methods, such as Chio’s condensation, QR decomposition method, Cholesky decomposition, Dodson’s condensation method, Hajrizaj’s method, Salihu and Gjonbalaj’s method, and Sobamowo’s extension of Sarrus’ rule to $4\times 4$ matrices [2,4,9,10,11]. Other methods are proposed on the basis of reducing built-in computational errors and specialized matrix forms to include division-free algorithms to compute the determinants of quasi-tridiagonal matrices [12,13], as well as develop determinant formulas for special matrices involving symmetry, such as block matrices [14], break-down free algorithms for computing the determinants of periodic tridiagonal matrices [15], and block diagonalization-based algorithms of block k-tridiagonal matrices [16].

While computer programs provide the fastest, most efficient, and highly accurate platforms in computing determinants, there are significant reasons why manually computing the determinants is essential [4]. In low-technology environments or in controlled environments, such as licensure certifications, where the use of advanced technological tools is restricted, there can only be reliance on one’s computational fluency to carry out problem solving. Hence, familiarity and mastery of efficient methods in manual computations is paramount. The goal of this paper is to develop a straightforward and efficient algorithm for the manual computation of determinants with pedagogical implications in mind. Mathematics education entails the development of procedural knowledge or action sequences for solving problems, as well as the conceptual knowledge or understanding of principles that govern relationships between pieces of knowledge [17]. With an efficient method that promotes computational fluency through optimized instructional time and cognitive effort, greater appreciation and deeper understanding of underlying concepts is likely facilitated.

1.1. Determinants by Definition

Definition 1.

The determinant of an $n\times n$ matrix A denoted by $det \left[A\right]$ or |A| is defined as

$$\mathit{det}\left(A\right)=\sum \left(\pm \right)a_{1j_1}{a}_{2j_2}{a}_{3j_3}\dots a_{n{j}_n}$$

where  $j_1{j}_2{j}_3\dots j_n$ are column numbers derived from the permutations of the set  $S=\{1, 2, 3, \dots , n\}$.

The sign is taken as + for even permutations and − for odd permutations [5,7]. Accordingly, the determinants of simple matrices can be readily computed. The determinant of a single-entry matrix equals the entry itself, $\mathrm{d}\mathrm{e}\mathrm{t}\left[a_{11}\right]=a_{11}.$ The butterfly method for $2\times 2$ matrices is so-called because the movement of computation suggests butterfly wings, that is, $\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|=a_{11}{a}_{22}-a_{21}{a}_{12}$. By Definition 1, the determinant of a $3\times 3$ is expanded as

$$\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|=a_{11}{a}_{22}{a}_{33}+{a_{13}a}_{21}{a}_{32}+a_{12}{a}_{23}{a}_{31}-a_{13}{a}_{22}{a}_{31}-a_{12}{a}_{21}{a}_{33}-a_{11}{a}_{23}{a}_{32}.$$

The Sarrus’ rule is derived from this expansion. By extending the first two columns of the matrix to the right, the downward arrows give the positive terms and the upward arrows give the negative terms in the expansion of the formula.

Applying the definition when computing the determinant of $n\times n$ matrix requires $n!$ Terms, with each containing $n$ factors involving $n-1$ multiplications.

1.2. Determinants by Co-factor Expansion

In the computation of determinants by co-factor expansion, two concepts defined below are essential.

Definition 2.

The minor of an entry $a_{ij}$ of an $n\times n$ matrix A is the determinant of the $(n-1)\times (n-1)$ submatrix $M_{ij}$ of A obtained by deleting the $ith row and jth column$ of A. The co-factor of $a_{ij}$ is a real number denoted by $A_{ij}$ and is defined by

$$A_{ij}={(-1)}^{i+j} \mathit{det}\left(M_{ij}\right)$$

After computing the minor and its associated co-factor, we compute the determinant by expanding about an ith row (the row containing all $a_{ij}$):

$$\det \left(A\right)=a_{i1}{A}_{i1}+a_{i2}{A}_{i2}+a_{i3}{A}_{i3}+\dots +a_{in}{A}_{in}$$

or jth column:

$$\det \left(A\right)=a_{1j}{A}_{1j}+a_{2j}{A}_{2j}+a_{3j}{A}_{3j}+\dots +a_{nj}{A}_{nj}$$

In co-factor expansion, we “expand” the $n\times n$ matrix, about a certain row or column, into $n$ submatrices $M_{ij}$, each with $n-1$ rows and $n-1$ columns. This means that, at the first expansion, we compute $n$ minors and $n$ co-factors. For larger matrices, we may not readily compute the minors, but we have to perform a second expansion of each $M_{ij}$ into $n-1$ submatrices, each with $n-2$ rows and $n-2$ columns. The process is repeated over the resulting submatrices until smaller submatrices with readily computed determinants are obtained. The determinant of the matrix is then computed by working backward.

1.3. Determinants by Elementary Row (Column) Operations

The Gaussian method of computing the determinants employs elementary row (column) operations to put the matrix into upper or lower triangular form, in which the resulting entries in the main diagonal yield the determinant. The three elementary row (column) operations and their effects on the determinant of a given matrix are stated as follows. For proofs of the theorems, see [5,7].

• Type 1. Multiplying a row (column) through by a nonzero constant.
• Type 2. Interchanging two rows (columns).
• Type 3. Adding a constant times one row (column) to another.

Theorem 1.

If $B=\left[b_{ij}\right]$ is the matrix that results when a single row (column) of $A=\left[a_{ij}\right]$ is multiplied by a scalar $c,$ then $\mathit{det}\left(B\right)=c det\left(A\right)$.

Theorem 2.

If $B=\left[b_{ij}\right]$ is the matrix that results when two rows (columns) of $A=\left[a_{ij}\right]$ are interchanged, then $\mathit{det}\left(B\right)=-det\left(A\right)$.

Theorem 3.

If $B=\left[b_{ij}\right]$ is the matrix that results when a multiple of a row $R_r$ (column $C_r$) of $A=\left[a_{ij}\right]$ is added to another row $R_s$ (column $C_s)$ then $\mathit{det}\left(B\right)=det\left(A\right)$.

By reducing a matrix into upper or lower triangular form and applying co-factor expansion, the determinant of the matrix is determined by the product of the entries in the main diagonal, as stated in the following theorem.

Theorem 4.

If A = $\left[a_{ij}\right]$ is upper (lower) triangular, then det(A) = $a_{11}{a}_{22}{a}_{33}\dots a_{nn}$; that is, the determinant of a triangular matrix is the product of the elements on the main diagonal.

1.4. Dodgson’s Condensation Method

The Dodgson’s condensation method, attributed to the English clergyman Rev. Charles Lutwidge Dodgson (1832–1898), famously known as Lewis Caroll for his literary works Alice in Wonderland and Through the Looking Glass, is found to be more efficient than Leibniz’s definition, especially for larger matrices [2,18,19]. Dodgson’s method aims to ‘condense’ the determinant by producing an $\left(n-1\right)\times (n-1)$ matrix from an $n\times n$ matrix, followed by $(n-2)\times \left(n-2\right)$, until a $1\times 1$ matrix is obtained. The condensation method is founded on Jacobi’s theorem [19].

Theorem 5.

Jacobi’s Theorem. Let A be an $n\times n$ matrix, let $A_{ij}$ be an $m\times m$ minor of $A$, where $m<n$ , let ${[A\prime }_{ij}]$ be the corresponding $m\times m$ minor of $A\prime$, and let $\left[A_{ij}^{*}\right]$ be the complementary $(n-m)\times (n-m)$ minor of $A$. Then:

$$\mathit{det} {[A\prime }_{ij}]={det\left(A\right)}^{m-1}·\mathit{det}\left[A_{ij}^{*}\right].$$

With $m=2,$ Dodgson realized that the determinant of $A$ can be readily computed with

$$\det \left[A\right]=\frac{\mathrm{d}\mathrm{e}\mathrm{t}{[A\prime }_{ij}]}{\det \left[A_{ij}^{*}\right].}$$

His algorithm consists of the following steps.

