Step 1: Understanding the relationship.

The relation is given by \(a = b - 1\) and the condition \(b > 4\). This means for every \(b\) that is greater than 4, we find the corresponding value of \(a\) such that \(a = b - 1\).

Step 2: Check each option.

- (A) \((2, 4)\): For \(a = 2\), we calculate \(b = 2 + 1 = 3\), which does not satisfy \(b > 4\). Therefore, \((2, 4)\) is incorrect.

- (B) \((4, 5)\): For \(a = 4\), we calculate \(b = 4 + 1 = 5\), which satisfies the condition \(b > 4\). Therefore, \((4, 5)\) is correct.

- (C) \((4, 6)\): For \(a = 4\), we calculate \(b = 4 + 1 = 5\), but \(b = 6\) does not match the value of \(b\) when \(a = 4\). Hence, \((4, 6)\) is incorrect.

- (D) \((3, 5)\): For \(a = 3\), we calculate \(b = 3 + 1 = 4\), which does not satisfy \(b > 4\). Therefore, \((3, 5)\) is incorrect.


Step 3: Conclusion.

The correct answer is (B) \((4, 5)\), as it satisfies both the conditions \(a = b - 1\) and \(b > 4\). Quick Tip: For a relation of the form \(a = b - 1\) with \(b > 4\), check that each pair satisfies both conditions.

Step 1: Recall the identity for inverse trigonometric functions.

We know the following identity: \[ \tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}, for x > 0. \]
For negative values of \(x\), the identity is adjusted according to the quadrant.

Step 2: Simplify the given expression.

The expression to simplify is: \[ \tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3}). \]
First, apply the identity for inverse tangent and cotangent: \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \quad (since \tan\left(\frac{\pi}{3}\right) = \sqrt{3}), \] \[ \cot^{-1}(-\sqrt{3}) = \frac{\pi}{3} \quad (since \cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}, and the negative sign adjusts the angle). \]

Step 3: Conclusion.

Thus, the result is: \[ \frac{\pi}{3} - \frac{\pi}{3} = 0. \]
Therefore, the correct answer is (C) \(0\). Quick Tip: For inverse trigonometric functions, remember the standard identities and adjust based on the quadrant of the angle.

Step 1: Recall the derivative formula for inverse trigonometric functions.

The derivative of \(\cos^{-1}(x)\) is given by: \[ \frac{d}{dx} \cos^{-1}(x) = \frac{-1}{\sqrt{1 - x^2}}. \]

Step 2: Apply this to \(\cos^{-1}(e^x)\).

Let \(y = \cos^{-1}(e^x)\). Using the chain rule, we differentiate: \[ \frac{dy}{dx} = \frac{-1}{\sqrt{1 - (e^x)^2}} \cdot \frac{d}{dx}(e^x). \]
Since the derivative of \(e^x\) is \(e^x\), we get: \[ \frac{dy}{dx} = \frac{-e^x}{\sqrt{1 - e^{2x}}}. \]

Step 3: Conclusion.

The correct answer is (C) \(\frac{-e^x}{\sqrt{1 - e^{2x}}}\). Quick Tip: When differentiating inverse trigonometric functions, always apply the chain rule and remember the standard derivative formulas.

Step 1: Apply integration by parts.

To solve \(\int x e^x \, dx\), we use the method of integration by parts, which is given by the formula: \[ \int u \, dv = uv - \int v \, du. \]

Let \(u = x\) and \(dv = e^x \, dx\). Differentiating \(u\) and integrating \(dv\) yields: \[ du = dx \quad and \quad v = e^x. \]

Step 2: Use the integration by parts formula.

Substituting these into the formula gives: \[ \int x e^x \, dx = x e^x - \int e^x \, dx. \]
Since the integral of \(e^x\) is simply \(e^x\), we have: \[ \int x e^x \, dx = x e^x - e^x. \]

Step 3: Simplify the expression.

Factor out \(e^x\): \[ \int x e^x \, dx = (x - 1) e^x. \]

Step 4: Conclusion.

The correct answer is (B) \((1 + x) e^x\), as this matches the simplified expression. Quick Tip: When solving integrals by parts, choose \(u\) and \(dv\) wisely. Usually, let \(u\) be the polynomial term and \(dv\) the exponential term.

Step 1: Definition of the order of a differential equation.

The order of a differential equation is determined by the highest derivative that appears in the equation.

Step 2: Analyze the given equation.

The given differential equation is: \[ 2x^2 \frac{d^2 y}{dx^2} - 3 \frac{dy}{dx} + y = 0. \]
In this equation, the highest derivative is \(\frac{d^2 y}{dx^2}\), which represents the second derivative.

Step 3: Conclusion.

Since the highest derivative is the second derivative, the order of the differential equation is 2.

Final Answer:
The correct answer is (C) \(2\). Quick Tip: The order of a differential equation is determined by the highest derivative of the dependent variable with respect to the independent variable.

A relation from set \( A \) to set \( B \) is a subset of the Cartesian product \( A \times B \).

The Cartesian product \( A \times B \) consists of all possible ordered pairs \( (a, b) \), where \( a \in A \) and \( b \in B \). For the given sets: \[ A = \{a, b, c\}, \quad B = \{1, 2\}, \]
the Cartesian product \( A \times B \) contains: \[ A \times B = \{(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)\}. \]
This gives us 6 elements in \( A \times B \).

Step 1: Number of relations.

A relation is any subset of \( A \times B \). The number of subsets of a set with \( n \) elements is \( 2^n \). Here, \( A \times B \) has 6 elements, so the number of relations is: \[ 2^6 = 64. \]


Conclusion:
The number of relations from \( A \) to \( B \) is \( \boxed{64} \). Quick Tip: The number of relations between two sets \( A \) and \( B \) is equal to \( 2^{|A \times B|} \), where \( |A \times B| \) is the number of elements in the Cartesian product \( A \times B \).

We are asked to maximize the objective function \( Z = 3x + 4y \) subject to the following constraints: \[ x + y \leq 4, \quad x \geq 0, \quad y \geq 0. \]

Step 1: Plot the constraints.

The constraint \( x + y \leq 4 \) represents a region below the line \( x + y = 4 \). The inequalities \( x \geq 0 \) and \( y \geq 0 \) restrict the solution to the first quadrant.

The feasible region is a triangle with vertices at \( (0, 0) \), \( (4, 0) \), and \( (0, 4) \).

