Step 1: Theoretical Framework for Expansion

To evaluate a $3 \times 3$ determinant, we employ the Laplace Expansion method. We expand the determinant $A = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$ along the first row ($R_1$):

$\det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$

[Per the definition of determinant expansion by minors and cofactors].

Step 2: Assigning Given Matrix Elements

Given the matrix:

$A = \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}$

• $a_{11} = 3, a_{12} = -4, a_{13} = 5$
• $a_{21} = 1, a_{22} = 1, a_{23} = -2$
• $a_{31} = 2, a_{32} = 3, a_{33} = 1$

Step 3: Calculating Minors of Row 1

We calculate the $2 \times 2$ determinants (minors) corresponding to each element of the first row:

Step 4: Final Algebraic Summation

Summing the components derived from the expansion along the first row:

$\Delta = 3[(1)(1) - (-2)(3)] - (-4)[(1)(1) - (-2)(2)] + 5[(1)(3) - (1)(2)]$

$\Delta = 3[1 + 6] + 4[1 + 4] + 5[3 - 2]$

$\Delta = 3[7] + 4[5] + 5[1]$

$\Delta = 21 + 20 + 5$

$\Delta = 46$

Visual Representation of Matrix Indices

Final Solution: 46

Step 1: Evaluation of the Determinant on the Left-Hand Side (LHS)

For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is defined as $|A| = ad - bc$ [By the definition of the determinant of a $2 \times 2$ matrix].

Given the LHS: $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}$

Calculation: $(2 \times 1) - (4 \times 5)$

$= 2 - 20$

$= -18$

Step 2: Evaluation of the Determinant on the Right-Hand Side (RHS)

Given the RHS: $\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$

Calculation: $(2x \times x) - (4 \times 6)$

$= 2x^2 - 24$

Step 3: Equating LHS and RHS to Form the Algebraic Equation

According to the problem statement, the determinants are equal:

$-18 = 2x^2 - 24$

Add $24$ to both sides [Addition Property of Equality]:

$-18 + 24 = 2x^2$

$6 = 2x^2$

Step 4: Solving for $x$

Divide both sides by $2$:

$\frac{6}{2} = x^2$

$3 = x^2$

Apply the square root property: $x = \pm\sqrt{3}$

Final Solution: $x = \pm\sqrt{3}$

Step 1: Evaluation of the Determinant on the Left-Hand Side (LHS)

To evaluate a $2 \times 2$ determinant of the form $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$, we utilize the standard formula: $ad - bc$ [By definition of the determinant of a $2 \times 2$ matrix].

For the LHS determinant $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}$:

LHS $= (2 \times 5) - (3 \times 4)$

LHS $= 10 - 12$

LHS $= -2$

Step 2: Evaluation of the Determinant on the Right-Hand Side (RHS)

Applying the same determinant property to the RHS $\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$:

RHS $= (x \times 5) - (3 \times 2x)$

RHS $= 5x - 6x$

RHS $= -x$

Step 3: Equating LHS and RHS to solve for $x$

Given the equation: $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$

Substituting the results from Step 1 and Step 2 into the equation:

$-2 = -x$

Step 4: Algebraic Simplification

Multiply both sides of the equation by $-1$ [By the property of equality]:

$x = 2$

Final Solution: $x = 2$

Step 1: Definition of Given Matrix and Determinant Calculation

Given the matrix $A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$.

To calculate the determinant of $A$, denoted as $|A|$, we apply the formula for a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ where $|A| = ad - bc$ [Per the definition of the determinant of a square matrix].

$|A| = (1 \times 2) - (2 \times 4)$

$|A| = 2 - 8$

$|A| = -6$

Step 2: Calculating 4|A|

Multiply the result obtained in Step 1 by the scalar $4$:

$4|A| = 4 \times (-6)$

$4|A| = -24$

Step 3: Evaluating the Matrix 2A

To find $2A$, we multiply every element of matrix $A$ by the scalar $2$ [Per the definition of scalar multiplication of a matrix]:

$2A = 2 \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 \times 1 & 2 \times 2 \\ 2 \times 4 & 2 \times 2 \end{bmatrix}$

$2A = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$

Step 4: Calculating the Determinant |2A|

Using the determinant formula for the resultant matrix $2A$:

$|2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix}$

$|2A| = (2 \times 4) - (4 \times 8)$

$|2A| = 8 - 32$

$|2A| = -24$

Step 5: Verification and Conclusion

Comparing the results from Step 2 and Step 4:

Since $|2A| = -24$ and $4|A| = -24$, it is demonstrated that $|2A| = 4|A|$. This is a specific application of the property $|kA| = k^n|A|$, where $n$ is the order of the square matrix [Per the properties of determinants for a matrix of order $n=2$].