• Check the interior of $A$ for a zero entry. The interior of A is an $(n-2)\times \left(n-2\right)$ matrix that remains when the first row, last row, first column, and last column of A are deleted. We perform elementary row operations to remove all zeros from the interior of A.
• Compute the determinant of every four adjacent terms to form a new $(n-1)\times \left(n-1\right)$ matrix B.
• Repeat Step 2 to produce an $(n-2)\times \left(n-2\right)$ matrix. We then divide each term by the corresponding entry in the interior of the original matrix A to obtain matrix C.
• We continue the process of condensation with the succeeding matrices until a $1\times 1$ matrix is obtained which computes detA.
• Note that Dodgson’s method employs division of entries in succeeding matrices beginning with an $(n-2)\times \left(n-2\right)$ matrix. This step which may magnify computational error, especially for matrices with non-integral entries.

2. Development of Proposed Algorithm

2.1. Reduction by Cross-Multiplication

We start with an $n\times n$ matrix denoted by $A_n$ with real entries

$$A_n=\left[\begin{array}{ccc}{a}_{11}& a_{12}& \begin{array}{cc}\dots & a_{1n}\end{array}\\ a_{21}& a_{22}& \begin{array}{cc}\dots & a_{2n}\end{array}\\ \begin{array}{c}\begin{array}{c}.\\ .\end{array}\\ a_{n-1 1}\\ a_{n1}\end{array}& \begin{array}{c}\begin{array}{c}.\\ .\end{array}\\ a_{n-1 2}\\ a_{n2}\end{array}& \begin{array}{c}\begin{array}{cc}\begin{array}{c}\dots \\ \dots \end{array}& \begin{array}{c}.\\ .\end{array}\end{array}\\ \begin{array}{cc}\dots & a_{n-1 n}\end{array}\\ \begin{array}{cc}\dots & a_{nn}\end{array}\end{array}\end{array}\right]$$

where the first entries are non-zero real numbers to be transformed into triangular form to apply Theorem 4. A more straightforward way of zeroing out entries under $a_{11}$ in $A_n$, starting from the bottom row $R_n$, is through $a_{n-1 1}{R}_n-a_{n1}{R}_{n-1}$. By Theorem 3, the process introduces the factor $a_{n-1 1}$ into the determinant. Working our way up, we zero out the first entry of $R_{n-1}$ through $a_{n-2 1}{R}_{n-1}-a_{n-1 1}{R}_{n-2}$, which then introduces the factor $a_{n-2 1}$ into the determinant. By repeating the process upward until $R_2$, we have introduced the following factors into the determinant of $A_n$:$a_{11, }{a}_{21}, \dots , a_{n-2 1}$, $a_{n-1 1}.$ At this stage, we obtain an $(n-1)\times (n-1)$ submatrix $A_{n-1 }$ by excluding the first column at the left with zero entries.

$A_{n-1 }$	$a_{11}{a}_{22}-a_{21}{a}_{12}$	$a_{11}{a}_{23}-a_{21}{a}_{13}$	…	$a_{11}{a}_{2n}-a_{21}{a}_{1n}$
	$a_{21}{a}_{32}-a_{31}{a}_{22}$	$a_{21}{a}_{33}-a_{31}{a}_{23}$	…	$a_{21}{a}_{3n}-a_{31}{a}_{2n}$
	…	…	…	…
	$a_{n-11}{a}_{n2}-a_{n1}{a}_{n-12}$	$a_{n-11}{a}_{n3}-a_{n1}{a}_{n-13}$	…	$a_{n-11}{a}_{nn}-a_{n1}{a}_{n-1n}$

For the purpose of this algorithm, we do away with the usual matrix notation and we put the rows of $A_{n-1 }$ under $A_n$. We denote the entries in succeeding matrices by $a_{ij}^{\left(m\right)}=a_{i1}^{(m-1)}{a}_{i+1 j}^{(m-1)}-a_{i+11}^{(m-1)}{a}_{i j}^{(m-1)}$; $\left(m\right)$ = number of reductions performed and $a_{ij}^{\left(0\right)}=a_{ij}$. In determinant form, we can also denote $a_{ij}^{\left(m\right)}=\left|\begin{array}{cc}{a}_{i1}^{(m-1)}& a_{i j}^{(m-1)}\\ a_{i+11}^{(m-1)}& a_{i+1 j}^{(m-1)}\end{array}\right|$, which provides a symmetric (butterfly) movement of the algorithm, hence the term cross-multiplication.

$A_n$	$a_{11}$	$a_{12}$		…	$a_{1n}$
	$a_{21}$	$a_{22}$		…	$a_{2n}$
	.	.		…	.
	.	.		…	.
	$a_{n-1 1}$	$a_{n-1 2}$			$a_{n-1 n}$
	$a_{n1}$	$a_{n2}$		…	$a_{nn}$
$A_{n-1}$		$a_{11}^{\left(1\right)}$	$a_{12}^{\left(1\right)}$		$a_{1n-1}^{\left(1\right)}$
		$a_{21}^{\left(1\right)}$	$a_{22}^{\left(1\right)}$		$a_{2 n-1}^{\left(1\right)}$
		.		…	.
		$a_{n-2 1}^{\left(1\right)}$	$a_{n-2 2}^{\left(1\right)}$	…	$a_{n-2 n-1}^{\left(1\right)}$
		$a_{n-1 1}^{\left(1\right)}$	$a_{n-1 2}^{\left(1\right)}$	…	$a_{n-1 n-1}^{\left(1\right)}$

The next stage is obtaining the rows for $A_{n-2}$, which is equivalent to zeroing out entries under $a_{11}^{\left(1\right)}$ by following a similar process to obtaining the rows for $A_{n-1}$. The process introduces the following factors into the determinant: $a_{11}^{\left(2\right)}$, $a_{21}^{\left(2\right)}$, …, $a_{n-3 1}^{\left(2\right)}$, $a_{n-2 1}^{\left(2\right)}$. Assuming no first entries of each $A_i$ are zero, continuing the process of reduction leads to a $2\times 2 \mathrm{s}\mathrm{u}\mathrm{b}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{x}$ $A_2$ with an introduced factor of $a_{1 1}^{(n-2)}$ and, eventually, $1\times 1 \mathrm{s}\mathrm{u}\mathrm{b}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{x}$ $A_1$ with entry

$$a_{1 1}^{(n-1)}.$$

By taking the first row from each submatrix, we can reconstruct an $n\times n$ matrix $B$ in upper triangular form that is a row equivalent to $A_n$.

$a_{11}$	$a_{12}$		…				$a_{1n}$
0	$a_{11}^{\left(1\right)}$	$a_{13}^{\left(1\right)}$					$a_{1n-1}^{\left(1\right)}$
0	0	$a_{11}^{\left(2\right)}$	$a_{14}^{\left(2\right)}$				$a_{1 n-2}^{\left(2\right)}$
0	0	0	.	.	…
0	0	0		.	…
0	0	0		.	…
0	0	0		.	…	$a_{11}^{(n-2)}$	$a_{12}^{(n-2)}$
0	0	0	0	0	…	0	$a_{1 1}^{(n-1)}$

By Theorem 4

$$\det \left(B_n\right)=a_{11}·a_{11}^{\left(1\right)}· a_{11}^{\left(2\right)}·\dots · a_{1 1}^{\left(n-2\right)}·a_{11}^{\left(n-1\right)}.$$

