Problem 44 - Precalculus

Question

Apply Cramer's rule to solve each system of equations, if possible. $$\begin{array}{rr} -x+y+z= & -4 \\ x+y-z= & 0 \\ x+y+z= & 2 \end{array}$$

Step-by-Step Solution

Answer

The solution is \( x = 3 \), \( y = -2 \), \( z = 1 \).

Step 1: Write the System in Matrix Form

Given the system of equations: \( -x + y + z = -4 \), \( x + y - z = 0 \), and \( x + y + z = 2 \), we can express this system in the matrix form \( AX = B \) where:\[A = \begin{bmatrix} -1 & 1 & 1 \ 1 & 1 & -1 \ 1 & 1 & 1 \end{bmatrix}, \, X = \begin{bmatrix} x \ y \ z \end{bmatrix}, \, B = \begin{bmatrix} -4 \ 0 \ 2 \end{bmatrix}\]

Step 2: Find the Determinant of Matrix A

The determinant of matrix \( A \) is calculated as follows:\[\text{det}(A) = \begin{vmatrix} -1 & 1 & 1 \ 1 & 1 & -1 \ 1 & 1 & 1 \end{vmatrix} = -1 \begin{vmatrix} 1 & -1 \ 1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \ 1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \ 1 & 1 \end{vmatrix}\]Calculate the minors and simplify:\[\text{det}(A) = -1(1(1) - (-1)(1)) - 1(1(1) - (-1)(1)) + 1(1(1) - 1(1))\]\[= -1(2) - 1(2) + 1(0)\]\[= -2 - 2 + 0 = -4\]

Step 3: Check if the Determinant is Zero

We determine that \( \text{det}(A) = -4 \). Since the determinant is not zero, Cramer's rule can be applied as the matrix \( A \) is invertible.

Step 4: Calculate the Determinant of Matrices A_x, A_y, A_z

To find the solutions using Cramer's rule, compute the determinants of \( A_x, A_y, A_z \). These matrices are formed by replacing the respective columns of \( A \) with \( B \).- For \( A_x \):\[A_x = \begin{bmatrix} -4 & 1 & 1 \ 0 & 1 & -1 \ 2 & 1 & 1 \end{bmatrix}, \]\[\text{det}(A_x) = \begin{vmatrix} -4 & 1 & 1 \ 0 & 1 & -1 \ 2 & 1 & 1 \end{vmatrix}\]\[= -4 \begin{vmatrix} 1 & -1 \ 1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 0 & -1 \ 2 & 1 \end{vmatrix} + 1 \begin{vmatrix} 0 & 1 \ 2 & 1 \end{vmatrix}\]Calculate:\[= -4(1 + 1) - 1(0 - (-2)) + 1(0 - 2)\]\[= -8 - 2 - 2 = -12\]- For \( A_y \):\[A_y = \begin{bmatrix} -1 & -4 & 1 \ 1 & 0 & -1 \ 1 & 2 & 1 \end{bmatrix}, \]\[\text{det}(A_y) = \begin{vmatrix} -1 & -4 & 1 \ 1 & 0 & -1 \ 1 & 2 & 1 \end{vmatrix}\]\[= -1 \begin{vmatrix} 0 & -1 \ 2 & 1 \end{vmatrix} + 4 \begin{vmatrix} 1 & -1 \ 1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 0 \ 1 & 2 \end{vmatrix}\]Calculate:\[= -1(0 - (-2)) + 4(1 + 1) + 1(2 - 0)\]\[= -2 + 8 + 2 = 8\]- For \( A_z \):\[A_z = \begin{bmatrix} -1 & 1 & -4 \ 1 & 1 & 0 \ 1 & 1 & 2 \end{bmatrix}, \]\[\text{det}(A_z) = \begin{vmatrix} -1 & 1 & -4 \ 1 & 1 & 0 \ 1 & 1 & 2 \end{vmatrix}\]\[= -1 \begin{vmatrix} 1 & 0 \ 1 & 2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 0 \ 1 & 2 \end{vmatrix} + 4 \begin{vmatrix} 1 & 1 \ 1 & 1 \end{vmatrix}\]Calculate:\[= -1(2 - 0) - 1(2 - 0) + 4(1 - 1)\]\[= -2 - 2 + 0 = -4\]

Step 5: Compute the Solutions for x, y, z

Using Cramer's Rule, solve for each variable: \( x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-12}{-4} = 3 \) \( y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{8}{-4} = -2 \) \( z = \frac{\text{det}(A_z)}{\text{det}(A)} = \frac{-4}{-4} = 1 \)

Key Concepts

Systems of Equations

A system of equations is a set of multiple equations working together to find common solutions. In mathematics, these equations often contain several unknown variables that need to be determined.
To solve a system of equations, we are tasked with finding values for the variables that satisfy all equations simultaneously.

There are various ways to solve a system of equations, including:

• Substitution Method: Solve one equation for one variable and substitute it into the other equations.
• Elimination Method: Add or subtract equations to eliminate one variable, making it easier to solve.
• Cramer's Rule: A technique using determinants to find the solution of a system of linear equations.

Cramer's Rule is particularly used when the number of equations equals the number of variables. It provides a straightforward approach but requires that the determinant of the coefficients matrix is non-zero, ensuring that the system has a unique solution.

Determinants

Determinants are special numbers that can be calculated from a square matrix. They are fundamental in linear algebra and useful in solving systems of linear equations. When dealing with a 3x3 matrix, as in the exercise provided, the determinant formula involves minors, which are smaller matrices obtained by removing one row and one column.
Calculating a determinant of a 3x3 matrix involves:

• Taking the sum of the products of the diagonal elements from one vertex of the matrix.
• Subtracting the sum of the products of the diagonal elements from the opposite vertex.

For a matrix \(A\) with elements \(a, b, c, d, e, f, g, h, i\), the determinant is given by:\[\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\]
This determinant helps tell if a system of equations has a unique solution, no solution, or infinitely many solutions. If the determinant is zero, the system may not have a unique solution.

Matrix Algebra

Matrix algebra involves performing operations with matrices, which are organized arrays of numbers. This mathematical tool is especially helpful in various applications including physics, computer graphics, and statistics.

Key concepts in matrix algebra include:

• Matrix Addition: Adding corresponding elements of two matrices of the same dimension.
• Matrix Multiplication: Combining rows of the first matrix with columns of the second, useful for transformations like translating systems of equations to matrix form.
• Inverse Matrices: The equivalent of division in matrix algebra, finding a matrix \( A^{-1} \) such that \( A \times A^{-1} = I \), where \( I \) is the identity matrix.

Matrices are essential in using Cramer's Rule, as we convert systems of equations into a matrix form. By working with the coefficient matrix and its determinant, we can find solutions to the system effectively. Each transformation manipulates the structure of the equations to simplify the process of finding variable values, illustrating the power of matrix algebra in solving complex problems.