Step 1: Setting up the Determinant Expansion

To evaluate the determinant $D = \begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}$, we perform expansion along the first row ($R_1$).

According to the Laplace expansion formula: $D = a_{11}(C_{11}) + a_{12}(C_{12}) + a_{13}(C_{13})$, where $C_{ij}$ is the cofactor of element $a_{ij}$.

Step 2: Calculating Minor Cofactors

The expansion is given by:

$D = (\cos\alpha\cos\beta) \begin{vmatrix} \cos\beta & 0 \\ \sin\alpha\sin\beta & \cos\alpha \end{vmatrix} - (\cos\alpha\sin\beta) \begin{vmatrix} -\sin\beta & 0 \\ \sin\alpha\cos\beta & \cos\alpha \end{vmatrix} + (-\sin\alpha) \begin{vmatrix} -\sin\beta & \cos\beta \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta \end{vmatrix}$

Step 3: Solving 2x2 Sub-determinants

Evaluating each minor [By determinant definition $ad-bc$]:

• Minor 1: $(\cos\beta)(\cos\alpha) - (0)(\sin\alpha\sin\beta) = \cos\alpha\cos\beta$
• Minor 2: $(-\sin\beta)(\cos\alpha) - (0)(\sin\alpha\cos\beta) = -\cos\alpha\sin\beta$
• Minor 3: $(-\sin\beta)(\sin\alpha\sin\beta) - (\cos\beta)(\sin\alpha\cos\beta) = -\sin\alpha\sin^2\beta - \sin\alpha\cos^2\beta$

Step 4: Algebraic Simplification

Substituting these back into the expression:

$D = (\cos\alpha\cos\beta)(\cos\alpha\cos\beta) - (\cos\alpha\sin\beta)(-\cos\alpha\sin\beta) - (\sin\alpha)(-\sin\alpha\sin^2\beta - \sin\alpha\cos^2\beta)$

$D = \cos^2\alpha\cos^2\beta + \cos^2\alpha\sin^2\beta + \sin^2\alpha(\sin^2\beta + \cos^2\beta)$

Step 5: Applying Trigonometric Identities

[By Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$]:

$D = \cos^2\alpha(\cos^2\beta + \sin^2\beta) + \sin^2\alpha(1)$

$D = \cos^2\alpha(1) + \sin^2\alpha$

$D = 1$

Final Solution: 1

Step 1: Computing the Determinant of Matrix $A$

Given the matrix $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$.

The determinant $|A|$ is calculated by expanding along the first row [Laplace Expansion Theorem]:

$|A| = 1(3(5) - 1(1)) - (-2)((-2)(5) - 1(1)) + 1((-2)(1) - 3(1))$

$|A| = 1(15 - 1) + 2(-10 - 1) + 1(-2 - 3)$

$|A| = 1(14) + 2(-11) + 1(-5) = 14 - 22 - 5 = -13$

Since $|A| \neq 0$, the inverse $A^{-1}$ exists.

Step 2: Computing the Adjoint of Matrix $A$

The adjoint, $adj(A)$, is the transpose of the cofactor matrix $C = [C_{ij}]$.

• $C_{11} = +(15-1) = 14$; $C_{12} = -(-10-1) = 11$; $C_{13} = +(-2-3) = -5$
• $C_{21} = -(-10-1) = 11$; $C_{22} = +(5-1) = 4$; $C_{23} = -(1+2) = -3$
• $C_{31} = +(-2-3) = -5$; $C_{32} = -(1+2) = -3$; $C_{33} = +(3-4) = -1$

$adj(A) = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}^T = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}$

Step 3: Calculating $[adj A]^{-1}$

Let $B = adj(A)$. To find $B^{-1} = \frac{1}{|B|} adj(B)$, we first find $|B|$:

$|B| = 14(-4 - 9) - 11(-11 - 15) - 5(-33 + 20) = 14(-13) - 11(-26) - 5(-13)$

$|B| = -182 + 286 + 65 = 169$

Note that $|adj(A)| = |A|^{n-1} = (-13)^{3-1} = (-13)^2 = 169$ [Property: $|adj(A)| = |A|^{n-1}$].