Since $A_n$ is row equivalent to $B_n$ and these factors

$$a_{11}{a}_{21}{a}_{31}\dots a_{n-1 1}·a_{11}^1{a}_{21}^{\left(1\right)}\dots a_{n-21}^{\left(1\right)}{a}_{n-1 1}^{\left(1\right)}·\dots ·a_{11}^{(n-2)}$$

are introduced into the determinant of $A_n$ in the process of forming $B_n$, then

$$\mathrm{d}\mathrm{e}\mathrm{t}\left(B_n\right)=a_{11}{a}_{21}{a}_{31}\dots a_{n-1 1}· a_{11}^1{a}_{21}^{\left(1\right)}\dots a_{n-2 1}^{\left(1\right)}{a}_{n-1 1}^{\left(1\right)}·\dots ·a_{11}^{\left(n-2\right)}\det \left(A_n\right)$$

from which

$$\det \left(A_n\right)=\frac{\mathrm{d}\mathrm{e}\mathrm{t}\left(B_n\right)}{a_{11}{a}_{21}{a}_{31}\dots a_{n-1 1}· a_{11}^1{a}_{21}^{\left(1\right)}\dots a_{n-2 1}^{\left(1\right)}{a}_{n-1 1}^{\left(1\right)}·\dots ·a_{11}^{\left(n-2\right)}}$$

$$\det \left(A_n\right)=\frac{a_{11}·a_{12}^{\left(1\right)}· a_{13}^{\left(2\right)}·\dots · a_{11}^{\left(n-2\right)}·a_{11}^{\left(n-1\right)}}{a_{11}{a}_{21}{a}_{31}\dots a_{n-1 1}· a_{11}^1{a}_{21}^{\left(1\right)}\dots a_{n-2 1}^{\left(1\right)}{a}_{n-1 1}^{\left(1\right)}·\dots ·a_{11}^{\left(n-2\right)}}$$

$$\det \left(A_n\right)=\frac{\prod _{m=0}^{n-1}{a}_{11}^{\left(m\right)}}{\prod _{m=0}^{n-2}{\prod }_{j=1}^{n-m-1}{a}_{j1}^{\left(m\right)}}$$

(1)

2.2. Alternate Proof of Cross-Multiplication Method

Suppose that B is an $n\times n$ matrix where all entries under the first element of the first row are zeros,

$$B_n=\left[\begin{array}{ccc}{b}_{11}& b_{12}& \begin{array}{cc}\dots & b_{1n}\end{array}\\ 0& b_{22}& \begin{array}{cc}\dots & b_{2n}\end{array}\\ \begin{array}{c}\begin{array}{c}.\\ .\end{array}\\ 0\\ 0\end{array}& \begin{array}{c}\begin{array}{c}.\\ .\end{array}\\ b_{n-1 2}\\ b_{n 2}\end{array}& \begin{array}{c}\begin{array}{cc}\begin{array}{c}\dots \\ \dots \end{array}& \begin{array}{c}.\\ .\end{array}\end{array}\\ \begin{array}{cc}\dots & b_{n-1 n}\end{array}\\ \begin{array}{cc}\dots & b_{n n}\end{array}\end{array}\end{array}\right].$$

By co-factor expansion, $\det \left(B_n\right)=b_{11}{(-1)}^{1+1}\left[\begin{array}{ccc}{b}_{22}& \dots & b_{2n}\\ .& \dots & .\\ \begin{array}{c}{b}_{n-12}\\ b_{n2}\end{array}& \begin{array}{c}\dots \\ \dots \end{array}& \begin{array}{c}{b}_{n-1n}\\ b_{nn}\end{array}\end{array}\right]$. If $n>4$, then computing the determinant of the resulting $(n-1)\times (n-1)$ submatrix requires a full co-factor expansion or elementary row operations. In deriving the proposed method, we first transform $A_n$ into $B_n$ through elementary row operations; then, we apply the definition of co-factor expansion by expanding the first column. We then repeat the process with the succeeding submatrices until a relatively smaller matrix is obtained where we can readily compute the determinant. By introducing the cross-multiplication, we only form one $(n-1)\times (n-1)$ submatrix instead of $n$ $(n-1)\times (n-1)$ submatrices in co-factor expansion.

Let $A$ be an $n\times n$ matrix with real entries, where $a_{11},a_{21},a_{n-1 1},a_{n1}\ne 0.$

$$A_n=\left[\begin{array}{ccc}{a}_{11}& a_{12}& \begin{array}{cc}\dots & a_{1n}\end{array}\\ a_{21}& a_{22}& \begin{array}{cc}\dots & a_{2n}\end{array}\\ \begin{array}{c}\begin{array}{c}.\\ .\end{array}\\ a_{n-1 1}\\ a_{n1}\end{array}& \begin{array}{c}\begin{array}{c}.\\ .\end{array}\\ a_{n-1 2}\\ a_{n2}\end{array}& \begin{array}{c}\begin{array}{cc}\begin{array}{c}\dots \\ \dots \end{array}& \begin{array}{c}.\\ .\end{array}\end{array}\\ \begin{array}{cc}\dots & a_{n-1 n}\end{array}\\ \begin{array}{cc}\dots & a_{nn}\end{array}\end{array}\end{array}\right]$$

We zero out entries under $a_{11}$ through the process illustrated in this paper: with two successive rows, $R_i$ and $R_{i+1}$, with first entries, $a_{i1}$ and $a_{i+1 1}$, respectively, and by $a_{i1}{R}_{i+1}-a_{i+1 1}{R}_i$, a matrix $B_1$ is produced.

$$B_1=\left[\begin{array}{ccc}\begin{array}{c}{a}_{11}\\ 0\end{array}& \begin{array}{c}{a}_{12}\\ a_{11}^{\left(1\right)}\end{array}& \begin{array}{cc}\dots & \begin{array}{c}{a}_{1n}\\ a_{1n-1}^{\left(1\right)}\end{array}\end{array}\\ .& .& \begin{array}{cc}\dots & .\end{array}\\ \begin{array}{c}.\\ .\\ 0\end{array}& \begin{array}{c}.\\ .\\ a_{n-1 1}^{\left(1\right)}\end{array}& \begin{array}{c}\begin{array}{cc}\dots & .\end{array}\\ \begin{array}{cc}\dots & .\end{array}\\ \begin{array}{cc}\dots & a_{n-1 n-1}^{\left(1\right)}\end{array}\end{array}\end{array}\right]$$

In the process, the factors $a_{11}, a_{21}, a_{31}, \dots , a_{n-1 1}$ are introduced into $\det \left(A_n\right).$ Thus, by Theorem 1,

$$\det {(B}_1)=a_{11}, a_{21}, a_{31}, \dots , a_{n-1 1}\mathrm{d}\mathrm{e}\mathrm{t}\left(A_n\right)$$

By co-factor expansion,

$$\det {(B}_1)=a_{11}\left|\begin{array}{ccc}{a}_{11}^{\left(1\right)}& \dots & a_{1n-1}^{\left(1\right)}\\ \dots & \dots & \dots \\ a_{n-1 1}^{\left(1\right)}& \dots & a_{n-1 n-1}^{\left(1\right)}\end{array}\right|=a_{11}\det {(A}_{n-1})$$

where $A_{n-1}=\left[\begin{array}{ccc}{a}_{11}^{\left(1\right)}& \dots & a_{1n-1}^{\left(1\right)}\\ \dots & \dots & \dots \\ a_{n-11}^{\left(1\right)}& \dots & a_{n-1n-1}^{\left(1\right)}\end{array}\right]$.

