Theoretical Foundation: Laplace Expansion of a Determinant

The value of a determinant of order 3, denoted as $\Delta$, is defined as the sum of the products of the elements of any row (or column) and their corresponding cofactors. Let $A_{ij}$ represent the cofactor of the element $a_{ij}$. The cofactor $A_{ij}$ is defined as $(-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$.

Step 1: Formal Definition of the Determinant

By the Laplace Expansion theorem, the value of the determinant $\Delta$ can be computed by expanding along any row $i$ or any column $j$. Specifically, for any row $i$ ($i=1, 2, 3$):

$\Delta = \sum_{j=1}^{3} a_{ij} A_{ij}$

Step 2: Explicit Expansion for Each Row

Based on the structure of the determinant $\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$, the value is derived as follows:

Analytical Note: Properties of Cofactors

It is important to distinguish the expansion formula from the property of cofactors where the sum of products of elements of one row with the cofactors of another row is zero [Per the properties of determinants in Linear Algebra]. Specifically, if $i \neq k$, then $a_{i1}A_{k1} + a_{i2}A_{k2} + a_{i3}A_{k3} = 0$.

Visual Representation of Matrix Indices

Final Solution: The value of $\Delta$ is given by the sum of the product of elements of any row (or column) and their corresponding cofactors: $\Delta = a_{i1}A_{i1} + a_{i2}A_{i2} + a_{i3}A_{i3}$ for any row $i = 1, 2, \text{ or } 3$.

Given Determinant and Structural Identification

The given determinant is $A = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = \begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$.

The elements are identified as follows:

• $a_{11} = 2$
• $a_{12} = -4$
• $a_{21} = 0$
• $a_{22} = 3$

Step 1: Calculating Minors ($M_{ij}$)

The minor $M_{ij}$ of an element $a_{ij}$ is the determinant of the $1 \times 1$ matrix obtained by deleting the $i^{th}$ row and $j^{th}$ column [Per definition of minors in determinants].

• $M_{11} = |a_{22}| = |3| = 3$
• $M_{12} = |a_{21}| = |0| = 0$
• $M_{21} = |a_{12}| = |-4| = -4$
• $M_{22} = |a_{11}| = |2| = 2$

Step 2: Calculating Cofactors ($C_{ij}$)

The cofactor $C_{ij}$ of an element $a_{ij}$ is defined as $C_{ij} = (-1)^{i+j} M_{ij}$ [By definition of cofactor expansion].

• $C_{11} = (-1)^{1+1} M_{11} = (-1)^2 (3) = 1 \times 3 = 3$
• $C_{12} = (-1)^{1+2} M_{12} = (-1)^3 (0) = -1 \times 0 = 0$
• $C_{21} = (-1)^{2+1} M_{21} = (-1)^3 (-4) = -1 \times (-4) = 4$
• $C_{22} = (-1)^{2+2} M_{22} = (-1)^4 (2) = 1 \times 2 = 2$

Visual Representation of Matrix Indices

Data Synthesis

Final Solution: The minors are $M_{11}=3, M_{12}=0, M_{21}=-4, M_{22}=2$ and the corresponding cofactors are $C_{11}=3, C_{12}=0, C_{21}=4, C_{22}=2$.

Given Determinant and Structural Analysis

We are given the determinant:

$\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$

The objective is to evaluate $\Delta$ by expanding along the third column ($C_3$). The elements of $C_3$ are $a_{13} = yz$, $a_{23} = zx$, and $a_{33} = xy$.

Step 1: Calculating Cofactors of the Third Column

The cofactor $C_{ij}$ of an element $a_{ij}$ is defined as $(-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor obtained by deleting the $i$-th row and $j$-th column. [Per definition of Cofactor in Determinant Theory].

For $a_{13} = yz$:

$C_{13} = (-1)^{1+3} \begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} = (1)(z - y) = z - y$

For $a_{23} = zx$:

$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} = -(1)(z - x) = x - z$

For $a_{33} = xy$:

$C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} = (1)(y - x) = y - x$

Step 2: Expansion along the Third Column

The determinant is evaluated as $\Delta = a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33}$. [Per Laplace Expansion Theorem].