Computing the adjugate of $B$ (cofactors of $adj(A)$):

$[adj(adj(A))] = \begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix} \implies [adj A]^{-1} = \frac{1}{169} \begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix} = \begin{bmatrix} -1/13 & 2/13 & -1/13 \\ 2/13 & -3/13 & -1/13 \\ -1/13 & -1/13 & -5/13 \end{bmatrix}$

Step 4: Calculating $adj(A^{-1})$

Using the property $A^{-1} = \frac{1}{|A|} adj(A) = -\frac{1}{13} \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}$:

$A^{-1} = \begin{bmatrix} -14/13 & -11/13 & 5/13 \\ -11/13 & -4/13 & 3/13 \\ 5/13 & 3/13 & 1/13 \end{bmatrix}$

The inverse of a matrix $X$ is $\frac{1}{|X|}adj(X)$. For a $3 \times 3$ matrix, $adj(A^{-1}) = |A^{-1}| (A^{-1})^{-1}$.

Since $(A^{-1})^{-1} = A$ and $|A^{-1}| = \frac{1}{|A|} = -\frac{1}{13}$:

$adj(A^{-1}) = -\frac{1}{13} A = -\frac{1}{13} \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix} = \begin{bmatrix} -1/13 & 2/13 & -1/13 \\ 2/13 & -3/13 & -1/13 \\ -1/13 & -1/13 & -5/13 \end{bmatrix}$

Comparison Table

Element	$[adj A]^{-1}$	$adj(A^{-1})$
Result	$\begin{bmatrix} -1/13 & 2/13 & -1/13 \\ 2/13 & -3/13 & -1/13 \\ -1/13 & -1/13 & -5/13 \end{bmatrix}$	$\begin{bmatrix} -1/13 & 2/13 & -1/13 \\ 2/13 & -3/13 & -1/13 \\ -1/13 & -1/13 & -5/13 \end{bmatrix}$

Final Solution: Since the matrices are identical, the identity $[adj A]^{-1} = adj (A^{-1})$ is verified.

Step 1: Analyzing the Matrix Structure

The given matrix $A$ is a diagonal matrix of the form:

$A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$

where $x, y, z \neq 0$. [By Definition of a Diagonal Matrix, all non-diagonal elements are zero].

Step 2: Calculating the Determinant

To determine if the inverse exists, we calculate the determinant $|A|$ [Per the Invertibility Criterion, $A^{-1}$ exists if $|A| \neq 0$].

$|A| = \begin{vmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{vmatrix}$

Expanding along the first row:

$|A| = x(yz - 0) - 0(0 - 0) + 0(0 - 0) = xyz$

Since $x, y, z \neq 0$, it follows that $xyz \neq 0$, thus $A$ is non-singular and invertible.

Step 3: Finding the Adjoint Matrix

The adjoint of a matrix, $\text{adj}(A)$, is the transpose of the cofactor matrix $C_{ij}$, where $C_{ij} = (-1)^{i+j}M_{ij}$ [By Definition of Adjoint].

The cofactors are:

• $C_{11} = +(yz - 0) = yz$
• $C_{12} = -(0 - 0) = 0$
• $C_{13} = +(0 - 0) = 0$
• $C_{21} = -(0 - 0) = 0$
• $C_{22} = +(xz - 0) = xz$
• $C_{23} = -(0 - 0) = 0$
• $C_{31} = +(0 - 0) = 0$
• $C_{32} = -(0 - 0) = 0$
• $C_{33} = +(xy - 0) = xy$

Therefore, the cofactor matrix is:

$C = \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}$

Since $C$ is symmetric, $\text{adj}(A) = C^T = \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}$.

Step 4: Computing the Inverse Matrix

The inverse is calculated using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$ [By Inverse Matrix Theorem].