We repeat the process of reduction on $A_{n-1}$ to form the matrix

$$B_2=\left[\begin{array}{ccc}\begin{array}{c}{a}_{11}^{\left(1\right)}\\ 0\end{array}& \begin{array}{c}{a}_{12}^{\left(1\right)}\\ a_{11}^{\left(2\right)}\end{array}& \begin{array}{cc}\dots & \begin{array}{c}{a}_{1n-1}^{\left(1\right)}\\ a_{1n-2}^{\left(2\right)}\end{array}\end{array}\\ .& .& \begin{array}{cc}\dots & .\end{array}\\ \begin{array}{c}.\\ .\\ 0\end{array}& \begin{array}{c}.\\ .\\ a_{n-2 1}^{\left(2\right)}\end{array}& \begin{array}{c}\begin{array}{cc}\dots & .\end{array}\\ \begin{array}{cc}\dots & .\end{array}\\ \begin{array}{cc}\dots & a_{n-2 n-2}^{\left(1\right)}\end{array}\end{array}\end{array}\right]$$

and which

$$\det {(B}_2)=a_{11}^{\left(1\right)}{a}_{21}^{\left(1\right)}{a}_{31}^{\left(1\right)}\dots a_{\mathrm{n}-21}^{\left(1\right)}\mathrm{d}\mathrm{e}\mathrm{t}(A_{n-1})$$

$$\det {(B}_2)=a_{11}^{\left(1\right)}\mathrm{d}\mathrm{e}\mathrm{t}\left(A_{n-2}\right) \mathrm{where} A_{n-2}=\left[\begin{array}{ccc}{a}_{11}^{\left(2\right)}& \dots & a_{1n-2}^{\left(2\right)}\\ \dots & \dots & \dots \\ a_{n-21}^{\left(2\right)}& \dots & a_{n-2n-2}^{\left(2\right)}\end{array}\right]$$

Continuing the reduction on subsequent submatrices, we summarize the following results

$$\det {(B}_1)=a_{11}, a_{21}, a_{31}, \dots ,a_{n-1 1}\det \left(A_n\right)=a_{11}\det {(A}_{n-1})$$

$$\det {(B}_2)=a_{11}^{\left(1\right)}{a}_{21}^{\left(1\right)}{a}_{31}^{\left(1\right)}\dots a_{n-21}^{\left(1\right)}\det {(A}_{n-1})=a_{11}^{\left(1\right)}\mathrm{d}\mathrm{e}\mathrm{t}{(A}_{n-2})$$

$$\det {(B}_3)=a_{11}^{\left(2\right)}{a}_{21}^{\left(2\right)}{a}_{31}^{\left(2\right)}\dots a_{n-31}^{\left(2\right)}\det {(A}_{n-2})=a_{11}^{\left(2\right)}\mathrm{d}\mathrm{e}\mathrm{t}{(A}_{n-3})$$

$$\dots$$

$$\det {(B}_{n-2})=a_{11}^{(n-3)}{a}_{21}^{(n-3)}\det {(A}_2)=a_{11}^{(n-3)}\mathrm{d}\mathrm{e}\mathrm{t}{(A}_1)$$

$$\det {(B}_{n-1})=a_{11}^{(n-2)}\det {(A}_2)=a_{11}^{(n-2)}\mathrm{d}\mathrm{e}\mathrm{t}{(A}_1)=a_{11}^{(n-1)}$$

Working backwards leads to

$$\det \left(A_n\right)=\frac{a_{11}}{a_{11}, a_{21}, a_{31}, \dots ,a_{n-1 1}}·\frac{a_{11}^{\left(1\right)}}{a_{11}^{\left(1\right)}{a}_{21}^{\left(1\right)}{a}_{31}^{\left(1\right)}\dots a_{n-21}^{\left(1\right)}}\dots \frac{a_{11}^{\left(n-3\right)}}{a_{11}^{\left(n-3\right)}{a}_{21}^{\left(n-3\right)}}·\frac{a_{11}^{\left(n-2\right)}}{a_{11}^{\left(n-2\right)}} a_{11}^{\left(n-1\right)}.$$

$$\det \left(A_n\right)=\frac{a_{11}·a_{12}^{\left(1\right)}· a_{13}^{\left(2\right)}·\dots · a_{11}^{\left(n-2\right)}·a_{11}^{\left(n-1\right)}}{a_{11}{a}_{21}{a}_{31}\dots a_{n-1 1}· a_{11}^1{a}_{21}^{\left(1\right)}\dots a_{n-2 1}^{\left(1\right)}{a}_{n-1 1}^{\left(1\right)}·\dots ·a_{11}^{\left(n-2\right)}}$$

$$\det \left(A_n\right)=\frac{\prod _{m=0}^{n-1}{a}_{11}^{\left(m\right)}}{\prod _{m=0}^{n-2}{\prod }_{j=1}^{n-m-1}{a}_{j1}^{\left(m\right)}}$$

2.3. Matrices with Zero First Entries

To generalize the determinant formula, we consider cases where there are zero first entries in any of the submatrices $A_{n-i}$. If an $m$th row of $A_{n-i}$ has $a_{m1}^{\left(i\right)}=0$, the reduction $a_{m1}^{\left(i\right)}{R}_{m+1}-a_{m+1 1}^{\left(i\right)}{R}_m$ is not possible since this will introduce a zero factor to the divisor of Equation (1). We reduce the entire submatrix by first moving the rows with zero first entries to the bottom part; then, we perform reduction to rows with nonzero first entries. The rows with zero first entries are treated as “standby rows” and are, then, moved to the next submatrix in the same placement. Considering that $\det \left(A_{n-i}\right)=a_{11}^{\left(k\right)}\mathrm{d}\mathrm{e}\mathrm{t}(A_{n-i-1})$, by Theorem 2, interchanging $R_m$ and the last row will lead to $\det \left(A_{n-i}\right)={-a}_{11}^{\left(k\right)}\mathrm{d}\mathrm{e}\mathrm{t}(A_{n-i-1})$. We introduce the factor ${(-1)}^k$, where $k$ indicates the number of row interchanges.

$A_{n-i}$	$a_{11}^{\left(i\right)}$	$a_{11}^{\left(i\right)}$
	$a_{21}^{\left(i\right)}$	$a_{22}^{\left(i\right)}$
	…
	0	$a_{m-1 2}^{\left(i\right)}$
	0	$a_{m2}^i$
	$a_{m+1 1}^{\left(i\right)}$	$a_{m+1 2}^{\left(i\right)}$
	…
	$a_{n-i 1}^{\left(i\right)}$	$a_{n-i 1}^{\left(i\right)}$
$A_{n-i}$	$a_{11}^{\left(i\right)}$	$a_{11}^{\left(i\right)}$
	$a_{21}^{\left(i\right)}$	$a_{22}^{\left(i\right)}$
	…
	$a_{m+1 1}^{\left(i\right)}$	$a_{m+1 2}^{\left(i\right)}$
	$a_{n-i 1}^{\left(i\right)}$	$a_{n-i 1}^{\left(i\right)}$
	0	$a_{m-1 2}^{\left(i\right)}$
	0	$a_{m2}^{\left(i\right)}$
$A_{n-i-1}$		$a_{11}^{(i+1)}$	$a_{12}^{(i+1)}$
		…
		$a_{m-1 2}^{\left(i\right)}$
		$a_{m2}^i$

If column interchanges result in a matrix with nonzero first entries, we incorporate ${(-1)}^k$ to Equation (1) by applying Type 2 operation.

$$\det \left(A_n\right)=\frac{\prod _{m=0}^{n-1}{a}_{11}^{\left(m\right)}}{\prod _{m=0}^{n-2}{\prod }_{j=1}^{n-m-1}{a}_{j1}^{\left(m\right)}}{(-1)}^k$$

where k is the number of column interchanges.

If only one row of $A_{n-i}$ has a non-zero first entry, we place this row as $R_1$ with first entry $a_{11}^{(n-i)}$. No reduction is performed since the $A_{n-i}$ already takes the reduced form. By Theorem 4, $a_{11}^{(n-i)}$ is a diagonal entry; thus, no factor is introduced into the determinant.

If all rows of $A_{n-i}$ have zero first entries, then no reduction is performed, determining that $\det \left(A\right)=0.$

Taking into account cases where there are zero first entries, the general determinant formula can now be expressed as

$$\det \left(A_n\right)=\frac{\prod _{m=0}^{n-1}{a}_{11}^{\left(m\right)}}{\prod _{m=0}^{n-1}{\prod }_{j=1}^{r-1}{a}_{j1}^{\left(m\right)}}{(-1)}^k,$$

(2)

where $a_{j1}^{\left(0\right)}=a_{j1}$, r = number of rows with nonzero first entries in $A_{n-m}$, and k = number of row permutations.

If the first entries are nonzero (after permutating columns or applying Type 3 operations), we can reduce Equation (1) and incorporate ${(-1)}^k$ if column permutations are applied to get

$$\det \left(A_n\right)=\frac{a_{11}^{(n-1)}}{\prod _{m=0}^{n-2}{\prod }_{j=2}^{n-m-1}{a}_{j1}^{\left(m\right)}}{(-1)}^k$$

(3)

where k = number of column permutations.

