Determinant Evaluation, Circle Equations, And Finding Function Values

Determinants are a fundamental concept in linear algebra, playing a crucial role in various applications, including solving systems of linear equations, finding eigenvalues, and determining the invertibility of matrices. Evaluating determinants involves applying specific rules and techniques, depending on the size of the matrix. For a 3x3 matrix, we can utilize the expansion by minors method, which systematically breaks down the determinant calculation into smaller, manageable steps.

The given determinant is:

$$\begin{vmatrix}3 & -1 & 1 \\-3 & 2 & 0 \\1 & 4 & -2\end{vmatrix}$$

To evaluate this determinant, we can expand along the first row. The formula for expanding a 3x3 determinant along the first row is:

$$|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$$

where $a_{ij}$ are the elements of the matrix and $C_{ij}$ are the cofactors, calculated as $C_{ij} = (-1)^{i+j}M_{ij}$, with $M_{ij}$ being the minor (the determinant of the submatrix obtained by removing the i-th row and j-th column).

Let's apply this to our matrix:

1. Element $a_{11} = 3$: The minor $M_{11}$ is the determinant of the submatrix formed by removing the first row and first column:

   $$\begin{vmatrix}2 & 0 \\4 & -2\end{vmatrix} = (2 imes -2) - (0 imes 4) = -4$$

   The cofactor $C_{11} = (-1)^{1+1}M_{11} = 1 imes -4 = -4$.

2. Element $a_{12} = -1$: The minor $M_{12}$ is the determinant of the submatrix formed by removing the first row and second column:

   $$\begin{vmatrix}-3 & 0 \\1 & -2\end{vmatrix} = (-3 imes -2) - (0 imes 1) = 6$$

   The cofactor $C_{12} = (-1)^{1+2}M_{12} = -1 imes 6 = -6$.

3. Element $a_{13} = 1$: The minor $M_{13}$ is the determinant of the submatrix formed by removing the first row and third column:

   $$\begin{vmatrix}-3 & 2 \\1 & 4\end{vmatrix} = (-3 imes 4) - (2 imes 1) = -12 - 2 = -14$$

   The cofactor $C_{13} = (-1)^{1+3}M_{13} = 1 imes -14 = -14$.

Now, we can calculate the determinant:

$$|A| = (3 imes -4) + (-1 imes -6) + (1 imes -14) = -12 + 6 - 14 = -20$$

Therefore, the determinant of the given matrix is -20. Understanding the process of evaluating determinants is crucial for various mathematical and engineering applications. This methodical approach ensures accuracy and provides a clear understanding of the determinant's significance. The determinant's value can indicate properties of the matrix, such as its invertibility and the volume scaling factor of a linear transformation. Proficiency in calculating determinants is essential for advanced mathematical studies and practical problem-solving.

Defining Circles and Determining Their Equations

Circles are fundamental geometric shapes defined by a set of points equidistant from a central point. Defining a circle involves specifying its center and radius, which are essential parameters for describing its position and size. The equation of a circle provides a concise algebraic representation of this geometric concept, enabling us to perform various analytical operations and solve related problems. Understanding the standard form equation of a circle is crucial for both theoretical and practical applications.

The general equation of a circle with center (h, k) and radius r in the Cartesian plane is given by:

$$(x - h)^2 + (y - k)^2 = r^2$$

This equation stems from the Pythagorean theorem, where the distance between any point (x, y) on the circle and the center (h, k) is equal to the radius r. Determining the equation of a circle involves substituting the known values of the center and radius into this standard form. Given the center (-2, 1) and radius 3, we can directly apply these values to construct the circle's equation.

Substituting h = -2, k = 1, and r = 3 into the general equation, we get:

$$(x - (-2))^2 + (y - 1)^2 = 3^2$$

Simplifying this, we have:

$$(x + 2)^2 + (y - 1)^2 = 9$$

Expanding the equation, we obtain:

$$(x^2 + 4x + 4) + (y^2 - 2y + 1) = 9$$

Combining like terms, the equation becomes:

$$x^2 + y^2 + 4x - 2y + 5 = 9$$

Finally, subtracting 9 from both sides, we get the standard form of the circle's equation:

$$x^2 + y^2 + 4x - 2y - 4 = 0$$

This equation represents the circle with center (-2, 1) and radius 3 in the Cartesian plane. Defining a circle through its equation allows for precise geometric representation and facilitates the solution of problems involving circles, such as finding points on the circle, determining intersections with lines, and calculating tangents. The ability to derive and manipulate circle equations is essential in fields like geometry, calculus, and engineering, where circles play a significant role in various applications.

Finding the Value of k for a Function

Functions are a fundamental concept in mathematics, representing a relationship between input values (x) and output values (f(x)). Finding the value of k in a function often involves solving an equation where the function's value is given for a specific input. This process typically requires substituting the given input into the function and then solving for the unknown parameter k. Understanding how to manipulate function equations and solve for unknown variables is a crucial skill in mathematics.

Let's consider the function f(x) = (x^2 + 3x + k) / (x - 2). To find the value of k, we need additional information, such as the function's value at a particular point. Suppose we are given that f(1) = 2. This means that when x = 1, the function's output is 2. We can use this information to solve for k.

First, substitute x = 1 into the function:

$$f(1) = \frac{(1^2 + 3(1) + k)}{(1 - 2)}$$

We know that f(1) = 2, so we can set up the equation:

$$2 = \frac{(1 + 3 + k)}{(-1)}$$

Simplify the equation:

$$2 = \frac{(4 + k)}{-1}$$

Multiply both sides by -1:

$$-2 = 4 + k$$

Now, solve for k:

$$k = -2 - 4$$

$$k = -6$$

Therefore, the value of k is -6. This process of finding the value of k demonstrates how function equations can be used to determine unknown parameters based on given conditions. By substituting known values and solving the resulting equation, we can gain deeper insights into the behavior and properties of the function. This skill is essential in various mathematical contexts, including calculus, algebra, and differential equations, where functions are used to model real-world phenomena and solve complex problems. The ability to find the value of k is particularly useful in applications where functions need to be calibrated or adjusted to meet specific requirements or conditions.