$A^{-1} = \frac{1}{xyz} \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}$

Distributing the scalar $\frac{1}{xyz}$ into the matrix:

$A^{-1} = \begin{bmatrix} \frac{yz}{xyz} & 0 & 0 \\ 0 & \frac{xz}{xyz} & 0 \\ 0 & 0 & \frac{xy}{xyz} \end{bmatrix} = \begin{bmatrix} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{bmatrix}$

Final Solution: $A^{-1} = \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}$

Step 1: Establishing the Determinant Expression

To analyze the properties of the matrix $A$, we compute its determinant $|A|$ using expansion along the first row. The matrix is given by:

$A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}$

The determinant is expressed as: $|A| = 1 \begin{vmatrix} 1 & \sin\theta \\ -\sin\theta & 1 \end{vmatrix} - \sin\theta \begin{vmatrix} -\sin\theta & \sin\theta \\ -1 & 1 \end{vmatrix} + 1 \begin{vmatrix} -\sin\theta & 1 \\ -1 & -\sin\theta \end{vmatrix}$

Step 2: Algebraic Expansion and Simplification

We evaluate each minor determinant individually [By Laplace Expansion Theorem]:

• Minor 1: $(1)(1) - (\sin\theta)(-\sin\theta) = 1 + \sin^2\theta$
• Minor 2: $(-\sin\theta)(1) - (\sin\theta)(-1) = -\sin\theta + \sin\theta = 0$
• Minor 3: $(-\sin\theta)(-\sin\theta) - (1)(-1) = \sin^2\theta + 1$

Substituting these back into the determinant expression:

$|A| = 1(1 + \sin^2\theta) - \sin\theta(0) + 1(\sin^2\theta + 1)$

$|A| = 1 + \sin^2\theta + \sin^2\theta + 1 = 2 + 2\sin^2\theta = 2(1 + \sin^2\theta)$

Step 3: Interval Analysis of the Determinant

Given the domain $0 \leq \theta \leq 2\pi$, the range of $\sin\theta$ is $[-1, 1]$. Consequently, the range of $\sin^2\theta$ is $[0, 1]$.

Function	Minimum Value	Maximum Value
$\sin^2\theta$	$0$	$1$
$|A| = 2(1 + \sin^2\theta)$	$2(1+0) = 2$	$2(1+1) = 4$

Step 4: Geometric Representation of the Determinant Range

Per the evaluation of the function $f(\theta) = 2 + 2\sin^2\theta$ within the provided trigonometric domain, the determinant values are constrained strictly between $2$ and $4$ inclusive.

Final Solution: $|A| \in [2, 4]$

Step 1: Theoretical Foundation

To verify the property $(A^{-1})^{-1} = A$, we must first determine the inverse of matrix $A$ using the adjoint method, defined as:

$A^{-1} = \frac{1}{|A|} \text{adj}(A)$

where $|A| \neq 0$ [Condition for invertibility].

Step 2: Calculate the Determinant $|A|$

Given $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$

Expanding along the first row:

$|A| = 1(3 \cdot 5 - 1 \cdot 1) - (-2)(-2 \cdot 5 - 1 \cdot 1) + 1(-2 \cdot 1 - 3 \cdot 1)$

$|A| = 1(14) + 2(-11) + 1(-5) = 14 - 22 - 5 = -13$

Since $|A| = -13 \neq 0$, $A^{-1}$ exists.

Step 3: Calculate the Adjoint of A

Find the matrix of cofactors $C_{ij} = (-1)^{i+j}M_{ij}$:

$C_{11} = +(15-1) = 14$	$C_{12} = -(-10-1) = 11$	$C_{13} = +(-2-3) = -5$
$C_{21} = -(-10-1) = 11$	$C_{22} = +(5-1) = 4$	$C_{23} = -(1+2) = -3$
$C_{31} = +(-2-3) = -5$	$C_{32} = -(1+2) = -3$	$C_{33} = +(3-4) = -1$

The adjoint is the transpose of the cofactor matrix:

$\text{adj}(A) = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}$

Step 4: Compute $A^{-1}$

$A^{-1} = \frac{1}{-13} \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix} = \begin{bmatrix} -14/13 & -11/13 & 5/13 \\ -11/13 & -4/13 & 3/13 \\ 5/13 & 3/13 & 1/13 \end{bmatrix}$

Step 5: Verify $(A^{-1})^{-1} = A$

Let $B = A^{-1}$. We find $B^{-1} = \frac{1}{|B|} \text{adj}(B)$.