It can be noted that the factors in the denominators are the in-between first entries in the submatrices, as reflected in the matrices below.

$A_n$	$a_{11}$	$a_{12}$		…			$a_{1n}$
	$a_{21}$	$a_{22}$		…			$a_{2n}$
	.	.		…			.
	.	.		…			.
	$a_{n-1 1}$	$a_{n-1 2}$					$a_{n-1 n}$
	$a_{n1}$	$a_{n2}$		…			$a_{nn}$
$A_{n-1}$		$a_{11}^{\left(1\right)}$	$a_{12}^{\left(1\right)}$				$a_{1n-1}^{\left(1\right)}$
		$a_{21}^{\left(1\right)}$	$a_{22}^{\left(1\right)}$				$a_{2 n-1}^{\left(1\right)}$
		.		…			.
		$a_{n-2 1}^{\left(1\right)}$	$a_{n-2 2}^{\left(1\right)}$	…			$a_{n-2 n-1}^{\left(1\right)}$
		$a_{n-1 1}^{\left(1\right)}$	$a_{n-1 2}^{\left(1\right)}$				$a_{n-1 n-1}^{\left(1\right)}$
$A_{n-2}$			$a_{11}^{\left(2\right)}$	$a_{12}^{\left(2\right)}$			$a_{1 n-2}^{\left(2\right)}$
			$a_{21}^{\left(2\right)}$	$a_{22}^{\left(2\right)}$			$a_{2 n-2}^{\left(2\right)}$
			.	.			.
			$a_{n-3 1}^{\left(2\right)}$	$a_{n-3 2}^{\left(2\right)}$			$a_{n-3 n-2}^{\left(2\right)}$
			$a_{n-2 1}^{\left(2\right)}$	$a_{n-2 2}^{\left(2\right)}$			$a_{n-2 n-2}^{\left(1\right)}$
…	…	…	…	…		…	…
$A_3$					$a_{1 1}^{(n-3)}$	$a_{1 2}^{(n-3)}$	$a_{1 3}^{(n-3)}$
					$a_{2 1}^{(n-3)}$	$a_{22}^{(n-3)}$	$a_{23}^{(n-3)}$
					$a_{3 1}^{(n-3)}$	$a_{3 2}^{(n-3)}$	$a_{3 3}^{(n-3)}$
$A_2$						$a_{1 1}^{(n-2)}$	$a_{12}^{(n-2)}$
						$a_{21 }^{(n-2)}$	$a_{22}^{(n-2)}$
$A_1$							$a_{1 1}^{(n-1)}$

3. Cross-Multiplication and Dodgson’s Condensation Method in 3 × 3 Matrices

We now show the equivalence of Dodgson’s condensation method and cross-multiplication.

Given that $A_3=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$, where $a_{11}, a_{21}, a_{31} and a_{22}$ are nonzero, by Dodgson’s condensation method, we take four adjacent entries at a time and compute their determinants to form a $2\times 2$ matrix.

$$\left[\begin{array}{cc}{a}_{11}{a}_{22}-a_{21}{a}_{12}& a_{12}{a}_{23}-a_{22}{a}_{13}\\ a_{21}{a}_{32}-a_{31}{a}_{22}& a_{22}{a}_{33}-a_{32}{a}_{23}\end{array}\right].$$

We then compute the determinant of the resulting matrix to obtain a $1\times 1$ matrix and divide the resulting entry by the corresponding entry of the interior of $A_3$, which is $a_{22}.$

$$\mathrm{d}\mathrm{e}\mathrm{t}(B_3)=\left[\begin{array}{c}{a}_{11}{a}_{22}{a}_{22}{a}_{33}-a_{11}{a}_{22}{a}_{32}{a}_{23}-a_{21}{a}_{12}{a}_{22}{a}_{33}\\ +a_{21}{a}_{32}{a}_{22}{a}_{13}+a_{31}{a}_{22}{a}_{12}{a}_{23}-a_{31}{a}_{22}{a}_{22}{a}_{13}\end{array}\right]$$

$$\mathrm{d}\mathrm{e}\mathrm{t} (B_3)=a_{22}\left[\begin{array}{c}{a}_{11}{a}_{22}{a}_{33}-a_{11}{a}_{32}{a}_{23}-a_{21}{a}_{12}{a}_{33}\\ +a_{21}{a}_{32}{a}_{13}+a_{31}{a}_{12}{a}_{23}-a_{31}{a}_{22}{a}_{13}\end{array}\right].$$

By Definition 1, ${\mathrm{d}\mathrm{e}\mathrm{t}(A}_3)$ = |$a_{11}{a}_{12}{a}_{33}-a_{11}{a}_{32}{a}_{23}-a_{21}{a}_{12}{a}_{33}+a_{21}{a}_{32}{a}_{13}+a_{31}{a}_{12}{a}_{23}-a_{31}{a}_{22}{a}_{13}|$.

Thus, ${\mathrm{d}\mathrm{e}\mathrm{t}(A}_3)=\frac{\mathrm{d}\mathrm{e}\mathrm{t}\left(B_3\right)}{a_{22}}$.

By exchanging columns 1 and 2 of $A_3,$ we have $A_3\mathrm{\prime }=\left[\begin{array}{ccc}{a}_{12}& a_{11}& a_{13}\\ a_{22}& a_{21}& a_{23}\\ a_{32}& a_{31}& a_{33}\end{array}\right]$, and it can be shown that $\mathrm{d}\mathrm{e}\mathrm{t}(A_3)=-\frac{\mathrm{d}\mathrm{e}\mathrm{t}\left({B^{\mathrm{\prime }}}_3\right)}{a_{21}}$.

By cross-multiplication,

$A_3$	$a_{11}$	$a_{12}$	$a_{13}$
	$a_{21}$	$a_{22}$	$a_{23}$
	$a_{31}$	$a_{32}$	$a_{33}$
$A_2$		$a_{11}{a}_{22}-a_{21}{a}_{12}$	$a_{11}{a}_{23}-a_{21}{a}_{13}$
		$a_{21}{a}_{32}-a_{31}{a}_{22}$	$a_{21}{a}_{33}-a_{31}{a}_{23}$
$A_1$			$\begin{array}{cc}\hfill a_{11}{a}_{22}{a}_{21}{a}_{33}-& a_{11}{a}_{22}{a}_{31}{a}_{23}\hfill \\ & -a_{21}{a}_{12}{a}_{21}{a}_{33}\hfill \\ & +a_{21}{a}_{12}{a}_{31}{a}_{23}\hfill \\ & -(a_{21}{a}_{32}{a}_{11}{a}_{23}\hfill \\ & -a_{21}{a}_{32}{a}_{21}{a}_{13}\hfill \\ & -a_{31}{a}_{22}{a}_{11}{a}_{23}\hfill \\ & +a_{31}{a}_{22}{a}_{21}{a}_{13})\hfill \end{array}$

Here, $a_{21}$ is the common factor; hence, $\mathrm{d}\mathrm{e}\mathrm{t}(A_3)=\frac{\mathrm{d}\mathrm{e}\mathrm{t}\left(A_1\right)}{a_{21}}$, which is consistent with Equation (1).

Similar to Dodgson’s approach, $a_{ij}^{\left(m\right)}$ is the determinant of the $2\times 2$ matrix formed by the first entries of two adjacent rows and the pair of corresponding entries in the $j\mathrm{t}\mathrm{h}$ column, that is, $a_{ij}^{\left(m\right)}=\left|\begin{array}{cc}{a}_{i1}^{(m-1)}& a_{i j}^{(m-1)}\\ a_{i+11}^{(m-1)}& a_{i+1 j}^{(m-1)}\end{array}\right|$. Here, we fix the first entries of two adjacent rows as the pivot in computing the determinants of consecutive $2\times 2$ matrices. By doing so, $a_{i+11}$ is the factor introduced into the original determinant, which is analogous to the corresponding entry of the interior matrix in the condensation method. By fixing the pivot at the first two entries, we can limit the number of divisions to be performed in preserving the determinant.

