Problem 11 - Intermediate Algebra

Question

For Problems \(11-30\), use Cramer's rule to find the solution set of each system. (Objective 2) $$ \left(\begin{array}{rr} 2 x-y+3 z= & -10 \\ x+2 y-3 z= & 2 \\ 3 x-2 y+5 z= & -16 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(x = -15\), \(y = 141\), \(z = -1\).

Step 1: Write Down the System of Equations

We have the system of equations given as: \[\begin{align*} 2x - y + 3z &= -10, \ x + 2y - 3z &= 2, \ 3x - 2y + 5z &= -16. \end{align*}\] These equations will be used to create the matrices needed for Cramer's Rule.

Step 2: Identify the Coefficient Matrix and the Determinant

The coefficient matrix \(A\) for the system is \[ A = \begin{pmatrix} 2 & -1 & 3 \ 1 & 2 & -3 \ 3 & -2 & 5 \end{pmatrix}. \] We need to calculate the determinant of this matrix, \(\det(A)\).

Step 3: Compute the Determinant of the Coefficient Matrix

Calculate \(\det(A)\) as follows: \[\det(A) = \begin{vmatrix} 2 & -1 & 3 \ 1 & 2 & -3 \ 3 & -2 & 5 \end{vmatrix} = 2(2 \cdot 5 - (-3)(-2)) + 1(1 \cdot 5 - (-3) \cdot 3) + 3(1 \cdot (-2) - 2 \cdot 3).\] Simplify to find: \(\det(A) = 2(10 - 6) + 1(5 + 9) + 3(-2 - 6) = 8 + 14 - 24 = -2.\)

Step 4: Create Matrices for Numerators in Cramer's Rule

For Cramer's Rule, replace each column of \(A\) with the constants on the right side to form matrices for finding \(x\), \(y\), and \(z\).

Step 5: Calculate Determinant for X (\(D_x\))

Replace the first column of \(A\) with the constants \([-10, 2, -16]\), giving us \(A_x\): \[ A_x = \begin{pmatrix} -10 & -1 & 3 \ 2 & 2 & -3 \ -16 & -2 & 5 \end{pmatrix}. \] Calculate \(\det(A_x)\): \[\det(A_x) = -10(10 - 6) + 1(2 \cdot 5 - (-3)(16)) + 3(2(-2) - 2(-3)). \] Which simplifies to \(-10 \times 4 + 1 \times 58 + 3 \times 4 = -40 + 58 + 12 = 30.\)

Step 6: Calculate Determinant for Y (\(D_y\))

Replace the second column of \(A\) with the constants \([-10, 2, -16]\), yielding \(A_y\): \[ A_y = \begin{pmatrix} 2 & -10 & 3 \ 1 & 2 & -3 \ 3 & -16 & 5 \end{pmatrix}. \] Calculate \(\det(A_y)\): \[\det(A_y) = 2(2 \cdot 5 - (-3)(-16)) - 10(1 \cdot 5 - (-3 \times 3)) + 3(1 \cdot (-16) - 2 \times 3).\] Simplifies to \(2(10 - 48) - 10(5 + 9) + 3(-16 - 6) = -76 - 140 - 66 = -282.\)

Step 7: Calculate Determinant for Z (\(D_z\))

Replace the third column of \(A\) with the constants \([-10, 2, -16]\), giving us \(A_z\): \[ A_z = \begin{pmatrix} 2 & -1 & -10 \ 1 & 2 & 2 \ 3 & -2 & -16 \end{pmatrix}. \] Calculate \(\det(A_z)\): \[\det(A_z) = 2(2(-16) - 2(-2)) + 1(1(-16) - 2 \times 3) + (-10)(1(-2) - 2 \times 3).\] This simplifies to \(2(-32 + 4) + 1(-16 - 6) + (-10)(-2 - 6) = -56 + -22 + 80 = 2.\)

Step 8: Apply Cramer's Rule

Using Cramer's Rule, find each variable: \[ x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)}, \quad z = \frac{\det(A_z)}{\det(A)}. \] Substituting in we get: \[ x = \frac{30}{-2}, \quad y = \frac{-282}{-2}, \quad z = \frac{2}{-2}. \] Calculate to find \(x = -15\), \(y = 141\), \(z = -1\).

Key Concepts

Determinant Calculation

In linear algebra, the determinant is a crucial value that helps us determine many properties of a matrix. When you're dealing with a system of equations, the determinant signifies whether a unique solution exists. Let's consider a 3x3 matrix, which is common when using Cramer's Rule for systems of three equations. We denote the matrix as \(A\) and calculate its determinant, \(\det(A)\), using the formula for a 3x3 matrix:\[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]where \(a, b, c,\hdots, i\) are the elements of the matrix. This determinant value informs us about the system:

• If \(\det(A) eq 0\), the system has a unique solution.
• If \(\det(A) = 0\), the system may have no solution or infinitely many solutions.

Calculating the determinant can initially seem complex, but with practice, it becomes a straightforward process. Always remember to follow the formula carefully to avoid mistakes.

System of Equations

A system of equations involves multiple equations that share common variables. These equations can describe anything from the intersection of lines to multi-variable problems in physics and engineering.In mathematics, solving a system of equations means finding values of the variables that satisfy all equations simultaneously. For instance, if you have equations in terms of \(x\), \(y\), and \(z\), a solution would be a trio \((x, y, z)\) that holds true for all the given equations.There are several ways to solve such systems:

• Substitution and Elimination: Classical methods manipulating equations algebraically to isolate each variable.
• Graphical Solutions: Where lines or curves are drawn, and the intersection points give the solutions.
• Matrix Methods: Involves matrices and operations like finding determinants to solve the system, which includes methods like Cramer's Rule.

When we apply Cramer's Rule, we're using the determinant of the coefficient matrix to find each variable, provided the determinant is not zero. Understanding the structure and type of system you're working with is key to choosing the best method for finding a solution.

Linear Algebra

Linear algebra is a branch of mathematics focused on vectors, matrices, and linear transformations. It's foundational for fields such as computer science, physics, engineering, and more. Here are key components of linear algebra:

• Vectors: Objects representing magnitude and direction.
• Matrices: Arrays of numbers arranged in rows and columns, used to express systems of linear equations and transformations.
• Determinants: Scalar values that can inform about the properties of matrices, like whether they are invertible or whether a system of equations has a unique solution (as used in Cramer's Rule).
• Eigenvalues and Eigenvectors: Special sets of scalars and vectors associated with matrices that are crucial in various applications, such as stability analysis and facial recognition algorithms.

What makes linear algebra so powerful is its ability to describe and solve complex multi-variable problems efficiently. By representing systems, transformations, and spaces in a structured way, it opens the door to solutions that aren't immediately obvious when using basic arithmetic or algebraic methods alone. Understanding these concepts is essential for anyone delving into scientific or technical fields.