First, $|B| = |A^{-1}| = \frac{1}{|A|} = -\frac{1}{13}$.

Next, find $\text{adj}(B)$. The cofactors of $B$ are calculated similarly. Given the scalar factor $-1/13$, the adjoint properties dictate $\text{adj}(kA) = k^{n-1} \text{adj}(A)$.

Here, $k = -1/13$ and $n=3$, so $\text{adj}(B) = (-1/13)^2 \cdot \text{adj}(\text{adj}(A))$.

Since $\text{adj}(\text{adj}(A)) = |A|^{n-2} A = (-13)^1 A = -13A$:

$\text{adj}(B) = \frac{1}{169} \cdot (-13)A = -\frac{1}{13}A$

$B^{-1} = \frac{1}{-1/13} \cdot (-\frac{1}{13}A) = A$

Final Solution: Through matrix inversion, we have demonstrated that applying the inversion operator twice returns the identity transformation of the matrix, confirming $(A^{-1})^{-1} = A$.

Step 1: Establishing the Initial Determinant

Let the given determinant be $\Delta$.

$$\Delta = \begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}$$

Step 2: Applying Property of Determinants for Simplification

We perform the column operation $C_1 \rightarrow C_1 + C_2 + C_3$ to simplify the first column [Per the property that the value of a determinant remains invariant under such elementary column transformations].

Summing the elements of each row for the first column:

• $C_{11}: x + y + (x+y) = 2(x+y)$
• $C_{21}: y + (x+y) + x = 2(x+y)$
• $C_{31}: (x+y) + x + y = 2(x+y)$

The determinant becomes:

$$\Delta = \begin{vmatrix} 2(x+y) & y & x+y \\ 2(x+y) & x+y & x \\ 2(x+y) & x & y \end{vmatrix}$$

Step 3: Factoring out the Common Constant

Extract $2(x+y)$ as a common factor from the first column [By the scalar multiplication property of determinants].

$$\Delta = 2(x+y) \begin{vmatrix} 1 & y & x+y \\ 1 & x+y & x \\ 1 & x & y \end{vmatrix}$$

Step 4: Row Reduction

Perform row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$ to introduce zeros in the first column.

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

Simplifying the entries:

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

Step 5: Expansion along the First Column

Expand along $C_1$ [By Laplace expansion]:

$$\Delta = 2(x+y) [1(x(-x-y) - (-y)(x-y)) - 0 + 0]$$

Expand the $2 \times 2$ determinant:

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

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

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

Step 6: Final Algebraic Consolidation

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

Final Solution: $-2(x+y)(x^2 + y^2)$

Step 1: Variable Substitution

To linearize the system of equations, let $u = \frac{1}{x}$, $v = \frac{1}{y}$, and $w = \frac{1}{z}$. Substituting these into the given equations yields the following linear system:

• $2u + 3v + 10w = 4$ [Eq. 1]
• $4u - 6v + 5w = 1$ [Eq. 2]
• $6u + 9v - 20w = 2$ [Eq. 3]

Step 2: Representation via Matrix Equation

The system can be expressed as $AX = B$, where:

$A = \begin{pmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{pmatrix}$, $X = \begin{pmatrix} u \\ v \\ w \end{pmatrix}$, $B = \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix}$

Step 3: Calculating the Determinant of A

Evaluating $|A|$ using expansion along the first row [Per Laplace Expansion]:

$|A| = 2((-6)(-20) - (5)(9)) - 3((4)(-20) - (5)(6)) + 10((4)(9) - (-6)(6))$

$|A| = 2(120 - 45) - 3(-80 - 30) + 10(36 + 36)$

$|A| = 2(75) - 3(-110) + 10(72)$

$|A| = 150 + 330 + 720 = 1200$

Since $|A| \neq 0$, the system has a unique solution given by $X = A^{-1}B$.

