Problem 24 - Algebra for College Students

Question

Use Cramer's rule to find the solution set for each system. If the equations are dependent, simply indicate that there are infinitely many solutions. \(\left(\begin{array}{c}2 x-y+3 z=-5 \\ 3 x+4 y-2 z=-25 \\\ -x+z=6\end{array}\right)\)

Step-by-Step Solution

Answer

The solution set is \( x = \frac{53}{19}, y = -\frac{42}{19}, z = \frac{25}{19} \).

Step 1: Identify Matrix Components

First, we'll write the given system of equations in the standard form. The system is given as \(2x - y + 3z = -5 \3x + 4y - 2z = -25 \-x + z = 6\).From this, we identify the coefficient matrix \(A\), the variable matrix \(X\), and the constant matrix \(B\).The coefficient matrix \(A\) is:\[A = \begin{bmatrix}2 & -1 & 3 \3 & 4 & -2 \-1 & 0 & 1\end{bmatrix}\]The variable matrix \(X\) is:\[X = \begin{bmatrix}x \y \z\end{bmatrix}\]The constant matrix \(B\) is:\[B = \begin{bmatrix}-5 \-25 \6\end{bmatrix}\]

Step 2: Calculate Determinant of Coefficient Matrix

To apply Cramer's rule, we need to calculate the determinant of the coefficient matrix \(A\), which is denoted as \(\det(A)\).\[\det(A) = \begin{vmatrix}2 & -1 & 3 \3 & 4 & -2 \-1 & 0 & 1\end{vmatrix}\]Summing the products of the diagonals, we have:\[\det(A) = 2 \begin{vmatrix} 4 & -2 \ 0 & 1 \end{vmatrix} + 1 \begin{vmatrix} 3 & -2 \ -1 & 1 \end{vmatrix} + 3 \begin{vmatrix} 3 & 4 \ -1 & 0 \end{vmatrix}\]This simplifies to:\[\det(A) = 2(4 \cdot 1 - (-2) \cdot 0) - (3 \cdot 1 - (-2) \cdot -1) + 3(3 \cdot 0 - 4 \cdot (-1))\]\[= 2(4) - (3 - 2) + 3(0 + 4) = 8 - 1 + 12 = 19 \]Therefore, \(\det(A) = 19\). Since the determinant is non-zero, the system has a unique solution.

Step 3: Solve for x Using Cramer's Rule

To find \(x\) we replace the first column of \(A\) with \(B\) and call the new matrix \(A_x\). Then, calculate \(\det(A_x)\).\[A_x = \begin{bmatrix}-5 & -1 & 3 \-25 & 4 & -2 \6 & 0 & 1\end{bmatrix}\]Calculate the determinant:\[\det(A_x) = -5 \begin{vmatrix} 4 & -2 \ 0 & 1 \end{vmatrix} + 1 \begin{vmatrix} -25 & -2 \ 6 & 1 \end{vmatrix} + 3 \begin{vmatrix} -25 & 4 \ 6 & 0 \end{vmatrix}\]\[= -5(4 \cdot 1 - (-2) \cdot 0) - 1(-25 \cdot 1 - (-2) \cdot 6) + 3(-25 \cdot 0 - 4 \cdot 6)\]\[= -20 + 25 + 48 = 53 \]By Cramer's rule, \(x = \frac{\det(A_x)}{\det(A)}\)\( = \frac{53}{19} = \frac{53}{19}\).

Step 4: Solve for y Using Cramer's Rule

Next, replace the second column of \(A\) with \(B\) to find \(y\). Call this new matrix \(A_y\) and calculate \(\det(A_y)\).\[A_y = \begin{bmatrix}2 & -5 & 3 \3 & -25 & -2 \-1 & 6 & 1\end{bmatrix}\]Calculate the determinant:\[\det(A_y) = 2 \begin{vmatrix} -25 & -2 \ 6 & 1 \end{vmatrix} + 5 \begin{vmatrix} 3 & -2 \ -1 & 1 \end{vmatrix} + 3 \begin{vmatrix} 3 & -25 \ -1 & 6 \end{vmatrix}\]\[= 2(-25 \cdot 1 - (-2) \cdot 6) + 5(3 \cdot 1 - (-2) \cdot -1) + 3(3 \cdot 6 - (-25) \cdot -1)\]\[= 2(-25 + 12) + 5(3 - 2) + 3(18 - 25)\]\[= 2(-13) + 5 + 3(-7) = -26 + 5 - 21 = -42 \]By Cramer's rule, \(y = \frac{\det(A_y)}{\det(A)}\)\( = \frac{-42}{19} = -\frac{42}{19}\).