$\Delta = (yz)(z - y) + (zx)(x - z) + (xy)(y - x)$

Step 3: Algebraic Simplification

Expanding the terms:

$\Delta = yz^2 - y^2z + zx^2 - z^2x + xy^2 - x^2y$

Group terms to facilitate factorization:

$\Delta = -(x^2y - xy^2) + (zx^2 - z^2x) + (yz^2 - y^2z)$

$\Delta = -xy(x - y) + zx(x - z) + yz(z - y)$

Continuing the expansion to reach the standard cyclic form:

$\Delta = yz^2 - y^2z + zx^2 - z^2x + xy^2 - x^2y$

Rearranging for factoring: $(x-y)(y-z)(z-x)$

$\Delta = x^2y - x^2z - xy^2 + y^2z + xz^2 - yz^2$

Factoring step-by-step results in: $\Delta = -(x - y)(y - z)(z - x)$

Visual Representation of Matrix Components

Final Solution: $\Delta = -(x - y)(y - z)(z - x) = (x - y)(y - z)(x - z)$ or $(x - y)(y - z)(z - x)$ expressed as $-(x-y)(y-z)(z-x)$

Initial Matrix Definition

Let the given determinant be denoted by $A$:

$A = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$

Theoretical Framework

• Minor ($M_{ij}$): The determinant of the sub-matrix obtained by deleting the $i$-th row and $j$-th column of the original determinant. [Per definition of Matrix Minors]
• Cofactor ($C_{ij}$): Defined as $C_{ij} = (-1)^{i+j} M_{ij}$. [Per Laplace expansion criteria]

Step 1: Calculating Minors ($M_{ij}$)

Step 2: Calculating Cofactors ($C_{ij}$)

Applying $C_{ij} = (-1)^{i+j} M_{ij}$:

• $C_{11} = (-1)^{1+1} (1) = 1(1) = 1$
• $C_{12} = (-1)^{1+2} (0) = -1(0) = 0$
• $C_{13} = (-1)^{1+3} (0) = 1(0) = 0$
• $C_{21} = (-1)^{2+1} (0) = -1(0) = 0$
• $C_{22} = (-1)^{2+2} (1) = 1(1) = 1$
• $C_{23} = (-1)^{2+3} (0) = -1(0) = 0$
• $C_{31} = (-1)^{3+1} (0) = 1(0) = 0$
• $C_{32} = (-1)^{3+2} (0) = -1(0) = 0$
• $C_{33} = (-1)^{3+3} (1) = 1(1) = 1$

Visualization of Identity Determinant Structure

Final Solution: The set of Minors is $\{1, 0, 0, 0, 1, 0, 0, 0, 1\}$ and the set of Cofactors is $\{1, 0, 0, 0, 1, 0, 0, 0, 1\}$.

Given Determinant Setup

Let the matrix $A$ be defined as:

Definitions

• Minor ($M_{ij}$): The determinant of the sub-matrix obtained by deleting the $i^{th}$ row and $j^{th}$ column. [Definition: $M_{ij} = \det(A_{ij})$]
• Cofactor ($C_{ij}$): Defined as $(-1)^{i+j} M_{ij}$. [Definition: $C_{ij} = (-1)^{i+j} M_{ij}$]

Step 1: Calculating Minors ($M_{ij}$)

For each element $a_{ij}$, we compute the $2 \times 2$ determinant:

• $M_{11} = \begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix} = (5)(2) - (-1)(1) = 10 + 1 = 11$
• $M_{12} = \begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix} = (3)(2) - (-1)(0) = 6 - 0 = 6$
• $M_{13} = \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = (3)(1) - (5)(0) = 3 - 0 = 3$
• $M_{21} = \begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix} = (0)(2) - (4)(1) = 0 - 4 = -4$
• $M_{22} = \begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix} = (1)(2) - (4)(0) = 2 - 0 = 2$
• $M_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1 - 0 = 1$
• $M_{31} = \begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix} = (0)(-1) - (4)(5) = 0 - 20 = -20$
• $M_{32} = \begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (4)(3) = -1 - 12 = -13$
• $M_{33} = \begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix} = (1)(5) - (0)(3) = 5 - 0 = 5$

Step 2: Calculating Cofactors ($C_{ij}$)

Applying the formula $C_{ij} = (-1)^{i+j} M_{ij}$:

Final Solution: The set of Minors is $\{11, 6, 3, -4, 2, 1, -20, -13, 5\}$ and the set of Cofactors is $\{11, -6, 3, 4, 2, -1, -20, 13, 5\}$.