Step 4: Finding the Inverse Matrix $A^{-1}$

We calculate the cofactor matrix $C_{ij}$ and transpose it to obtain $adj(A)$.

Co-factors:

• $C_{11} = 120 - 45 = 75$; $C_{12} = -(-80 - 30) = 110$; $C_{13} = 36 + 36 = 72$
• $C_{21} = -(-60 - 90) = 150$; $C_{22} = -40 - 60 = -100$; $C_{23} = -(18 - 18) = 0$
• $C_{31} = 15 + 60 = 75$; $C_{32} = -(10 - 40) = 30$; $C_{33} = -12 - 12 = -24$

$adj(A) = \begin{pmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{pmatrix}$

Step 5: Solving for $u, v, w$

$X = \frac{1}{1200} \begin{pmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{pmatrix} \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix}$

$u = \frac{1}{1200} (75 \times 4 + 150 \times 1 + 75 \times 2) = \frac{300 + 150 + 150}{1200} = \frac{600}{1200} = \frac{1}{2}$

$v = \frac{1}{1200} (110 \times 4 - 100 \times 1 + 30 \times 2) = \frac{440 - 100 + 60}{1200} = \frac{400}{1200} = \frac{1}{3}$

$w = \frac{1}{1200} (72 \times 4 + 0 \times 1 - 24 \times 2) = \frac{288 - 48}{1200} = \frac{240}{1200} = \frac{1}{5}$

Step 6: Converting back to $x, y, z$

$x = \frac{1}{u} = 2$, $y = \frac{1}{v} = 3$, $z = \frac{1}{w} = 5$.

Final Solution: x = 2, y = 3, z = 5

Step 1: Setting up the Determinant Matrix

Let the given determinant be $\Delta$. We are tasked to evaluate:

$\Delta = \begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$

Step 2: Applying Row Operations for Simplification

To simplify the determinant, we utilize the property that the value of a determinant remains unchanged under the operation $R_i \rightarrow R_i - R_j$ [Property of Determinants: Invariance under Elementary Row Operations].

Perform the following operations:

• $R_2 \rightarrow R_2 - R_1$
• $R_3 \rightarrow R_3 - R_1$

Applying $R_2 \rightarrow R_2 - R_1$:

$R_2 = [1 - 1, (x+y) - x, y - y] = [0, y, 0]$

Applying $R_3 \rightarrow R_3 - R_1$:

$R_3 = [1 - 1, x - x, (x+y) - y] = [0, 0, x]$

The updated determinant is:

$\Delta = \begin{vmatrix} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{vmatrix}$

Step 3: Expansion along the First Column

Since the first column ($C_1$) contains two zeros, expanding along $C_1$ is the most efficient method [Laplace Expansion Theorem].

$\Delta = 1 \cdot \begin{vmatrix} y & 0 \\ 0 & x \end{vmatrix} - 0 \cdot \begin{vmatrix} x & y \\ 0 & x \end{vmatrix} + 0 \cdot \begin{vmatrix} x & y \\ y & 0 \end{vmatrix}$

Step 4: Evaluating the 2x2 Determinant

For a $2 \times 2$ matrix $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$, the determinant is given by $ad - bc$.

$\Delta = 1 \cdot (y \cdot x - 0 \cdot 0)$

$\Delta = xy$

Visual Representation of Matrix Structure

Final Solution: $xy$

Given Determinant and Expansion Strategy

Let the determinant be denoted by $\Delta$:

$\Delta = \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$

To prove that $\Delta$ is independent of $\theta$, we perform a cofactor expansion along the first row [Per Laplace Expansion Theorem].

