Problem 65 - A Graphical Approach to Precalculus with Limits

Question

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to complete the solution. $$\begin{aligned}&4 x-y+3 z=-3\\\&3 x+y+z=0\\\&2 x-y+4 z=0\end{aligned}$$

Step-by-Step Solution

Answer

\(x = -1, y = \frac{12}{5}, z = 1\)

Step 1: Write the system in matrix form

The given system of equations can be expressed in matrix form as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the column matrix of variables \((x, y, z)\), and \(B\) is the column matrix of constants. \[ A = \begin{bmatrix} 4 & -1 & 3 \ 3 & 1 & 1 \ 2 & -1 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad B = \begin{bmatrix} -3 \ 0 \ 0 \end{bmatrix} \]

Step 2: Calculate the determinant of A (D)

The determinant of matrix \(A\) is calculated using the rule of Sarrus or the expansion by minors for 3x3 matrices:\[D = \begin{vmatrix} 4 & -1 & 3 \ 3 & 1 & 1 \ 2 & -1 & 4 \end{vmatrix} = 4(1 \cdot 4 - (-1) \cdot 1) - (-1)(3 \cdot 4 - 1 \cdot 2) + 3(3 \cdot (-1) - 1 \cdot 2)\]Calculate each component to find \(D\).

Step 3: Compute the determinant D

Calculate the determinant:\[D = 4(4 + 1) + 1(12 - 2) + 3(-3 - 2) = 4 \times 5 + 10 - 15 = 20 + 10 - 15 = 15\]Here, \(D = 15\), which is non-zero, so we can proceed with Cramer's Rule.

Step 4: Calculate determinant D_x

To find \(D_x\), replace the first column of \(A\) with \(B\) and find the determinant:\[D_x = \begin{vmatrix} -3 & -1 & 3 \ 0 & 1 & 1 \ 0 & -1 & 4 \end{vmatrix} = -3(1 \times 4 - (-1) \times 1) - (-1)(0 - 0) + 3(0 - 0) = -3 \times 5 = -15\]

Step 5: Calculate determinant D_y

To find \(D_y\), replace the second column of \(A\) with \(B\) and find the determinant:\[D_y = \begin{vmatrix} 4 & -3 & 3 \ 3 & 0 & 1 \ 2 & 0 & 4 \end{vmatrix} = 4(0 \times 4 - 0 \times 1) - (-3)(3 \times 4 - 1 \times 2) + 3(3 \times 0 - 0 \times 2)\]Simplify to find \(D_y = 0 + 36 + 0 = 36\).

Step 6: Calculate determinant D_z

To find \(D_z\), replace the third column of \(A\) with \(B\) and find the determinant:\[D_z = \begin{vmatrix} 4 & -1 & -3 \ 3 & 1 & 0 \ 2 & -1 & 0 \end{vmatrix} = 4(1 \cdot 0 - 0 \cdot (-1)) - (-1)(3 \cdot 0 - 0 \cdot 2) - 3(3 \times (-1) - 1 \cdot 2)\]Simplify to find \(D_z = 0 + 0 - 3(-5) = 15\).

Step 7: Solve for x, y, and z using Cramer's Rule

Using Cramer's rule, solve for \(x\), \(y\), and \(z\):\[x = \frac{D_x}{D} = \frac{-15}{15} = -1, \quad y = \frac{D_y}{D} = \frac{36}{15} = \frac{12}{5}, \quad z = \frac{D_z}{D} = \frac{15}{15} = 1\]

Key Concepts

Determinant Calculation

The determinant of a matrix is a special number that can be calculated from its elements. It is crucial in solving systems of equations using matrix methods like Cramer's Rule. Determinants help us understand if a system of equations has a unique solution or not.
The calculation of determinants follows specific rules depending on the size of the matrix:

• For a 2x2 matrix, the determinant is found by cross-multiplying and subtracting the products.
• For a 3x3 matrix, as in our exercise, you can use the rule of Sarrus or the expansion by minors.

For our matrix \(A\), the 3x3 determinant is calculated as follows: We arrange all elements, calculate the products along the diagonals, and apply the signs according to the rule. Calculating all components, we found that \(D=15\). Since it is non-zero, the system has a unique solution.

Systems of Equations

A system of equations consists of multiple equations that are solved together because they share common variables. The solutions to the system are the values of the variables that satisfy all equations simultaneously.
In our exercise, we have three equations involving three variables \(x\), \(y\), and \(z\). The system is represented as follows:

• First equation: \(4x - y + 3z = -3\)
• Second equation: \(3x + y + z = 0\)
• Third equation: \(2x - y + 4z = 0\)

The goal here is to find a common set of values \((x, y, z)\), which will satisfy all three equations simultaneously. With the help of Cramer's Rule, which uses determinant calculations, we can efficiently solve such systems when they involve multiple equations and unknowns.

Matrix Algebra

Matrix algebra is a collection of mathematical operations involving matrices. Understanding matrix algebra is key to solving systems of equations using matrix methods like Cramer's Rule.
Here’s a brief look at some of the core concepts:

• A matrix is a rectangular arrangement of numbers in rows and columns.
• The coefficients of the system of equations are represented in matrix form, known as the coefficient matrix \(A\).
• Matrix multiplication is a fundamental operation that allows us to express the system of equations as \(AX = B\), where \(X\) is the column matrix of variables, and \(B\) is the constants matrix.

In our exercise, multiplying the coefficient matrix by the variables' matrix gives the constants matrix, making it easier to manipulate and solve the system using linear algebra techniques.

Linear Algebra Methods

Linear algebra provides various powerful methods for solving systems of linear equations. Among these, Cramer's Rule is a straightforward approach when the coefficient matrix has a non-zero determinant.
Cramer's Rule relies on these steps:

• Compute the determinant \(D\) of the coefficient matrix.
• For each variable, replace the respective column in the coefficient matrix with the constants matrix, compute the determinant of this new matrix, and solve for the variable using \(\frac{D}{D}\).

In our given system, we calculated determinants \(D_x\), \(D_y\), and \(D_z\) for solving \(x\), \(y\), and \(z\) respectively.
If \(D\) had been zero, we would need to apply different techniques such as Gaussian elimination or matrix inverses, which are other linear algebra methods that effectively handle such scenarios.