4. Verifications of Proposed Methods

We provide illustrative examples for the proposed methods, and we verify the results using co-factor expansion, Dodgson’s condensation method, and Matlab (See Appendix A).

Illustration 1.

All first entries are nonzeros ($4\times 4$ matrix)

$$A=\left[\begin{array}{ccc}2& 1& \begin{array}{cc}5& 2\end{array}\\ 2& 3& \begin{array}{cc}2& 3\end{array}\\ \begin{array}{c}1\\ 1\end{array}& \begin{array}{c}-1\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}4\\ 4\end{array}& \begin{array}{c}2\\ 1\end{array}\end{array}\end{array}\right]$$

The solution is found by the cross-multiplication method,

$A_4$	2	1	5	2
	2	3	2	3
	1	−1	4	2
	1	2	4	1
$A_3$		4	−6	2
		−5	6	1
		3	0	−1
$A_2$			−6	14
			−18	2
$A_1$				240	240/(2)(1)(−5)
det(A)				−24

by Dodgson’s condensation method,

A	2	1	5	2
	2	3	2	3
	1	−1	4	2
	1	2	4	1
B		4	−13	11
		−5	14	−8
		3	−12	−4
C			−9	50	Division by the interior in A
			18	−152
D			−3	−25
			−18	−38
E				−24

and by co-factor expansion.

$$\left|\begin{array}{ccc}2& 1& \begin{array}{cc}5& 2\end{array}\\ 2& 3& \begin{array}{cc}2& 3\end{array}\\ \begin{array}{c}1\\ 1\end{array}& \begin{array}{c}-1\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}4\\ 4\end{array}& \begin{array}{c}2\\ 1\end{array}\end{array}\end{array}\right|=2\left|\begin{array}{ccc}3& 2& 3\\ -1& 4& 2\\ 2& 4& 1\end{array}\right|-2\left|\begin{array}{ccc}1& 5& 2\\ -1& 4& 2\\ 2& 4& 1\end{array}\right|+\left|\begin{array}{ccc}1& 5& 2\\ 3& 2& 3\\ 2& 4& 1\end{array}\right|-\left|\begin{array}{ccc}1& 5& 2\\ 3& 2& 3\\ -1& 4& 2\end{array}\right|=-24$$

Illustration 2.

All first entries are nonzeros ($5\times 5$ matrix)

$$B=\left[\begin{array}{ccc}2& 2& \begin{array}{cc}1& \begin{array}{cc}3& 1\end{array}\end{array}\\ 1& 3& \begin{array}{cc}-1& \begin{array}{cc}1& 2\end{array}\end{array}\\ \begin{array}{c}1\\ \begin{array}{c}2\\ 1\end{array}\end{array}& \begin{array}{c}2\\ \begin{array}{c}2\\ 3\end{array}\end{array}& \begin{array}{cc}\begin{array}{c}4\\ \begin{array}{c}3\\ 2\end{array}\end{array}& \begin{array}{c}\begin{array}{cc}-2& 3\end{array}\\ \begin{array}{cc}\begin{array}{c}2\\ 1\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\end{array}\end{array}\right]$$

Computation of determinants is found by the cross-multiplication method,

$B_5$	2	2	1	3	1
	1	3	−1	1	2
	1	2	4	−2	3
	2	2	3	2	1
	1	3	2	1	5
$B_4$		4	−3	−1	3
		−1	5	−3	1
		−2	−5	6	−5
		4	1	0	9
$B_3$			17	−13	7
			15	−12	7
			18	−24	2
$B_2$				−9	14
				−144	−96
$B_1$					2880	2880/(1)(1)(2)(−1)(−2)(15)
det(B)					48

Dodgson’s condensation method,

A	2	2	1	3	1
	1	3	−1	1	2
	1	2	4	−2	3
	2	2	3	2	1
	1	3	2	1	5
B		4	−5	4	5
		−1	14	−2	7
		−2	−2	14	−8
		4	−5	−1	9
C			51	−46	38
			30	192	−82
			18	72	118
D			17	46	38	Division by interior entries in A
			15	48	41
			9	24	59
E				126	62
				−72	1848
F				9	−31	Division by interior entries in B
				36	132
G					48

and with verification of the results by co-factor expansion.

$$\begin{array}{cc}\hfill \det \left[A\right]& =\left|\begin{array}{ccc}2& 2& \begin{array}{cc}1& \begin{array}{cc}3& 1\end{array}\end{array}\\ 1& 3& \begin{array}{cc}-1& \begin{array}{cc}1& 2\end{array}\end{array}\\ \begin{array}{c}1\\ \begin{array}{c}2\\ 1\end{array}\end{array}& \begin{array}{c}2\\ \begin{array}{c}2\\ 3\end{array}\end{array}& \begin{array}{cc}\begin{array}{c}4\\ \begin{array}{c}3\\ 2\end{array}\end{array}& \begin{array}{c}\begin{array}{cc}-2& 3\end{array}\\ \begin{array}{cc}\begin{array}{c}2\\ 1\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\end{array}\end{array}\right|\hfill \\ & =2\left|\begin{array}{ccc}3& -1& \begin{array}{cc}1& 2\end{array}\\ 2& 4& \begin{array}{cc}-2& 3\end{array}\\ \begin{array}{c}2\\ 3\end{array}& \begin{array}{c}3\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}2\\ 1\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\right|-\left|\begin{array}{ccc}2& 1& \begin{array}{cc}3& 1\end{array}\\ 2& 4& \begin{array}{cc}-2& 3\end{array}\\ \begin{array}{c}2\\ 3\end{array}& \begin{array}{c}3\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}2\\ 1\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\right|+\left|\begin{array}{ccc}2& 1& \begin{array}{cc}3& 1\end{array}\\ 3& -1& \begin{array}{cc}1& 2\end{array}\\ \begin{array}{c}2\\ 3\end{array}& \begin{array}{c}3\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}2\\ 1\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\right|-2\left|\begin{array}{ccc}2& 1& \begin{array}{cc}3& 1\end{array}\\ 3& -1& \begin{array}{cc}1& 2\end{array}\\ \begin{array}{c}2\\ 3\end{array}& \begin{array}{c}4\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}-2\\ 1\end{array}& \begin{array}{c}3\\ 5\end{array}\end{array}\end{array}\right|\hfill \\ & +\left|\begin{array}{ccc}2& 1& \begin{array}{cc}3& 1\end{array}\\ 3& -1& \begin{array}{cc}1& 2\end{array}\\ \begin{array}{c}2\\ 2\end{array}& \begin{array}{c}4\\ 3\end{array}& \begin{array}{cc}\begin{array}{c}-2\\ 2\end{array}& \begin{array}{c}3\\ 1\end{array}\end{array}\end{array}\right|\hfill \end{array}$$

We compute the determinant of each submatrix separately.