Step 1: Expanding the Determinant

Expanding along the first row ($R_1$):

$\Delta = x \begin{vmatrix} -x & 1 \\ 1 & x \end{vmatrix} - (\sin\theta) \begin{vmatrix} -\sin\theta & 1 \\ \cos\theta & x \end{vmatrix} + (\cos\theta) \begin{vmatrix} -\sin\theta & -x \\ \cos\theta & 1 \end{vmatrix}$

Step 2: Evaluating the $2 \times 2$ Minors

Calculate each determinant component individually:

• $M_{11} = (-x)(x) - (1)(1) = -x^2 - 1$
• $M_{12} = (-\sin\theta)(x) - (1)(\cos\theta) = -x\sin\theta - \cos\theta$
• $M_{13} = (-\sin\theta)(1) - (-x)(\cos\theta) = -\sin\theta + x\cos\theta$

Step 3: Algebraic Substitution and Simplification

Substitute the minors back into the expansion formula:

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

Distribute the terms:

$\Delta = -x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \sin\theta\cos\theta + x\cos^2\theta$

Step 4: Isolating and Eliminating $\theta$

Notice the cancellation of the cross-product terms ($\sin\theta\cos\theta - \sin\theta\cos\theta = 0$):

$\Delta = -x^3 - x + x\sin^2\theta + x\cos^2\theta$

Factor out $x$ from the trigonometric terms:

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

Apply the Fundamental Trigonometric Identity ($\sin^2\theta + \cos^2\theta = 1$):

$\Delta = -x^3 - x + x(1)$

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

$\Delta = -x^3$

Visual Representation of the Determinant Structure

Final Solution: The result of the determinant is $-x^3$, which contains no $\theta$ variable; therefore, the determinant is independent of $\theta$.

Step 1: Theoretical Foundation of Inverse of Matrix Products

To find the inverse of the product of two matrices, we utilize the property of reversal of order [Per the Property of Invertible Matrices]: $$(AB)^{-1} = B^{-1}A^{-1}$$ Given the variables: $A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$

Step 2: Calculating $B^{-1}$

The inverse of a matrix $B$ is given by $B^{-1} = \frac{1}{|B|} \text{adj}(B)$.

First, calculate $|B|$ expanding along the first row: $|B| = 1(3(1) - 0(-2)) - 2(-1(1) - 0(0)) + (-2)(-1(-2) - 3(0))$ $|B| = 1(3) - 2(-1) - 2(2) = 3 + 2 - 4 = 1$

Next, find the matrix of cofactors $C_{ij}$:

$C_{11} = +(3-0) = 3$	$C_{12} = -(-1-0) = 1$	$C_{13} = +(2-0) = 2$
$C_{21} = -(2-4) = 2$	$C_{22} = +(1-0) = 1$	$C_{23} = -(-2-0) = 2$
$C_{31} = +(0-(-6)) = 6$	$C_{32} = -(0-2) = 2$	$C_{33} = +(3-(-2)) = 5$

The Adjoint, $\text{adj}(B) = [C_{ij}]^T = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$. Since $|B| = 1$, $B^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$.

Step 3: Calculating $(AB)^{-1} = B^{-1}A^{-1}$

Perform the matrix multiplication: $(AB)^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$

Calculating each element $(i, j)$ using the dot product of the $i$-th row of $B^{-1}$ and $j$-th column of $A^{-1}$:

• Row 1: $(3)(3) + (2)(-15) + (6)(5) = 9 - 30 + 30 = 9$ $(3)(-1) + (2)(6) + (6)(-2) = -3 + 12 - 12 = -3$ $(3)(1) + (2)(-5) + (6)(2) = 3 - 10 + 12 = 5$
• Row 2: $(1)(3) + (1)(-15) + (2)(5) = 3 - 15 + 10 = -2$ $(1)(-1) + (1)(6) + (2)(-2) = -1 + 6 - 4 = 1$ $(1)(1) + (1)(-5) + (2)(2) = 1 - 5 + 4 = 0$
• Row 3: $(2)(3) + (2)(-15) + (5)(5) = 6 - 30 + 25 = 1$ $(2)(-1) + (2)(6) + (5)(-2) = -2 + 12 - 10 = 0$ $(2)(1) + (2)(-5) + (5)(2) = 2 - 10 + 10 = 2$

Visual Representation of Matrix Transformation

Final Solution: $(AB)^{-1} = \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}$