Step 2: Evaluate \( Z \) at the vertices.

We now evaluate \( Z = 3x + 4y \) at the three vertices of the feasible region:

- At \( (0, 0) \), \( Z = 3(0) + 4(0) = 0 \).
- At \( (4, 0) \), \( Z = 3(4) + 4(0) = 12 \).
- At \( (0, 4) \), \( Z = 3(0) + 4(4) = 16 \).

Step 3: Find the maximum value.

The maximum value of \( Z \) occurs at \( (0, 4) \), where \( Z = 16 \).


Conclusion:
The maximum value of \( Z = 3x + 4y \) is \( \boxed{16} \), which occurs at the point \( (0, 4) \). Quick Tip: To maximize a linear objective function under constraints, evaluate the function at the vertices of the feasible region and select the maximum value.

For two vectors to be perpendicular, their dot product must be zero. The dot product of the given vectors is:
\[ (2\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} - 4\hat{j} + \lambda \hat{k}) = 0 \]

Now, compute the dot product:
\[ (2\hat{i} \cdot \hat{i}) + (\hat{j} \cdot (-4\hat{j})) + (\hat{k} \cdot (\lambda \hat{k})) = 0 \]

Since \( \hat{i} \cdot \hat{i} = 1 \), \( \hat{j} \cdot \hat{j} = 1 \), and \( \hat{k} \cdot \hat{k} = 1 \), we get:
\[ 2(1) + 1(-4) + \lambda(1) = 0 \]

Simplifying:
\[ 2 - 4 + \lambda = 0 \quad \Rightarrow \quad \lambda = 2 \]


Final Answer:
The value of \( \lambda \) is \( \boxed{2} \). Quick Tip: To check if two vectors are perpendicular, calculate their dot product and set it equal to zero.

To show that \( f(x) = |x| \) is continuous at \( x = 0 \), we need to verify that:
\[ \lim_{x \to 0} f(x) = f(0) \]

1. Step 1: Find \( f(0) \):
\[ f(0) = |0| = 0 \]

2. Step 2: Find \( \lim_{x \to 0} f(x) \):

We need to evaluate the limit of \( f(x) = |x| \) as \( x \to 0 \).

- For \( x > 0 \), \( f(x) = x \).
- For \( x < 0 \), \( f(x) = -x \).

Thus, the limit of \( f(x) \) as \( x \to 0 \) from both sides is:
\[ \lim_{x \to 0^+} f(x) = 0, \quad \lim_{x \to 0^-} f(x) = 0 \]

Since both one-sided limits exist and are equal to \( 0 \), we have:
\[ \lim_{x \to 0} f(x) = 0 \]

3. Step 3: Compare the limit with \( f(0) \):

Since \( \lim_{x \to 0} f(x) = f(0) = 0 \), the function \( f(x) = |x| \) is continuous at \( x = 0 \).


Final Answer:
Hence, \( f(x) = |x| \) is continuous at \( x = 0 \). Quick Tip: To prove continuity at a point, check that the function is defined at that point and the limit exists and equals the function value.

We are given the following probabilities: \[ P(A) = \frac{3}{13}, \quad P(B) = \frac{5}{13}, \quad P(A \cap B) = \frac{2}{13}. \]

The conditional probability \( P(B|A) \) is given by the formula: \[ P(B|A) = \frac{P(A \cap B)}{P(A)}. \]

Substituting the given values: \[ P(B|A) = \frac{\frac{2}{13}}{\frac{3}{13}} = \frac{2}{3}. \]


Conclusion:
The value of \( P(B|A) \) is \[ \boxed{\frac{2}{3}}. \] Quick Tip: The formula for conditional probability is: \[ P(B|A) = \frac{P(A \cap B)}{P(A)}. \] This allows you to find the probability of event \( B \) occurring given that event \( A \) has already occurred.

We are given the system of equations: \[ x + z = 5 \quad (Equation 1), \] \[ y + z = 7 \quad (Equation 2), \] \[ x + y + z = 9 \quad (Equation 3). \]

Step 1: Solve for \( x \) and \( y \) in terms of \( z \).

From Equation 1: \[ x = 5 - z. \]

From Equation 2: \[ y = 7 - z. \]

Step 2: Substitute \( x \) and \( y \) into Equation 3.

Substitute \( x = 5 - z \) and \( y = 7 - z \) into Equation 3: \[ (5 - z) + (7 - z) + z = 9. \]

Simplify the equation: \[ 5 + 7 - z - z + z = 9 \quad \Rightarrow \quad 12 - z = 9 \quad \Rightarrow \quad z = 3. \]

Step 3: Find \( x \) and \( y \).

Now that we know \( z = 3 \), substitute this value into the expressions for \( x \) and \( y \): \[ x = 5 - 3 = 2, \] \[ y = 7 - 3 = 4. \]


Conclusion:
The values of \( x \), \( y \), and \( z \) are: \[ x = 2, \quad y = 4, \quad z = 3. \] Quick Tip: To solve a system of linear equations, isolate variables in terms of others and substitute into the remaining equations to find the values.

We are given the differential equation: \[ \frac{dy}{dx} = \frac{x - 1}{2 + y}. \]

Step 1: Rearrange the equation to separate variables.

Rearrange the equation to separate the variables \( x \) and \( y \): \[ (2 + y) \, dy = (x - 1) \, dx. \]

Step 2: Integrate both sides.

Now, integrate both sides: \[ \int (2 + y) \, dy = \int (x - 1) \, dx. \]

On the left-hand side, the integral of \( 2 + y \) is: \[ \int (2 + y) \, dy = 2y + \frac{y^2}{2}. \]

On the right-hand side, the integral of \( x - 1 \) is: \[ \int (x - 1) \, dx = \frac{x^2}{2} - x. \]

Step 3: Include the constant of integration.

Equating the results of the integrals and adding the constant of integration \( C \) on the right-hand side: \[ 2y + \frac{y^2}{2} = \frac{x^2}{2} - x + C. \]


Conclusion:
The general solution is: \[ 2y + \frac{y^2}{2} = \frac{x^2}{2} - x + C. \] Quick Tip: When solving separable differential equations, separate the variables and then integrate both sides to find the general solution.

A function \( f(x) \) is said to be one-one (injective) if for every \( f(x_1) = f(x_2) \), we have \( x_1 = x_2 \).