1. Definitions and Theoretical Foundation

For a determinant of order $n$, the minor $M_{ij}$ of an element $a_{ij}$ is the determinant of the $(n-1) \times (n-1)$ matrix obtained by deleting the $i$-th row and $j$-th column. The cofactor $C_{ij}$ is defined as $C_{ij} = (-1)^{i+j} M_{ij}$. [Per Definition of Determinant Minors and Cofactors].

2. Given Matrix Specification

Let the given determinant be $\Delta = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = \begin{vmatrix} a & c \\ b & d \end{vmatrix}$.

• $a_{11} = a$
• $a_{12} = c$
• $a_{21} = b$
• $a_{22} = d$

3. Calculation of Minors ($M_{ij}$)

Minors are calculated by excluding the corresponding row and column of the targeted element:

• $M_{11}$: Delete row 1, column 1. The remaining element is $d$. Thus, $M_{11} = d$.
• $M_{12}$: Delete row 1, column 2. The remaining element is $b$. Thus, $M_{12} = b$.
• $M_{21}$: Delete row 2, column 1. The remaining element is $c$. Thus, $M_{21} = c$.
• $M_{22}$: Delete row 2, column 2. The remaining element is $a$. Thus, $M_{22} = a$.

4. Calculation of Cofactors ($C_{ij}$)

Using the relation $C_{ij} = (-1)^{i+j} M_{ij}$:

• $C_{11} = (-1)^{1+1} M_{11} = (+1) \cdot d = d$
• $C_{12} = (-1)^{1+2} M_{12} = (-1) \cdot b = -b$
• $C_{21} = (-1)^{2+1} M_{21} = (-1) \cdot c = -c$
• $C_{22} = (-1)^{2+2} M_{22} = (+1) \cdot a = a$

5. Tabular Summary of Results

Final Solution: The minors are $M_{11}=d, M_{12}=b, M_{21}=c, M_{22}=a$ and the cofactors are $C_{11}=d, C_{12}=-b, C_{21}=-c, C_{22}=a$.

Step 1: Identifying the Determinant and Elements of the Second Row

The given determinant is:

$\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$

The elements of the second row are:

• $a_{21} = 2$
• $a_{22} = 0$
• $a_{23} = 1$

Step 2: Calculating Cofactors for the Second Row

The cofactor $A_{ij}$ of an element $a_{ij}$ is defined as $A_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$ [Definition of Cofactor].

For $a_{21} = 2$:

$A_{21} = (-1)^{2+1} \begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = -1 \times [(3)(3) - (8)(2)] = -1 \times [9 - 16] = -1 \times (-7) = 7$

For $a_{22} = 0$:

$A_{22} = (-1)^{2+2} \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = 1 \times [(5)(3) - (8)(1)] = 1 \times [15 - 8] = 7$

For $a_{23} = 1$:

$A_{23} = (-1)^{2+3} \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = -1 \times [(5)(2) - (3)(1)] = -1 \times [10 - 3] = -7$

Step 3: Evaluating the Determinant using Row 2 Cofactors

The value of a determinant is the sum of the products of elements of any row (or column) and their corresponding cofactors [Laplace Expansion Theorem]:

$\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}$

Substituting the calculated values:

$\Delta = (2)(7) + (0)(7) + (1)(-7)$

$\Delta = 14 + 0 - 7$

$\Delta = 7$

Verification via Geometric/Visual Representation

Final Solution: The value of the determinant $\Delta$ evaluated using the second row is 7.