$$\left|\begin{array}{ccc}3& -1& \begin{array}{cc}1& 2\end{array}\\ 2& 4& \begin{array}{cc}-2& 3\end{array}\\ \begin{array}{c}2\\ 3\end{array}& \begin{array}{c}3\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}2\\ 1\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\right|=3\left|\begin{array}{ccc}4& -2& 3\\ 3& 2& 1\\ 2& 1& 5\end{array}\right|-2\left|\begin{array}{ccc}-1& 1& 2\\ 3& 2& 1\\ 2& 1& 5\end{array}\right|+2\left|\begin{array}{ccc}-1& 1& 2\\ 4& -2& 3\\ 2& 1& 5\end{array}\right|-3\left|\begin{array}{ccc}-1& 1& 2\\ 4& -2& 3\\ 3& 2& 1\end{array}\right|=132$$

$$\left|\begin{array}{ccc}2& 1& \begin{array}{cc}3& 1\end{array}\\ 2& 4& \begin{array}{cc}-2& 3\end{array}\\ \begin{array}{c}2\\ 3\end{array}& \begin{array}{c}3\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}2\\ 1\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\right|=2\left|\begin{array}{ccc}4& -2& 3\\ 3& 2& 1\\ 2& 1& 5\end{array}\right|-2\left|\begin{array}{ccc}1& 3& 1\\ 3& 2& 1\\ 2& 1& 5\end{array}\right|+2\left|\begin{array}{ccc}1& 3& 1\\ 4& -2& 3\\ 2& 1& 5\end{array}\right|-3\left|\begin{array}{ccc}1& 3& 1\\ 4& -2& 3\\ 3& 2& 1\end{array}\right|=23$$

$$\left|\begin{array}{ccc}2& 1& \begin{array}{cc}3& 1\end{array}\\ 3& -1& \begin{array}{cc}1& 2\end{array}\\ \begin{array}{c}2\\ 3\end{array}& \begin{array}{c}3\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}2\\ 1\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\right|=2\left|\begin{array}{ccc}-1& 1& 2\\ 3& 2& 1\\ 2& 1& 5\end{array}\right|-3\left|\begin{array}{ccc}1& 3& 1\\ 3& 2& 1\\ 2& 1& 5\end{array}\right|+2\left|\begin{array}{ccc}1& 3& 1\\ -1& 1& 2\\ 2& 1& 5\end{array}\right|-3\left|\begin{array}{ccc}1& 3& 1\\ -1& 1& 2\\ 2& 1& 5\end{array}\right|=60$$

$$\left|\begin{array}{ccc}2& 1& \begin{array}{cc}3& 1\end{array}\\ 3& -1& \begin{array}{cc}1& 2\end{array}\\ \begin{array}{c}2\\ 3\end{array}& \begin{array}{c}4\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}-2\\ 1\end{array}& \begin{array}{c}3\\ 5\end{array}\end{array}\end{array}\right|=2\left|\begin{array}{ccc}-1& 1& 2\\ 4& -2& 3\\ 2& 1& 5\end{array}\right|-3\left|\begin{array}{ccc}1& 3& 1\\ 4& -2& 3\\ 2& 1& 5\end{array}\right|+2\left|\begin{array}{ccc}1& 3& 1\\ -1& 1& 2\\ 2& 1& 5\end{array}\right|-3\left|\begin{array}{ccc}1& 3& 1\\ -1& 1& 2\\ 4& -2& 3\end{array}\right|=111$$

$$\left|\begin{array}{ccc}2& 1& \begin{array}{cc}3& 1\end{array}\\ 3& -1& \begin{array}{cc}1& 2\end{array}\\ \begin{array}{c}2\\ 2\end{array}& \begin{array}{c}4\\ 3\end{array}& \begin{array}{cc}\begin{array}{c}-2\\ 2\end{array}& \begin{array}{c}3\\ 1\end{array}\end{array}\end{array}\right|=2\left|\begin{array}{ccc}-1& 1& 2\\ 4& -2& 3\\ 3& 2& 1\end{array}\right|-3\left|\begin{array}{ccc}1& 3& 1\\ 4& -2& 3\\ 3& 2& 1\end{array}\right|+2\left|\begin{array}{ccc}1& 3& 1\\ -1& 1& 2\\ 3& 2& 1\end{array}\right|-2\left|\begin{array}{ccc}1& 3& 1\\ -1& 1& 2\\ 4& -2& 3\end{array}\right|=-31$$

Thus, $\det \left[A\right]=2\left(132\right)-23+60-2\left(111\right)-31=48$.

Illustration 3.

Some first entries are zeros

$$D=\left[\begin{array}{ccc}2& 1& \begin{array}{cc}5& 2\end{array}\\ 0& 3& \begin{array}{cc}2& 3\end{array}\\ \begin{array}{c}1\\ 0\end{array}& \begin{array}{c}-1\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}4\\ 4\end{array}& \begin{array}{c}2\\ 1\end{array}\end{array}\end{array}\right]$$

$D_4$	2	1	5	2
	0	3	2	3
	1	−1	4	2
	0	2	4	1
$D_4$	2	1	5	2
	1	−1	4	2	Interchanged row (−1)
	0	3	2	3	Standby row
	0	2	4	1	Standby row
$D_3$		−3	3	2
		3	2	3	Transferred standby row
		2	4	1	Transferred standby row
$D_2$			−15	−15
			8	−3
$D_1$				165	165/(3)(−1)
det(D)				−55

We apply Type 3 column operations by interchanging column 1 and column 2,

$D_4$	2	1	5	2
	0	3	2	3
	1	−1	4	2
	0	2	4	1
$D_4$	1	2	5	2	Column 1 and column 2 interchange (−1)
	3	0	2	3
	−1	1	4	2
	2	0	4	1
$D_3$		−6	−13	−3
		3	14	9
		−2	−12	−5
$D_2$			−45	−45
			−8	3
$D_1$				−495	−495(−1)/(3)(−1)(3)
det(D)				−55	−495(−1)/(3)(−1)(3)

Dodgson’s condensation method,

A	2	1	5	2
	0	3	2	3
	1	−1	4	2
	0	2	4	1
B		6	−13	11
		−3	14	−8
		2	−12	−4
C			45	−50
			8	−152
D			15	−25	Division by interior entries of A
			−8	−38
E				−55

and by co-factor expansion, where we can expand about column 1.

$$D=\left[\begin{array}{ccc}2& 1& \begin{array}{cc}5& 2\end{array}\\ 0& 3& \begin{array}{cc}2& 3\end{array}\\ \begin{array}{c}1\\ 0\end{array}& \begin{array}{c}-1\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}4\\ 4\end{array}& \begin{array}{c}2\\ 1\end{array}\end{array}\end{array}\right]$$

$$\det \left(D\right)=2\left|\begin{array}{ccc}3& 2& 3\\ -1& 4& 2\\ 2& 4& 1\end{array}\right|-0+2\left|\begin{array}{ccc}1& 5& 2\\ 3& 2& 1\\ 2& 4& 1\end{array}\right|=-55$$

Illustration 4.

Only one row has a nonzero first entry

$$E=\left[\begin{array}{ccc}0& 2& \begin{array}{cc}1& \begin{array}{cc}3& 1\end{array}\end{array}\\ 0& 0& \begin{array}{cc}-2& \begin{array}{cc}1& 1\end{array}\end{array}\\ \begin{array}{c}3\\ \begin{array}{c}0\\ 0\end{array}\end{array}& \begin{array}{c}3\\ \begin{array}{c}2\\ 3\end{array}\end{array}& \begin{array}{cc}\begin{array}{c}4\\ \begin{array}{c}5\\ 2\end{array}\end{array}& \begin{array}{c}\begin{array}{cc}1& 5\end{array}\\ \begin{array}{cc}\begin{array}{c}2\\ 2\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\end{array}\end{array}\right]$$

$E_5$	0	2	1	3	1
	0	0	−2	1	1
	3	3	4	1	5
	0	2	5	2	1
	0	3	2	2	5
$E_5$	3	3	4	1	5	Interchanged row (−1)
	0	0	−2	1	1	Standby row
	0	2	1	3	1	Interchanged row
	0	2	5	2	1	Standby row
	0	3	2	2	5	Standby row
$E_4$		0	−2	1	1	Transferred/Standby row
		2	1	3	1	Transferred row
		2	5	2	1	Transferred row
		3	2	2	5	Transferred row
$E_3$			−2	1	1	Transferred row
			8	−2	0
			−11	−2	7
$E_2$				−4	−8
				−38	56
$E_1$					528	528(3)/(2)(8)(−1)
det(A)					−99

We interchange column 1 and column 3, and we proceed with usual cross-multiplication.

$E_5$	0	2	1	3	1
	0	0	−2	1	1
	3	3	4	1	5
	0	2	5	2	1
	0	3	2	2	5
$E_5$	1	2	0	3	1	Column 1 and column 3 exchange (−1)
	−2	0	0	1	1
	4	3	3	1	5
	5	2	0	2	1
	2	3	0	2	5
$E_4$		4	0	7	3
		−6	−6	−6	−14
		−7	−15	3	−21
		11	0	6	23
$E_3$			−24	18	−38
			48	−60	28
			165	−75	70
$E_2$				576	1152
				6300	−1260
$E_1$					−7,983,360	−7,983,360(−1)/(−2)(4)(5)(48)
det(E)					−99

By Dodgson’s condensation, we interchange row 1 and row 5.