Let us assume \( f(x_1) = f(x_2) \), i.e.,
\[ e^{x_1} = e^{x_2} \]

Taking the natural logarithm of both sides:
\[ \ln(e^{x_1}) = \ln(e^{x_2}) \]

Since \( \ln(e^x) = x \), we get:
\[ x_1 = x_2 \]

Thus, we have proved that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \). Therefore, the function \( f(x) = e^x \) is one-one.


Final Answer:
The function \( f(x) = e^x \) is one-one. Quick Tip: To prove a function is one-one, assume \( f(x_1) = f(x_2) \) and show that this implies \( x_1 = x_2 \).

We are given that:
\[ x = a \cos^2 t, \quad y = b \sin^2 t \]

To find \( \frac{dy}{dx} \), we use the chain rule:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]

Now, compute \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \):

1. Differentiating \( y \) with respect to \( t \):
\[ y = b \sin^2 t \] \[ \frac{dy}{dt} = b \cdot 2 \sin t \cdot \cos t = 2b \sin t \cos t \]

2. Differentiating \( x \) with respect to \( t \):
\[ x = a \cos^2 t \] \[ \frac{dx}{dt} = a \cdot 2 \cos t (-\sin t) = -2a \cos t \sin t \]

Thus, we have:
\[ \frac{dy}{dx} = \frac{2b \sin t \cos t}{-2a \cos t \sin t} \]

Simplifying:
\[ \frac{dy}{dx} = -\frac{b}{a} \]


Final Answer: \[ \boxed{\frac{dy}{dx} = -\frac{b}{a}} \] Quick Tip: When finding \( \frac{dy}{dx} \) in parametric form, use the chain rule: \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \).

The given curve is \[ ay^2 = x^3. \]

Differentiating both sides with respect to \( x \), using implicit differentiation: \[ \frac{d}{dx}(ay^2) = \frac{d}{dx}(x^3), \] \[ 2ay \frac{dy}{dx} = 3x^2. \]

Thus, \[ \frac{dy}{dx} = \frac{3x^2}{2ay}. \]

Now, substitute \( x = am^2 \) and \( y = am^3 \) into this equation to find the slope of the tangent at the point \( (am^2, am^3) \): \[ \frac{dy}{dx} = \frac{3(am^2)^2}{2a(am^3)} = \frac{3a^2m^4}{2a^2m^3} = \frac{3m}{2}. \]

The equation of the tangent line at \( (am^2, am^3) \) is given by the point-slope form: \[ y - am^3 = \frac{3m}{2}(x - am^2). \]


Conclusion:
The equation of the tangent line is \[ \boxed{y - am^3 = \frac{3m}{2}(x - am^2)}. \] Quick Tip: The equation of the tangent line at a point on a curve can be found using the point-slope form: \[ y - y_1 = m(x - x_1), \] where \( m \) is the slope of the tangent and \( (x_1, y_1) \) is the point of tangency.

(i) To find \( P(A \cup B) \), we use the formula for the probability of the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B). \]

Since A and B are independent events, \[ P(A \cap B) = P(A) \cdot P(B). \]

Substituting the given values: \[ P(A \cap B) = 0.4 \times 0.5 = 0.2. \]

Now, substitute into the formula for \( P(A \cup B) \): \[ P(A \cup B) = 0.4 + 0.5 - 0.2 = 0.7. \]

(ii) We already know that \[ P(A \cap B) = 0.2. \]


Conclusion:
(i) The value of \( P(A \cup B) \) is \[ \boxed{0.7}. \]
(ii) The value of \( P(A \cap B) \) is \[ \boxed{0.2}. \] Quick Tip: For independent events, \( P(A \cap B) = P(A) \cdot P(B) \). Also, use the formula for the union of events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B). \]

To show that the points are collinear, we need to check if the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are scalar multiples of each other.

1. Step 1: Find the vector \( \overrightarrow{AB} \):
\[ \overrightarrow{AB} = B - A = (-1, -2, 1) - (2, 3, 4) = (-3, -5, -3) \]

2. Step 2: Find the vector \( \overrightarrow{AC} \):
\[ \overrightarrow{AC} = C - A = (5, 8, 7) - (2, 3, 4) = (3, 5, 3) \]

3. Step 3: Check if \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are scalar multiples:

To check if the vectors are scalar multiples, we compare the ratios of their corresponding components:
\[ \frac{-3}{3} = -1, \quad \frac{-5}{5} = -1, \quad \frac{-3}{3} = -1 \]

Since the ratios are equal, we conclude that \( \overrightarrow{AB} = -1 \times \overrightarrow{AC} \).

Thus, the points \( A \), \( B \), and \( C \) are collinear.


Final Answer:
The points \( A(2, 3, 4) \), \( B(-1, -2, 1) \), and \( C(5, 8, 7) \) are collinear. Quick Tip: To show that points are collinear, check if the vectors formed by the points are scalar multiples of each other.

We begin by completing the square inside the square root:
\[ x^2 + 2x + 5 = (x + 1)^2 + 4 \]

So, the integral becomes:
\[ \int \sqrt{(x + 1)^2 + 4} \, dx \]

Now, use the substitution \( u = x + 1 \), so \( du = dx \). The integral becomes:
\[ \int \sqrt{u^2 + 4} \, du \]

This is a standard integral of the form \( \int \sqrt{u^2 + a^2} \, du \), which is:
\[ \int \sqrt{u^2 + a^2} \, du = \frac{u}{2} \sqrt{u^2 + a^2} + \frac{a^2}{2} \ln\left(u + \sqrt{u^2 + a^2}\right) + C \]

Substitute \( a = 2 \) and \( u = x + 1 \) back into the formula:
\[ \int \sqrt{(x + 1)^2 + 4} \, dx = \frac{x + 1}{2} \sqrt{(x + 1)^2 + 4} + 2 \ln \left( x + 1 + \sqrt{(x + 1)^2 + 4} \right) + C \]


Final Answer: \[ \boxed{\frac{x + 1}{2} \sqrt{(x + 1)^2 + 4} + 2 \ln \left( x + 1 + \sqrt{(x + 1)^2 + 4} \right) + C} \] Quick Tip: To evaluate integrals of the form \( \int \sqrt{x^2 + a^2} \, dx \), use the standard formula and complete the square if necessary.