Final Solution: $|2A| = 4|A| = -24$

Step 1: Calculating the Determinant of Matrix $A$

Given the matrix $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$. We calculate the determinant $|A|$ by expanding along the first column $C_1$ because it contains the most zeros [Per Laplace Expansion Theorem].

$|A| = 1 \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} - 0 \begin{vmatrix} 0 & 1 \\ 0 & 4 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix}$

$|A| = 1((1)(4) - (2)(0)) - 0 + 0 = 4$

Thus, $27|A| = 27 \times 4 = 108$.

Step 2: Scalar Multiplication of Matrix $A$ by 3

To find $3A$, we multiply every entry of matrix $A$ by the scalar $3$ [Per Scalar Multiplication Law].

$3A = 3 \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}$

Step 3: Calculating the Determinant of Matrix $3A$

Now, we compute $|3A|$ using the determinant of the resulting matrix from Step 2. We expand along the first column $C_1$.

$|3A| = \begin{vmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{vmatrix}$

$|3A| = 3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} - 0 \begin{vmatrix} 0 & 3 \\ 0 & 12 \end{vmatrix} + 0 \begin{vmatrix} 0 & 3 \\ 3 & 6 \end{vmatrix}$

$|3A| = 3((3)(12) - (6)(0)) = 3(36) = 108$

Step 4: Theoretical Verification

The property $|kA| = k^n|A|$ holds for any $n \times n$ matrix, where $n$ is the order of the matrix [Per Determinant Property Theorem]. Here, $n=3$ and $k=3$.

$|3A| = 3^3 |A| = 27 |A|$

Substituting the values derived:

Final Solution: Since $|3A| = 108$ and $27|A| = 27(4) = 108$, it is proven that $|3A| = 27|A|$.

Step 1: Evaluation of the Determinants

To solve for the unknown variable $x$, we must apply the definition of a $2 \times 2$ determinant. For any matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is given by $|A| = ad - bc$ [Definition of Determinant for $2 \times 2$ matrices].

For the Left-Hand Side (LHS):

$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = (x)(x) - (18)(2)$

$= x^2 - 36$ [Algebraic simplification]

For the Right-Hand Side (RHS):

$\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} = (6)(6) - (18)(2)$

$= 36 - 36$

$= 0$

Step 2: Constructing the Algebraic Equation

Equating the results from Step 1 as per the problem statement:

$x^2 - 36 = 0$

Step 3: Solving for $x$

To isolate $x$, we apply the addition property of equality:

$x^2 = 36$

Taking the square root on both sides requires considering both positive and negative roots [Principle of Square Roots]:

$x = \pm \sqrt{36}$

$x = \pm 6$

Visual Representation

Step 4: Verification

Final Solution: $x = \pm 6$

Step 1: Theoretical Foundation of a $2 \times 2$ Determinant

For a square matrix $A$ of order $2 \times 2$ defined as:

$A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$

The determinant of $A$, denoted as $|A|$ or $\det(A)$, is calculated using the following algebraic axiom [Per the definition of the determinant of a $2 \times 2$ matrix]:

$\det(A) = (a_{11} \cdot a_{22}) - (a_{12} \cdot a_{21})$

Step 2: Mapping Given Values to Variables

Given the determinant $\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$, we identify the elements:

• $a_{11} = 2$
• $a_{12} = 4$
• $a_{21} = -5$
• $a_{22} = -1$

Step 3: Execution of Calculation

Applying the formula derived in Step 1:

$\det = (2 \times -1) - (4 \times -5)$

Perform the multiplication operations:

$\det = (-2) - (-20)$

Applying the sign property of subtraction of negative integers [By the additive inverse principle]:

$\det = -2 + 20$

$\det = 18$

Visual Representation of Determinant Calculation

Final Solution: 18

Step 1: Establishing the Determinant Formula

To find the determinant of a $3 \times 3$ matrix $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$, we use the expansion by the first row [Per Laplace's expansion theorem]:

$|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$

Step 2: Mapping Given Matrix Elements

From the given matrix $A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$, we identify:

• $a_{11} = 1, a_{12} = 1, a_{13} = -2$
• $a_{21} = 2, a_{22} = 1, a_{23} = -3$
• $a_{31} = 5, a_{32} = 4, a_{33} = -9$

Step 3: Calculating the Determinant

Expanding along the first row:

$|A| = 1 \begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} - 1 \begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} + (-2) \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix}$

Applying the $2 \times 2$ determinant formula $(ad - bc)$:

Term 1: $1 \times [(1)(-9) - (-3)(4)] = 1 \times [-9 + 12] = 1 \times 3 = 3$

Term 2: $-1 \times [(2)(-9) - (-3)(5)] = -1 \times [-18 + 15] = -1 \times [-3] = 3$

Term 3: $-2 \times [(2)(4) - (1)(5)] = -2 \times [8 - 5] = -2 \times 3 = -6$

Step 4: Summation of Terms

$|A| = 3 + 3 - 6$

$|A| = 6 - 6$

$|A| = 0$

Final Solution: |A| = 0

Step 1: Theoretical Foundation of a $2 \times 2$ Determinant

For any square matrix of order $2$, defined as $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is calculated using the formula:
$\det(A) = |A| = (ad) - (bc)$
[Per the definition of the determinant of a $2 \times 2$ matrix].

Step 2: Assigning Variables from the Given Matrix

Given the determinant $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$, we identify the elements as follows:

• $a = \cos\theta$
• $b = -\sin\theta$
• $c = \sin\theta$
• $d = \cos\theta$

Step 3: Execution of the Determinant Calculation

Substituting the variables into the formula $(ad - bc)$:

$\Delta = (\cos\theta)(\cos\theta) - (-\sin\theta)(\sin\theta)$

$\Delta = \cos^2\theta - (-\sin^2\theta)$

$\Delta = \cos^2\theta + \sin^2\theta$

Step 4: Simplification via Trigonometric Identity

Applying the fundamental Pythagorean identity of trigonometry:
$\cos^2\theta + \sin^2\theta = 1$
[Per the Fundamental Trigonometric Identity].

Step 5: Visual Representation

Final Solution: 1

Step 1: Application of Determinant Definition

For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is defined as $|A| = ad - bc$ [By the definition of a second-order determinant].

Given the determinant $\Delta = \begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix}$, we identify the components:

• $a = x^2 - x + 1$
• $b = x - 1$
• $c = x + 1$
• $d = x + 1$

Step 2: Algebraic Expansion

Applying the formula $\Delta = (x^2 - x + 1)(x + 1) - (x - 1)(x + 1)$:

$\Delta = [(x^2 - x + 1)(x + 1)] - [(x - 1)(x + 1)]$

Step 3: Simplification via Factorization

We observe that $(x + 1)$ is a common factor in both terms. Factoring $(x + 1)$ out allows for easier simplification:

$\Delta = (x + 1) \cdot [(x^2 - x + 1) - (x - 1)]$

[By the Distributive Property of multiplication over subtraction: $a \cdot c - b \cdot c = c(a - b)$]

Step 4: Final Polynomial Reduction

Simplify the expression within the square brackets:

$\Delta = (x + 1) \cdot [x^2 - x + 1 - x + 1]$

$\Delta = (x + 1) \cdot [x^2 - 2x + 2]$

Now, perform the final multiplication:

$\Delta = x(x^2 - 2x + 2) + 1(x^2 - 2x + 2)$

$\Delta = x^3 - 2x^2 + 2x + x^2 - 2x + 2$

Group the like terms:

$\Delta = x^3 + (-2x^2 + x^2) + (2x - 2x) + 2$

$\Delta = x^3 - x^2 + 2$

Visual Representation of the Determinant Structure

Final Solution: x^3 - x^2 + 2

Step 1: Definition of the Determinant

To evaluate a $3 \times 3$ determinant of a matrix $A = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$, we expand along the first row [Per Laplace's Expansion Theorem]:

$\Delta = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$

Step 2: Identifying Matrix Elements

Given the determinant $\Delta = \begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix}$:

• $a_{11} = 2, a_{12} = -1, a_{13} = -2$
• $a_{21} = 0, a_{22} = 2, a_{23} = -1$
• $a_{31} = 3, a_{32} = -5, a_{33} = 0$

Step 3: Expansion Along the First Row

Applying the expansion formula:

$\Delta = 2 \begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} - (-1) \begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} + (-2) \begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix}$

Step 4: Calculating $2 \times 2$ Determinants

Utilizing the property $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$:

• First minor: $(2 \times 0) - (-1 \times -5) = 0 - 5 = -5$
• Second minor: $(0 \times 0) - (-1 \times 3) = 0 - (-3) = 3$
• Third minor: $(0 \times -5) - (2 \times 3) = 0 - 6 = -6$

Step 5: Final Computation

Substitute the calculated values back into the expanded expression:

$\Delta = 2(-5) + 1(3) - 2(-6)$

$\Delta = -10 + 3 + 12$ [By Arithmetic Addition]

$\Delta = 5$

Visual Representation of Matrix Indices

Final Solution: The value of the determinant is $5$.

Step 1: Establishing the Determinant Structure

To evaluate the determinant of a $3 \times 3$ matrix $A$, we apply the Laplace expansion method. Let $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$. The determinant $|A|$ is calculated by expanding along any row or column. We shall expand along the first row ($R_1$).

The formula for expansion along $R_1$ is: $|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$

Step 2: Identifying Matrix Elements

Given the determinant $\Delta = \begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}$, we identify the coefficients:

• $a_{11} = 0, a_{12} = 1, a_{13} = 2$
• $a_{21} = -1, a_{22} = 0, a_{23} = -3$
• $a_{31} = -2, a_{32} = 3, a_{33} = 0$

Step 3: Calculating Minors and Expansion

Substituting the values into the expansion formula [Per the Definition of Determinant of a Matrix]:

$\Delta = 0 \begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} - 1 \begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} + 2 \begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix}$

Calculating the $2 \times 2$ determinants:

• First term: $0 \cdot [(0)(0) - (-3)(3)] = 0 \cdot [0 - (-9)] = 0$
• Second term: $-1 \cdot [(-1)(0) - (-3)(-2)] = -1 \cdot [0 - (6)] = -1 \cdot (-6) = 6$
• Third term: $2 \cdot [(-1)(3) - (0)(-2)] = 2 \cdot [-3 - 0] = 2 \cdot (-3) = -6$

Step 4: Algebraic Summation

$\Delta = 0 + 6 + (-6)$

$\Delta = 6 - 6$

$\Delta = 0$

Final Solution: The value of the determinant is $0$.

Step 1: Theoretical Foundation of Determinant Expansion

To evaluate a $3 \times 3$ determinant $A = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$, we employ the Laplace expansion method. We may expand along any row or column. [Per the properties of determinants, expanding along a row or column containing the most zeros simplifies calculation].

Step 2: Selection of Expansion Path

Given the matrix $A = \begin{pmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{pmatrix}$, notice that Row 2 ($R_2$) contains two zeros. Expanding along $R_2$ significantly reduces computational complexity.

The expansion formula for $R_2$ is: $\det(A) = -a_{21}M_{21} + a_{22}M_{22} - a_{23}M_{23}$

Step 3: Calculating Minors and Determinant

Mapping the elements of $R_2$:

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

Applying the expansion formula: $\det(A) = -(0) \begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix} + (0) \begin{vmatrix} 3 & -2 \\ 3 & 0 \end{vmatrix} - (-1) \begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix}$

Step 4: Algebraic Simplification

The terms involving $0$ vanish. We are left with the final term:

$\det(A) = 0 + 0 + 1 \times \begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix}$

Applying the formula for a $2 \times 2$ determinant, $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = (ad - bc)$:

$\det(A) = 1 \times [(3 \times -5) - (-1 \times 3)]$

$\det(A) = 1 \times [-15 - (-3)]$

$\det(A) = 1 \times [-15 + 3]$

$\det(A) = -12$

Visual Representation of Matrix Indices

Final Solution: -12