Problem 14 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The solution is \((-\frac{23}{3}, \frac{19}{3}, -\frac{31}{3})\).

Step 1: Write the System in Matrix Form

First, represent the given system of equations in the form of matrices. The system is: \[ x - 2y + z = 3 \] \[ 3x + 2y + z = -3 \] \[ 2x - 3y - 3z = -5 \] This can be written as \( AX = B \), where:\[A = \begin{bmatrix} 1 & -2 & 1 \ 3 & 2 & 1 \ 2 & -3 & -3 \end{bmatrix}, \ X = \begin{bmatrix} x \ y \ z \end{bmatrix}, \ B = \begin{bmatrix} 3 \ -3 \ -5 \end{bmatrix}\]

Step 2: Calculate the Determinant of A

Calculate the determinant of matrix \( A \). Use the determinant formula for a 3x3 matrix:\[ \text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg) \]Applying it to matrix \( A \):\[\text{det}(A) = 1(2(-3) - 1(-3)) - (-2)(3(-3) - 1(2)) + 1(3(-3) - 2(2))\]\[\text{det}(A) = 1(-6 + 3) + 2(9 + 2) + 1(-9 - 4) = -3 + 22 - 13 = 6\]

Step 3: Calculate Determinant of A with Replaced Columns

For \( x \), replace the first column of \( A \) with \( B \):\[ A_x = \begin{bmatrix} 3 & -2 & 1 \ -3 & 2 & 1 \ -5 & -3 & -3 \end{bmatrix} \]Calculate \( \text{det}(A_x) \) using the determinant formula:\[ \text{det}(A_x) = 3(2(-3) - 1(-3)) + 2(-3(-3) - 1(-5)) + 1(-3(-5) - -3(2)) \]\[= 3(-6 + 3) + (-2)(9 + 5) + (-1)(15 - 6) = -9 - 28 - 9 = -46\]For \( y \), replace the second column:\[ A_y = \begin{bmatrix} 1 & 3 & 1 \ 3 & -3 & 1 \ 2 & -5 & -3 \end{bmatrix}\]Calculate \( \text{det}(A_y) \):\[\text{det}(A_y) = 1(-3(-3) - 1(-5)) - 3(3(-3) - 1(2)) + 1(3(-5) - (-3)(2))\]\[= 1(9 + 5) - 3(-9 - 2) + 1(-15 + 6) = 14 + 33 - 9 = 38\]For \( z \), replace the third column:\[ A_z = \begin{bmatrix} 1 & -2 & 3 \ 3 & 2 & -3 \ 2 & -3 & -5 \end{bmatrix}\]Calculate \( \text{det}(A_z) \):\[\text{det}(A_z) = 1(2(-5) - (-3)(-3)) + (-2)(3(-5) - (-3)(2)) + 3(3(-3) - 2(2)) \]\[= 1(-10 - 9) + 2(15 + 6) + 3(-9 - 4) = -19 - 42 - 39 = -62\]

Step 4: Solve for x, y, z using Cramer's Rule

Using Cramer's Rule, solve for \( x \), \( y \), and \( z \) as follows:For \( x \):\[ x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-46}{6} = -\frac{23}{3} \]For \( y \):\[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{38}{6} = \frac{19}{3} \]For \( z \):\[ z = \frac{\text{det}(A_z)}{\text{det}(A)} = \frac{-62}{6} = -\frac{31}{3} \]

Step 5: Write the Solution

The solution set for the system is:\[(x, y, z) = \left(-\frac{23}{3}, \frac{19}{3}, -\frac{31}{3}\right)\]

Key Concepts

Determinants and Their Use

A determinant is a special number that you can calculate from a square matrix. It plays a crucial role in various applications across mathematics, particularly in solving systems of equations using Cramer's Rule and in finding inverses of matrices. To calculate the determinant of a 3x3 matrix, you use the formula:

• det(A) = a(ei − fh) − b(di − fg) + c(dh − eg)

This formula might look complex, but it's simply a way to combine the elements of the matrix to produce a single number.
The determinant has several important properties:

• If the determinant of a matrix is zero, the matrix has no inverse, and the system of equations it represents has no unique solution.
• If the determinant is non-zero, the matrix is invertible, indicating a unique solution for the system of equations.

In the exercise above, we calculated the determinant of the given square matrix to ensure that a solution could indeed be found using Cramer's Rule.

System of Equations

A system of equations consists of two or more equations with the same set of unknowns. The goal is to find values for these unknowns that satisfy all the equations simultaneously. In mathematical terms:

• Each equation represents a linear equation in three-dimensional space.
• The solution to the system is the point where all the lines intersect, giving a common solution that satisfies each equation.

In our specific system:

1. The equations are written as:
   • \(x - 2y + z = 3\)
   • \(3x + 2y + z = -3\)
   • \(2x - 3y - 3z = -5\)
2. These represented a complex scenario where the lines intersect at a unique point, only solvable with methods like Cramer's Rule due to non-linear overlapping geometry.

Understanding and solving systems of equations is foundational in mathematics as it can apply to physics, engineering, economics, and numerous other fields where relationships between variables need to be established.

Matrix Algebra and Cramer's Rule

Matrix algebra is an essential tool in solving systems of equations through simplification, transformation, and calculation. Cramer's Rule is one method specifically designed for solving systems of linear equations using matrices:

• **Matrix Representation:** Convert the system into matrix form, \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constants matrix.
• **Calculate Determinants:** Compute the determinant of \(A\) to check the system is solvable (it must be non-zero for a unique solution).
• **Replace Columns:** Form new matrices by replacing each column of \(A\) with \(B\) one at a time to solve for each variable's determinant.

Finally, divide each new matrix's determinant by that of the original \(A\) to find the values of \(x\), \(y\), and \(z\):

• \(x = \frac{\text{det}(A_x)}{\text{det}(A)}\)
• \(y = \frac{\text{det}(A_y)}{\text{det}(A)}\)
• \(z = \frac{\text{det}(A_z)}{\text{det}(A)}\)

Matrix algebra not only simplifies these calculations, but it also provides a framework for handling larger, more complicated systems that would be difficult to solve otherwise.