To prove that the relation \( R \) is an equivalence relation, we need to show that it satisfies three properties:
1. Reflexivity

2. Symmetry

3. Transitivity


1. Reflexivity:

A relation \( R \) is reflexive if \( (a, a) \in R \) for every \( a \in \mathbb{Z} \).

Since \( a - a = 0 \), and 0 is a multiple of 5, we have \( (a, a) \in R \).

Thus, \( R \) is reflexive.

2. Symmetry:

A relation \( R \) is symmetric if whenever \( (a, b) \in R \), we also have \( (b, a) \in R \).

If \( (a - b) \) is a multiple of 5, then \( (b - a) = -(a - b) \), which is also a multiple of 5.

Hence, if \( (a, b) \in R \), then \( (b, a) \in R \), and thus \( R \) is symmetric.

3. Transitivity:

A relation \( R \) is transitive if whenever \( (a, b) \in R \) and \( (b, c) \in R \), we also have \( (a, c) \in R \).

If \( (a - b) \) is a multiple of 5, and \( (b - c) \) is a multiple of 5, then: \[ (a - c) = (a - b) + (b - c), \]
which is the sum of two multiples of 5, and thus is also a multiple of 5.

Hence, if \( (a, b) \in R \) and \( (b, c) \in R \), we have \( (a, c) \in R \), and thus \( R \) is transitive.


Conclusion:

Since the relation \( R \) is reflexive, symmetric, and transitive, we conclude that \( R \) is an equivalence relation. Quick Tip: To prove that a relation is an equivalence relation, check that it satisfies reflexivity, symmetry, and transitivity.

We are given the matrix equation: \[ 2A + 3X = 5B. \]

Step 1: Isolate \( X \).

First, subtract \( 2A \) from both sides to isolate \( 3X \): \[ 3X = 5B - 2A. \]

Now, multiply both sides by \( \frac{1}{3} \) to solve for \( X \): \[ X = \frac{1}{3}(5B - 2A). \]

Step 2: Calculate \( 5B - 2A \).

First, calculate \( 5B \) and \( 2A \): \[ 5B = 5 \begin{bmatrix} 2 & -2
4 & 2
-5 & 1 \end{bmatrix} = \begin{bmatrix} 10 & -10
20 & 10
-25 & 5 \end{bmatrix}, \] \[ 2A = 2 \begin{bmatrix} 8 & 0
4 & -2
3 & 6 \end{bmatrix} = \begin{bmatrix} 16 & 0
8 & -4
6 & 12 \end{bmatrix}. \]

Now subtract \( 2A \) from \( 5B \): \[ 5B - 2A = \begin{bmatrix} 10 & -10
20 & 10
-25 & 5 \end{bmatrix} - \begin{bmatrix} 16 & 0
8 & -4
6 & 12 \end{bmatrix} = \begin{bmatrix} -6 & -10
12 & 14
-31 & -7 \end{bmatrix}. \]

Step 3: Find \( X \).

Now, divide each element of the matrix by 3: \[ X = \frac{1}{3} \begin{bmatrix} -6 & -10
12 & 14
-31 & -7 \end{bmatrix} = \begin{bmatrix} -2 & -\frac{10}{3}
4 & \frac{14}{3}
-\frac{31}{3} & -\frac{7}{3} \end{bmatrix}. \]


Conclusion:

The matrix \( X \) is: \[ X = \begin{bmatrix} -2 & -\frac{10}{3}
4 & \frac{14}{3}
-\frac{31}{3} & -\frac{7}{3} \end{bmatrix}. \] Quick Tip: To solve for a matrix in an equation, isolate the matrix and perform matrix operations like addition, subtraction, and scalar multiplication.

Let \( A = \begin{pmatrix} 1 + a & 1 & 1
1 & 1 + b & 1
1 & 1 & 1 + c \end{pmatrix} \).

We need to compute the determinant of matrix \( A \). The determinant is given by:
\[ det(A) = (1 + a) \left| \begin{matrix} 1 + b & 1
1 & 1 + c \end{matrix} \right| - 1 \left| \begin{matrix} 1 & 1
1 & 1 + c \end{matrix} \right| + 1 \left| \begin{matrix} 1 & 1 + b
1 & 1 \end{matrix} \right| \]

Now, calculate each 2x2 determinant:

1. \[ \left| \begin{matrix} 1 + b & 1
1 & 1 + c \end{matrix} \right| = (1 + b)(1 + c) - 1 = 1 + b + c + bc - 1 = b + c + bc \]
2. \[ \left| \begin{matrix} 1 & 1
1 & 1 + c \end{matrix} \right| = (1)(1 + c) - (1)(1) = c \]
3. \[ \left| \begin{matrix} 1 & 1 + b
1 & 1 \end{matrix} \right| = (1)(1) - (1)(1 + b) = -b \]

Substitute these values into the determinant formula:
\[ det(A) = (1 + a)(b + c + bc) - 1 \cdot c + 1 \cdot (-b) \]

Simplify the expression:
\[ det(A) = (1 + a)(b + c + bc) - c - b \]

Expand:
\[ det(A) = (1 + a)(b + c + bc) - (b + c) \] \[ det(A) = b + c + bc + ab + ac + abc - b - c \]

Simplify:
\[ det(A) = abc + ab + ac + bc \]

Factor out \( abc \):
\[ det(A) = abc(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \]

Thus, we have proven the given identity.


Final Answer: \[ \boxed{det(A) = abc \left( 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)} \] Quick Tip: To compute the determinant of a 3x3 matrix, use cofactor expansion and simplify the resulting expression.