A	0	2	1	3	1
	3	3	4	1	5
	0	2	5	2	1
	0	3	2	2	5
	0	0	−2	1	1
B		−6	5	−11	14
		6	7	3	−9
		0	−11	6	8
		0	−6	6	−3
C			−72	92	57
			−66	75	78
			0	−30	−66
D			−24	23	57	Division by inner entries in A
			−33	15	39
			0	−15	−33
E				399	42
				495	90
F				57	14	Division by inner entries in B
				−45	15
G					99
					−99	Multiplication of −1 by an exchange of row

By co-factor expansion, we can expand about column 1:

$$E=\left[\begin{array}{ccc}0& 2& \begin{array}{cc}1& \begin{array}{cc}3& 1\end{array}\end{array}\\ 0& 0& \begin{array}{cc}-2& \begin{array}{cc}1& 1\end{array}\end{array}\\ \begin{array}{c}3\\ \begin{array}{c}0\\ 0\end{array}\end{array}& \begin{array}{c}3\\ \begin{array}{c}2\\ 3\end{array}\end{array}& \begin{array}{cc}\begin{array}{c}4\\ \begin{array}{c}5\\ 2\end{array}\end{array}& \begin{array}{c}\begin{array}{cc}1& 5\end{array}\\ \begin{array}{cc}\begin{array}{c}2\\ 2\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\end{array}\end{array}\right]$$

$$\det \left(E\right)=0-0+3\left|\begin{array}{ccc}2& 1& \begin{array}{cc}3& 1\end{array}\\ 0& -2& \begin{array}{cc}1& 1\end{array}\\ \begin{array}{c}2\\ 3\end{array}& \begin{array}{c}5\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}2\\ 2\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\right|-0+0$$

Then, we compute $\left|\begin{array}{ccc}2& 1& \begin{array}{cc}3& 1\end{array}\\ 0& -2& \begin{array}{cc}1& 1\end{array}\\ \begin{array}{c}2\\ 3\end{array}& \begin{array}{c}5\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}2\\ 2\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\right|$ by another co-factor expansion.

$$\left|\begin{array}{ccc}2& 1& \begin{array}{cc}3& 1\end{array}\\ 0& -2& \begin{array}{cc}1& 1\end{array}\\ \begin{array}{c}2\\ 3\end{array}& \begin{array}{c}5\\ 2\end{array}& \begin{array}{cc}\begin{array}{c}2\\ 2\end{array}& \begin{array}{c}1\\ 5\end{array}\end{array}\end{array}\right|=2\left|\begin{array}{ccc}-2& 1& 1\\ 5& 2& 1\\ 2& 2& 5\end{array}\right|-0+2\left|\begin{array}{ccc}1& 3& 1\\ -2& 1& 1\\ 2& 2& 5\end{array}\right|-3\left|\begin{array}{ccc}1& 3& 1\\ -2& 1& 1\\ 5& 2& 1\end{array}\right|=-99$$

The above illustrations suggest the following comparisons of the cross-multiplication method, Dodgson’s condensation, and co-factor expansion summarized in

Table 1. Comparisons of methods in the manual computation of determinants.

Criteria	Cross-Multiplication	Dodgson’s Condensation	Co-Factor Expansion
Matrices generated	Each reduction produces one matrix that is one row and one column less than the preceding matrix	Similar to cross-multiplication but doubles the number of matrices from the 3rd and succeeding condensations.	Each matrix expansion produces matrices of same number as the matrix size and with one row and one column less than the matrix expanded.
Divisions	$\frac{n^2-3n+2}{2}$ terminal divisions	$n^2-3n+1$ internal divisions	No divisions
$\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{operations} S_n$ (including divisions)	$S_n=(n+1)({n-1)}^2$	$\frac{2n^3-n^2-7n+2}{2}$	$S_n=n{S}_{n-1}+2n-1;$   $n\ge 3, S_2=3$
Adjustments with zero entries	Applying Type 2 row (column) operation- Rows with zero first entries are placed at the bottom of the submatrix, or first column with zero entries is interchange with other columns; applying Type 3 row (column) operations	Applying Type 2 or Type 3 row (column) operations.	Zero entries in a matrix decreases the number of co-factors in the expansion about the row or column with zero entries
Algorithmic movement	Symmetric; computation of entries traces butterfly movement (cross-multiplication)	Symmetric; computation of entries traces butterfly movement	Non-symmetric except in the use of Sarrus’ rule for determinants of order 3 and butterfly movement for order 2.
Generalizability	Represented by a formula	Algorithmic	Represented by a formula

. Operation and matrix counts are based on cases where there are no zero entries in the first rows for cross-multiplication, in the interior matrix for Dodgson’s condensation, and in an expanded row or column in co-factor expansion.

5. Conclusions

The cross-multiplication method of computing the determinants of $n\times n$ matrices is a strategic use of elementary row operations in reducing a matrix to upper triangular form by using the nonzero first entries of adjacent rows as row multipliers. This technique gives an expression for each entry of the reduced matrix

$$a_{ij}^{\left(m\right)}=a_{i1}^{(m-1)}{a}_{i+1 j}^{(m-1)}-a_{i+1 1}^{\left(m-1\right)}{a}_{i j}^{\left(m-1\right)};$$

where $\left(m\right)$ = number of reductions performed, and $a_{ij}^{\left(0\right)}=a_{ij}$, or in determinant form, $a_{ij}^{\left(m\right)}=\left|\begin{array}{cc}{a}_{i1}^{(m-1)}& a_{i j}^{(m-1)}\\ a_{i+1 1}^{(m-1)}& a_{i+1 j}^{(m-1)}\end{array}\right|$. This approach is consistent with Dodgson’s condensation method by taking the first entries of adjacent rows as pivot entries in the computation of the determinant of the adjacent $2\times 2 matrices.$ The algorithm proceeds by computing each entry to produce a submatrix that is one row and one column less than the preceding submatrix until a $1\times 1$ matrix with entry $a_{11}^{(n-1)}$ is obtained.

When some first entries in the rows are zero, any of the following adjustments can be performed: applying Type 2 row operation where rows with zero first entries are transferred to the bottom of the submatrix and, then, transferred to the next submatrix; applying Type 2 column operation by interchanging the first column with zero first entries and another column with nonzero entries; applying Type 3 row (column) operations to remove zero first entries. The determinant is then computed for the following cases.

1. The general determinant formula is given by

$$\det \left(A_n\right)=\frac{\prod _{m=0}^{n-1}{a}_{11}^{\left(m\right)}}{\prod _{m=0}^{n-1}{\prod }_{j=1}^{r-1}{a}_{j1}^{\left(m\right)}}{(-1)}^k,$$

where $a_{j1}^{\left(0\right)}=a_{j1}$, r = number of rows with nonzero first entries in $A_{n-m}$, and k = number of row (column) permutations.

2. If the first entries are nonzero, the determinant formula is given by

$$\det \left(A_n\right)=\frac{a_{11}^{(n-1)}}{\prod _{m=0}^{n-2}{\prod }_{j=2}^{n-m-1}{a}_{j1}^{\left(m\right)}}{(-1)}^k;$$

where k = number of column permutations, if applicable.

3. When all rows of a submatrix $A_{n-i}$ have zero first entries, then $\det \left(A_n\right)=0.$

Compared to Dodgson’s condensation and co-factor expansion, the cross-multiplication method is the most efficient in terms of the number of matrices generated and the operations employed.

Both the cross-multiplication and Dodgson’s condensation methods apply Type 2 or Type 3 row (column) operations when there are zeros in the first entries or in the interior matrix. These two methods have symmetric algorithmic movement, which makes the execution of operations relatively easier and faster, and at $n=3$, both methods are equivalent. Unlike the Dodgson’s condensation, the cross-multiplication method can be generalized through formulas.