Step 5: Solve for z Using Cramer's Rule

Finally, replace the third column of \(A\) with \(B\) to find \(z\). Call this new matrix \(A_z\) and calculate \(\det(A_z)\).\[A_z = \begin{bmatrix}2 & -1 & -5 \3 & 4 & -25 \-1 & 0 & 6\end{bmatrix}\]Calculate the determinant:\[\det(A_z) = 2 \begin{vmatrix} 4 & -25 \ 0 & 6 \end{vmatrix} + 1 \begin{vmatrix} 3 & -25 \ -1 & 6 \end{vmatrix} + 5 \begin{vmatrix} 3 & 4 \ -1 & 0 \end{vmatrix}\]\[= 2(4 \cdot 6 - (-25) \cdot 0) - 1(3 \cdot 6 - (-25) \cdot -1) + 5(3 \cdot 0 - 4 \cdot -1)\]\[= 2(24) - (18 + 25) + 5(4)\]\[= 48 - 43 + 20 = 25 \]By Cramer's rule, \(z = \frac{\det(A_z)}{\det(A)}\)\( = \frac{25}{19} = \frac{25}{19}\).

Step 6: Conclusion: Solution Set

By using Cramer's rule, we have:\(x = \frac{53}{19}\), \(y = -\frac{42}{19}\), \(z = \frac{25}{19}\). Thus, the solution set is \(\left(\frac{53}{19}, -\frac{42}{19}, \frac{25}{19}\right)\).

Key Concepts

Determinants

Determinants are a key concept in matrix algebra. A determinant is a scalar value that can be computed from the elements of a square matrix. It provides important information about the matrix, such as whether it is invertible. In the context of solving linear equations using Cramer's Rule, determinants play a crucial role in finding the unique solution. If the determinant of the coefficient matrix is zero, it implies that the matrix is singular and the system of equations does not have a unique solution.
For a 3x3 matrix, the determinant can be calculated using the rule of Sarrus or the general formula for determinants, which involves expanding along rows or columns and multiplying the corresponding minors by their cofactors.
In the provided solution, we calculated the determinant of the coefficient matrix (denoted as \( \det(A) \)) to determine if the system of linear equations could be solved using Cramer's Rule. By confirming that the determinant is non-zero (\( \det(A) = 19 \)), we ensured the system has a unique solution.

Matrix Algebra

Matrix algebra is a fundamental part of linear algebra and involves operations with matrices, such as addition, multiplication, and finding inverses and determinants. Matrices are rectangular arrays of numbers that can represent systems of linear equations. They are instrumental in efficiently organizing and solving large sets of linear equations.
Matrix operations follow specific rules. For example, matrix multiplication is not commutative, which means that \(AB eq BA\) in most cases. In Cramer's Rule, we primarily focus on finding determinants and creating modified matrices (replacing columns with the constant matrix) to find each variable in the system.

• Coefficient Matrix (\(A\)): This represents the coefficients from the system of equations.
• Variable Matrix (\(X\)): Contains the variables \(x, y, \) and \(z\) which are the unknowns we need to find.
• Constant Matrix (\(B\)): Represents the values on the right-hand side of the equations.

In the original step-by-step solution, we saw how these matrices were defined and manipulated to solve the system using Cramer's Rule.

Linear Equations

Linear equations are algebraic expressions where each term is either a constant or the product of a constant with a single variable. Solving a system of linear equations involves finding all possible values of the variables that satisfy all equations simultaneously.
In Cramer's Rule, linear equations are expressed using matrices to find precise solutions. The rule requires that the number of equations equals the number of unknowns, which is often represented by a square matrix.
Solving the system involves checking if the determinant of the coefficient matrix is non-zero. A non-zero determinant indicates that the system has a unique solution. If the determinant is zero, the system is either dependent or has no solution, indicating infinitely many solutions or an inconsistent system, respectively.
In our case, the given system of linear equations had a non-zero determinant, allowing us to use Cramer's Rule to find the unique values of \(x, y,\) and \(z\). By substituting the respective modified determinants, each variable could be calculated uniquely, demonstrating the power and simplicity of matrix methods in solving linear systems.