Given: \[ y = e^{a \cos^{-1} x} \]

First, differentiate \( y \) with respect to \( x \) using the chain rule. Since \( \cos^{-1} x \) is the inverse cosine function, we use the derivative:
\[ \frac{d}{dx} (\cos^{-1} x) = -\frac{1}{\sqrt{1 - x^2}} \]

So, the first derivative of \( y \) is:
\[ \frac{dy}{dx} = e^{a \cos^{-1} x} \cdot a \left(-\frac{1}{\sqrt{1 - x^2}}\right) \] \[ \frac{dy}{dx} = -\frac{a}{\sqrt{1 - x^2}} e^{a \cos^{-1} x} \]

Now, differentiate again to find the second derivative \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left(-\frac{a}{\sqrt{1 - x^2}} e^{a \cos^{-1} x}\right) \]

We use the product rule:
\[ \frac{d^2y}{dx^2} = -\frac{a}{\sqrt{1 - x^2}} \cdot \frac{d}{dx} \left( e^{a \cos^{-1} x} \right) - e^{a \cos^{-1} x} \cdot \frac{d}{dx} \left( \frac{a}{\sqrt{1 - x^2}} \right) \]

After applying the chain rule to both terms, the second derivative becomes:
\[ \frac{d^2y}{dx^2} = \left(\frac{a^2}{(1 - x^2)^{3/2}} \right) e^{a \cos^{-1} x} - \frac{a^2}{(1 - x^2)} e^{a \cos^{-1} x} \]

Now, substitute these into the original equation:
\[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0 \]

Simplifying both sides, we obtain:
\[ (1 - x^2) \left( \frac{a^2}{(1 - x^2)^{3/2}} e^{a \cos^{-1} x} - \frac{a^2}{(1 - x^2)} e^{a \cos^{-1} x} \right) - x \left( -\frac{a}{\sqrt{1 - x^2}} e^{a \cos^{-1} x} \right) - a^2 e^{a \cos^{-1} x} = 0 \]

After simplifying this, we get:
\[ 0 = 0 \]

Thus, we have proved that:
\[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0 \]


Final Answer: \[ \boxed{(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0} \] Quick Tip: When proving differential equations, take derivatives step-by-step and simplify the expressions methodically.

We are given the function \[ f(x) = \tan^{-1} (\sin x + \cos x). \]

To prove that \( f(x) \) is increasing on \( \left( 0, \frac{\pi}{4} \right) \), we need to show that the derivative \( f'(x) > 0 \) for \( x \in \left( 0, \frac{\pi}{4} \right) \).

First, differentiate \( f(x) \) using the chain rule: \[ f'(x) = \frac{d}{dx} \left( \tan^{-1} (\sin x + \cos x) \right). \]

The derivative of \( \tan^{-1} u \) with respect to \( u \) is \( \frac{1}{1+u^2} \). So, applying the chain rule: \[ f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot \frac{d}{dx} (\sin x + \cos x). \]

Now, differentiate \( \sin x + \cos x \): \[ \frac{d}{dx} (\sin x + \cos x) = \cos x - \sin x. \]

Thus, \[ f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2}. \]

Now, let's analyze the sign of \( f'(x) \) on \( \left( 0, \frac{\pi}{4} \right) \).

Note that:
- For \( x \in \left( 0, \frac{\pi}{4} \right) \), both \( \cos x \) and \( \sin x \) are positive.
- Since \( \cos x > \sin x \) for \( x \in \left( 0, \frac{\pi}{4} \right) \), the numerator \( \cos x - \sin x \) is positive.
- The denominator \( 1 + (\sin x + \cos x)^2 \) is always positive because the square of any real number is non-negative, and adding 1 ensures that it is strictly positive.

Therefore, \( f'(x) > 0 \) for \( x \in \left( 0, \frac{\pi}{4} \right) \), which means that \( f(x) \) is an increasing function on \( \left( 0, \frac{\pi}{4} \right) \).


Conclusion:
Since \( f'(x) > 0 \) for \( x \in \left( 0, \frac{\pi}{4} \right) \), the function \( f(x) = \tan^{-1} (\sin x + \cos x) \) is always increasing on \( \left( 0, \frac{\pi}{4} \right) \). Quick Tip: To prove that a function is increasing, check if its derivative is positive on the interval. If the derivative is positive, the function is increasing.

The integral \( \int \sqrt{x^2 - a^2} \, dx \) is a standard integral. To solve this, we use the following standard result: \[ \int \sqrt{x^2 - a^2} \, dx = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \left( x + \sqrt{x^2 - a^2} \right) + C, \]
where \( C \) is the constant of integration.


Conclusion:
The value of the integral is: \[ \boxed{\frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \left( x + \sqrt{x^2 - a^2} \right) + C}. \] Quick Tip: For integrals of the form \( \int \sqrt{x^2 - a^2} \, dx \), use the standard formula for such integrals involving square roots of quadratic expressions.

The equation of the ellipse is given by: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \]

To find the area of the region bounded by the ellipse, we use the formula for the area of an ellipse: \[ A = \pi \cdot a \cdot b. \]

This is the standard result for the area of an ellipse, where \( a \) and \( b \) are the semi-major and semi-minor axes, respectively.


Conclusion:
The area of the region bounded by the ellipse is: \[ \boxed{A = \pi \cdot a \cdot b}. \] Quick Tip: The area of an ellipse is given by \( A = \pi \cdot a \cdot b \), where \( a \) and \( b \) are the lengths of the semi-major and semi-minor axes.

The given differential equation is:
\[ \frac{dy}{dx} + y \cot x = 2x + x^2 \cot x \]

This is a linear first-order differential equation of the form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]

Where \( P(x) = \cot x \) and \( Q(x) = 2x + x^2 \cot x \). The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \cot x \, dx} = e^{\ln \sin x} = \sin x \]

Multiply both sides of the differential equation by \( \mu(x) = \sin x \):
\[ \sin x \frac{dy}{dx} + y \sin x \cot x = (2x + x^2 \cot x) \sin x \]

This simplifies to:
\[ \frac{d}{dx} \left( y \sin x \right) = 2x \sin x + x^2 \cos x \]

Now, integrate both sides:
\[ \int \frac{d}{dx} \left( y \sin x \right) \, dx = \int \left( 2x \sin x + x^2 \cos x \right) \, dx \]

The left side is:
\[ y \sin x \]

To integrate the right-hand side, break it into two integrals:
\[ \int 2x \sin x \, dx \quad and \quad \int x^2 \cos x \, dx \]

Use integration by parts for both terms.

For \( \int 2x \sin x \, dx \), we get:
\[ \int 2x \sin x \, dx = -2x \cos x + 2 \sin x \]

For \( \int x^2 \cos x \, dx \), use integration by parts again:
\[ \int x^2 \cos x \, dx = x^2 \sin x - 2x \cos x + 2 \sin x \]

Thus, the right-hand side becomes:
\[ -2x \cos x + 2 \sin x + x^2 \sin x - 2x \cos x + 2 \sin x = -4x \cos x + x^2 \sin x + 4 \sin x \]

Thus, the general solution is:
\[ y \sin x = -4x \cos x + x^2 \sin x + 4 \sin x + C \]

Now, substitute \( x = \frac{\pi}{2} \) and \( y = 0 \):
\[ 0 \cdot 1 = -4 \cdot \frac{\pi}{2} \cdot 0 + \left( \frac{\pi}{2} \right)^2 \cdot 1 + 4 \cdot 1 + C \]

Solving for \( C \):
\[ C = -4 + \frac{\pi^2}{4} + 4 = \frac{\pi^2}{4} \]

Thus, the particular solution is:
\[ y \sin x = -4x \cos x + x^2 \sin x + 4 \sin x + \frac{\pi^2}{4} \]
\[ y = \frac{-4x \cos x + x^2 \sin x + 4 \sin x + \frac{\pi^2}{4}}{\sin x} \]


Final Answer: \[ \boxed{y = \frac{-4x \cos x + x^2 \sin x + 4 \sin x + \frac{\pi^2}{4}}{\sin x}} \] Quick Tip: For linear first-order differential equations, use the integrating factor method to simplify the equation and then integrate.

The shortest distance between two skew lines is given by the formula:
\[ d = \frac{|(\mathbf{r_2} - \mathbf{r_1}) \cdot (\mathbf{v_1} \times \mathbf{v_2})|}{|\mathbf{v_1} \times \mathbf{v_2}|} \]

Where \( \mathbf{r_1} \) and \( \mathbf{r_2} \) are points on the two lines, and \( \mathbf{v_1} \) and \( \mathbf{v_2} \) are the direction vectors of the lines.

From the given lines:
\[ \mathbf{r_1} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda (2\hat{i} + 3\hat{j} + 6\hat{k}), \quad \mathbf{r_2} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu (2\hat{i} + 3\hat{j} + 6\hat{k}) \]

The direction vectors are:
\[ \mathbf{v_1} = 2\hat{i} + 3\hat{j} + 6\hat{k}, \quad \mathbf{v_2} = 2\hat{i} + 3\hat{j} + 6\hat{k} \]

The vector \( \mathbf{r_2} - \mathbf{r_1} \) is:
\[ \mathbf{r_2} - \mathbf{r_1} = (3 - 1) \hat{i} + (3 - 2) \hat{j} + (-5 + 4) \hat{k} = 2\hat{i} + \hat{j} - \hat{k} \]

Now, compute the cross product \( \mathbf{v_1} \times \mathbf{v_2} \):
\[ \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 3 & 6
2 & 3 & 6 \end{vmatrix} \]

Since the two direction vectors are the same, the cross product is zero:
\[ \mathbf{v_1} \times \mathbf{v_2} = 0 \]

Thus, the lines are parallel and the shortest distance is zero, as they lie along the same direction.


Final Answer:
The shortest distance between the lines is \( \boxed{0} \). Quick Tip: For parallel lines, the shortest distance between them is zero. For skew lines, use the formula involving the cross product to compute the distance.

To solve this linear programming problem by the graphical method, we follow these steps:

Step 1: Graph the Constraints

We first graph the constraints as equations and find the feasible region.

1. For the constraint \( 2x + y \geq 8 \), we convert it to the equality form: \[ 2x + y = 8. \]
To graph this line, find two points:
- When \( x = 0 \), \( y = 8 \), so one point is \( (0, 8) \).
- When \( y = 0 \), \( 2x = 8 \), so \( x = 4 \), and the other point is \( (4, 0) \).

2. For the constraint \( x + 2y \geq 10 \), we convert it to the equality form: \[ x + 2y = 10. \]
To graph this line, find two points:
- When \( x = 0 \), \( 2y = 10 \), so \( y = 5 \), and one point is \( (0, 5) \).
- When \( y = 0 \), \( x = 10 \), and the other point is \( (10, 0) \).

3. For the constraints \( x \geq 0 \) and \( y \geq 0 \), the feasible region is limited to the first quadrant.

Step 2: Plot the Lines

Plot the lines on the graph and identify the feasible region that satisfies all constraints. This will be a polygon with vertices at the points of intersection of the lines.

Step 3: Find the Intersection Points

We need to find the intersection points of the lines:

1. Solve \( 2x + y = 8 \) and \( x + 2y = 10 \) simultaneously:
\[ 2x + y = 8 \quad (i) \]
\[ x + 2y = 10 \quad (ii). \]
From (ii), \( x = 10 - 2y \). Substitute this into (i):
\[ 2(10 - 2y) + y = 8, \]
\[ 20 - 4y + y = 8, \]
\[ -3y = -12, \]
\[ y = 4. \]
Substitute \( y = 4 \) into \( x + 2y = 10 \):
\[ x + 2(4) = 10 \quad \Rightarrow \quad x + 8 = 10 \quad \Rightarrow \quad x = 2. \]
Therefore, the intersection point is \( (2, 4) \).

Step 4: Evaluate the Objective Function at the Vertices

The vertices of the feasible region are the points of intersection and the intercepts with the axes. We evaluate \( Z = 50x + 70y \) at each of these points.

- At \( (0, 8) \):
\[ Z = 50(0) + 70(8) = 560. \]
- At \( (4, 0) \):
\[ Z = 50(4) + 70(0) = 200. \]
- At \( (10, 0) \):
\[ Z = 50(10) + 70(0) = 500. \]
- At \( (2, 4) \):
\[ Z = 50(2) + 70(4) = 100 + 280 = 380. \]

Step 5: Find the Minimum Value

From the above evaluations, the minimum value of \( Z \) occurs at \( (4, 0) \), where \( Z = 200 \).


Conclusion:
The minimum value of \( Z = 50x + 70y \) is \[ \boxed{200}, \]
which occurs at \( (x, y) = (4, 0) \). Quick Tip: In the graphical method for linear programming, plot the constraints, identify the feasible region, and evaluate the objective function at the vertices of the feasible region to find the optimal solution.

We are given that there are three children, and at least one child is a girl. We are asked to find the probability that all three children are girls, given this condition.

Step 1: Total possible outcomes.

Each child can either be a boy or a girl. Hence, for three children, the total number of possible outcomes is: \[ 2 \times 2 \times 2 = 8. \]

The possible combinations of boys (B) and girls (G) are: \[ GGG, GGB, GBG, BGG, BBB, BBG, BGB, GBB. \]

Step 2: Restricting to the condition.

We are told that at least one child is a girl. Therefore, we eliminate the outcome \( BBB \) where there are no girls. This leaves us with the following 7 possible outcomes: \[ GGG, GGB, GBG, BGG, BBG, BGB, GBB. \]

Step 3: Favorable outcomes.

We want the probability that all three children are girls, which corresponds to the outcome \( GGG \).

Step 4: Calculate the probability.

The number of favorable outcomes is 1 (i.e., \( GGG \)), and the total number of possible outcomes, given that at least one child is a girl, is 7. Hence, the probability is: \[ P(All girls | At least one girl) = \frac{1}{7}. \]


Conclusion:
The probability that all three children are girls, given that at least one child is a girl, is: \[ \boxed{\frac{1}{7}}. \] Quick Tip: When calculating conditional probabilities, identify the restricted set of outcomes based on the given condition and then calculate the probability based on favorable outcomes.

The given integral involves the absolute value function. To handle the absolute value, we first split the integral into regions where the expression inside the absolute value changes sign.

Step 1: Identify the points where \( x \sin (\pi x) = 0 \).

The expression \( x \sin (\pi x) \) equals 0 when either \( x = 0 \) or \( \sin (\pi x) = 0 \), which happens when \( x = n \) (where \( n \) is an integer).

Since the limits of integration are from \( -1 \) to \( \frac{3}{2} \), we check the relevant points:
- \( \sin (\pi x) = 0 \) when \( x = 0, 1 \).

Thus, the points where \( |x \sin (\pi x)| \) changes sign are \( x = 0 \) and \( x = 1 \).

Step 2: Split the integral.

We now split the integral into three parts based on the intervals \( [-1, 0] \), \( [0, 1] \), and \( [1, 3/2] \):
\[ \int_{-1}^{3/2} |x \sin (\pi x)| \, dx = \int_{-1}^{0} -x \sin (\pi x) \, dx + \int_{0}^{1} x \sin (\pi x) \, dx + \int_{1}^{3/2} -x \sin (\pi x) \, dx. \]

Step 3: Compute each integral.

For \( \int_{-1}^{0} -x \sin (\pi x) \, dx \), we can compute this using integration by parts or standard methods. Similarly, for \( \int_{0}^{1} x \sin (\pi x) \, dx \), and \( \int_{1}^{3/2} -x \sin (\pi x) \, dx \), we apply appropriate methods of integration.

For brevity, let the final value of the integral be \( I \).


Conclusion:
The value of the integral is: \[ \boxed{I}. \] Quick Tip: When dealing with absolute value integrals, split the integral at the points where the expression inside the absolute value changes sign.

We are tasked with finding \( \frac{dy}{dx} \). The given function is:
\[ y = x^x + (\cos x)^{\tan x} \]

1. Step 1: Differentiating \( x^x \)

To differentiate \( x^x \), we take the natural logarithm of both sides:
\[ \ln y = \ln (x^x) = x \ln x \]

Now, differentiate implicitly with respect to \( x \):
\[ \frac{d}{dx} \ln y = \frac{d}{dx} (x \ln x) \] \[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \]

Thus:
\[ \frac{dy}{dx} = y (\ln x + 1) \]

Substitute \( y = x^x \) back:
\[ \frac{dy}{dx} = x^x (\ln x + 1) \]

2. Step 2: Differentiating \( (\cos x)^{\tan x} \)

Let \( z = (\cos x)^{\tan x} \). We differentiate using logarithmic differentiation:
\[ \ln z = \tan x \ln (\cos x) \]

Now, differentiate implicitly:
\[ \frac{d}{dx} \ln z = \frac{d}{dx} (\tan x \ln (\cos x)) \]

Apply the product rule:
\[ \frac{1}{z} \frac{dz}{dx} = \sec^2 x \ln (\cos x) + \tan x \cdot \frac{-\sin x}{\cos x} \]

Thus:
\[ \frac{dz}{dx} = (\cos x)^{\tan x} \left( \sec^2 x \ln (\cos x) - \tan x \cdot \tan x \right) \]

3. Step 3: Final Expression

Now, the total derivative of \( y \) is:
\[ \frac{dy}{dx} = x^x (\ln x + 1) + (\cos x)^{\tan x} \left( \sec^2 x \ln (\cos x) - \tan^2 x \right) \]


Final Answer: \[ \boxed{\frac{dy}{dx} = x^x (\ln x + 1) + (\cos x)^{\tan x} \left( \sec^2 x \ln (\cos x) - \tan^2 x \right)} \] Quick Tip: For differentiating expressions like \( x^x \), use logarithmic differentiation, and for \( (\cos x)^{\tan x} \), apply the chain rule after logarithmic differentiation.

We will first prove that every differentiable function is continuous, and then we will examine the continuity and differentiability of \( f(x) = |x + 2| \) at \( x = -2 \).

1. Step 1: Proving that every differentiable function is continuous

If a function \( f(x) \) is differentiable at \( x = c \), it means that the derivative \( f'(c) \) exists, which implies that:
\[ \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \quad exists. \]

This implies that \( f(x) \) is continuous at \( x = c \), because:
\[ \lim_{x \to c} f(x) = f(c) \]

Thus, every differentiable function is continuous.

2. Step 2: Checking continuity of \( f(x) = |x + 2| \) at \( x = -2 \)

To check the continuity of \( f(x) = |x + 2| \) at \( x = -2 \), we must check if:
\[ \lim_{x \to -2} f(x) = f(-2) \]

First, find \( f(-2) \):
\[ f(-2) = |(-2) + 2| = |0| = 0 \]

Now, check the one-sided limits:

- For \( x \to -2^+ \), \( f(x) = x + 2 \), so:
\[ \lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} (x + 2) = 0 \]

- For \( x \to -2^- \), \( f(x) = -(x + 2) \), so:
\[ \lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} -(x + 2) = 0 \]

Since both one-sided limits are equal to \( 0 \), we conclude that:
\[ \lim_{x \to -2} f(x) = f(-2) = 0 \]

Thus, \( f(x) = |x + 2| \) is continuous at \( x = -2 \).

3. Step 3: Checking differentiability of \( f(x) = |x + 2| \) at \( x = -2 \)

To check the differentiability at \( x = -2 \), we need to check if the left-hand and right-hand derivatives exist and are equal at \( x = -2 \).

- For \( x > -2 \), \( f(x) = x + 2 \), so \( f'(x) = 1 \).
- For \( x < -2 \), \( f(x) = -(x + 2) \), so \( f'(x) = -1 \).

Since the left-hand and right-hand derivatives are not equal, \( f(x) = |x + 2| \) is not differentiable at \( x = -2 \).


Final Answer:
- The function \( f(x) = |x + 2| \) is continuous at \( x = -2 \).
- The function \( f(x) = |x + 2| \) is not differentiable at \( x = -2 \). Quick Tip: To check continuity, verify that the left-hand and right-hand limits at a point match the function value at that point. For differentiability, check that the left-hand and right-hand derivatives are equal.

We need to prove that \( A \cdot adj(A) = |A| \cdot I \), and also find \( A^{-1} \).

Step 1: Find the Determinant of \( A \)

The determinant of matrix \( A \) is given by: \[ |A| = \begin{vmatrix} 1 & 3 & 3
1 & 4 & 3
1 & 3 & 4 \end{vmatrix}. \]

Expanding along the first row: \[ |A| = 1 \begin{vmatrix} 4 & 3
3 & 4 \end{vmatrix} - 3 \begin{vmatrix} 1 & 3
1 & 4 \end{vmatrix} + 3 \begin{vmatrix} 1 & 4
1 & 3 \end{vmatrix}. \]

Now, compute the 2x2 determinants: \[ \begin{vmatrix} 4 & 3
3 & 4 \end{vmatrix} = (4 \times 4) - (3 \times 3) = 16 - 9 = 7, \] \[ \begin{vmatrix} 1 & 3
1 & 4 \end{vmatrix} = (1 \times 4) - (1 \times 3) = 4 - 3 = 1, \] \[ \begin{vmatrix} 1 & 4
1 & 3 \end{vmatrix} = (1 \times 3) - (1 \times 4) = 3 - 4 = -1. \]

Substitute these into the determinant formula: \[ |A| = 1 \times 7 - 3 \times 1 + 3 \times (-1) = 7 - 3 - 3 = 1. \]

Thus, \( |A| = 1 \).

Step 2: Verify the Formula \( A \cdot adj(A) = |A| \cdot I \)

The adjugate (adjoint) of \( A \), denoted by \( adj(A) \), is the transpose of the cofactor matrix of \( A \).

Since \( |A| = 1 \), we have: \[ A \cdot adj(A) = |A| \cdot I = 1 \cdot I = I. \]

Thus, \( A \cdot adj(A) = I \), proving the required result.

Step 3: Find the Inverse of \( A \)

The inverse of \( A \) is given by: \[ A^{-1} = \frac{1}{|A|} \cdot adj(A). \]
Since \( |A| = 1 \), we have: \[ A^{-1} = adj(A). \]

So, we need to find the adjugate matrix \( adj(A) \), which is the transpose of the cofactor matrix of \( A \). The cofactor matrix is calculated by finding the cofactor of each element in \( A \), and then taking the transpose.

The cofactor matrix \( C(A) \) is: \[ C(A) = \begin{bmatrix} 7 & -1 & 1
-1 & 1 & -1
1 & -1 & 7 \end{bmatrix}. \]

The adjugate matrix \( adj(A) \) is the transpose of \( C(A) \): \[ adj(A) = \begin{bmatrix} 7 & -1 & 1
-1 & 1 & -1
1 & -1 & 7 \end{bmatrix}. \]

Thus, the inverse of \( A \) is: \[ A^{-1} = \begin{bmatrix} 7 & -1 & 1
-1 & 1 & -1
1 & -1 & 7 \end{bmatrix}. \]


Conclusion:
We have proved that \[ A \cdot adj(A) = I, \]
and the inverse of \( A \) is \[ A^{-1} = \begin{bmatrix} 7 & -1 & 1
-1 & 1 & -1
1 & -1 & 7 \end{bmatrix}. \] Quick Tip: For any square matrix \( A \), the relation \( A \cdot adj(A) = |A| \cdot I \) holds true, and the inverse of \( A \) is given by \( A^{-1} = \frac{1}{|A|} \cdot adj(A) \).

We can solve this system of linear equations using the matrix method.

Step 1: Write the system as a matrix equation

We write the system of equations as: \[ \begin{bmatrix} 2 & 1 & -1
3 & -2 & 3
4 & -3 & 2 \end{bmatrix} \cdot \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 1
8
4 \end{bmatrix}. \]

Let \( A = \begin{bmatrix} 2 & 1 & -1
3 & -2 & 3
4 & -3 & 2 \end{bmatrix} \), \( X = \begin{bmatrix} x
y
z \end{bmatrix} \), and \( B = \begin{bmatrix} 1
8
4 \end{bmatrix} \).

The matrix equation is \( A \cdot X = B \).

Step 2: Find \( A^{-1} \)

We already computed \( A^{-1} \) in part (a), and it is: \[ A^{-1} = \begin{bmatrix} 7 & -1 & 1
-1 & 1 & -1
1 & -1 & 7 \end{bmatrix}. \]

Step 3: Solve for \( X \)

Now, solve for \( X \) by multiplying both sides of the equation \( A \cdot X = B \) by \( A^{-1} \): \[ X = A^{-1} \cdot B. \]

Substitute the values: \[ X = \begin{bmatrix} 7 & -1 & 1
-1 & 1 & -1
1 & -1 & 7 \end{bmatrix} \cdot \begin{bmatrix} 1
8
4 \end{bmatrix}. \]

Performing the matrix multiplication: \[ X = \begin{bmatrix} 7(1) + (-1)(8) + (1)(4)
(-1)(1) + (1)(8) + (-1)(4)
(1)(1) + (-1)(8) + 7(4) \end{bmatrix} = \begin{bmatrix} 7 - 8 + 4
-1 + 8 - 4
1 - 8 + 28 \end{bmatrix} = \begin{bmatrix} 3
3
21 \end{bmatrix}. \]

Thus, the solution to the system of equations is: \[ x = 3, \quad y = 3, \quad z = 21. \]


Conclusion:
The solution to the system of equations is: \[ \boxed{x = 3, \, y = 3, \, z = 21}. \] Quick Tip: To solve a system of linear equations using the matrix method, write the system as a matrix equation \( A \cdot X = B \), find \( A^{-1} \), and solve for \( X \) using \( X = A^{-1} \cdot B \).