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Magazine Chapter 9: Problem 12
Use matrices to solve each system of linear equations. See Example 3. $$ \left\\{\begin{aligned} 4 y+3 z &=-2 \\ 5 x-4 y &=1 \\ -5 x+4 y+z &=-3 \end{aligned}\right. $$
Short Answer
Expert verified
The solution is \((x, y, z) = (1, 1, -2)\).
Step by step solution
01
Write the system in matrix form
The system of equations is:\[\begin{align*}4y + 3z &= -2 \ 5x - 4y &= 1 \ -5x + 4y + z &= -3\end{align*}\]Since the coefficients matrix must be consistent with the given system, write the system in the form of \(AX = B\), where \(A\) is the coefficients matrix, \(X\) is the column matrix of variables, and \(B\) is the constants matrix:\[A = \begin{bmatrix}0 & 4 & 3 \5 & -4 & 0 \-5 & 4 & 1\end{bmatrix}, \quadX = \begin{bmatrix}x \y \z\end{bmatrix}, \quadB = \begin{bmatrix}-2 \1 \-3\end{bmatrix}\]
02
Set up the augmented matrix
Write the augmented matrix \([A|B]\) from the matrices derived:\[\begin{bmatrix}0 & 4 & 3 & | & -2 \5 & -4 & 0 & | & 1 \-5 & 4 & 1 & | & -3\end{bmatrix}\]
03
Perform row operations to achieve row-echelon form
Use Gaussian elimination to simplify the matrix into row-echelon form.1. Swap row 1 with row 2 to get: \[ \begin{bmatrix} 5 & -4 & 0 & | & 1 \ 0 & 4 & 3 & | & -2 \ -5 & 4 & 1 & | & -3 \end{bmatrix} \]2. Add row 1 to row 3: \[ \begin{bmatrix} 5 & -4 & 0 & | & 1 \ 0 & 4 & 3 & | & -2 \ 0 & 0 & 1 & | & -2 \end{bmatrix} \]3. Multiply row 2 by 1/4 to simplify the second row: \[ \begin{bmatrix} 5 & -4 & 0 & | & 1 \ 0 & 1 & 3/4 & | & -1/2 \ 0 & 0 & 1 & | & -2 \end{bmatrix} \]
04
Back-substitute to find variable values
Using the row-echelon matrix, start from the last row to solve for the variables:1. From row 3, \(z = -2\).2. Substitute \(z = -2\) into row 2: \( y + \frac{3}{4}(-2) = -\frac{1}{2} \ y - \frac{3}{2} = -\frac{1}{2} \ y = 1 \)3. Substitute \(y = 1\) and \(z = -2\) into row 1: \( 5x - 4(1) = 1 \ 5x - 4 = 1 \ 5x = 5 \ x = 1 \)
05
Write the solution
The solution to the system of equations is the set \((x, y, z) = (1, 1, -2)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Representation of Linear Equations
Matrix representation of linear equations is a fundamental concept in linear algebra. It involves transforming a system of linear equations into a format using matrices, which can be solved more efficiently. When given a system of equations, each equation contains variables and constants. The first step is forming the coefficient matrix, variable matrix, and constants matrix.
- The coefficients from each equation form the coefficient matrix, denoted as \(A\).
- The variables form a single column matrix, denoted as \(X\).
- The constants from the right-hand side of each equation form the constants matrix, denoted as \(B\).
Gaussian Elimination
Gaussian elimination is a method used to solve systems of linear equations. It simplifies matrices through a series of steps to reveal the solution. The process involves transforming the matrix into a form where solving the equations becomes straightforward. Here鈥檚 how it works:
- Forming an Augmented Matrix: Combine the coefficient matrix and constants matrix to form an augmented matrix, \([A|B]\).
- Using Row Operations: Through simple operations like swapping rows, adding multiples of one row to another, and multiplying rows by a constant, you systematically reduce the matrix.
- Achieving Triangular Form: Repeated row operations reformat the matrix until the goal is reached, which is a triangular or row-echelon form.
Row-Echelon Form
Row-echelon form is a key outcome of Gaussian elimination. Once achieved, it simplifies the process of finding solutions to systems of equations. In this form:
- The first nonzero element in each row is 1, creating a staircase pattern.
- Each leading 1 is to the right of any leading 1s in the row above.
- Any rows of all zeros are at the bottom.
Matrix Row Operations
Matrix row operations are the tools used in Gaussian elimination to manipulate and simplify matrices, allowing solutions to be found efficiently. The operations consist of three allowable transformations:
Efficient use of these operations transforms an augmented matrix from its original state to row-echelon form efficiently, helping to solve for the unknown variables quickly and accurately.
- Row Swapping: Interchanging two rows to better position pivot elements.
- Row Multiplication: Multiplying all entries of a row by a nonzero constant to achieve desired leading coefficients.
- Row Addition: Adding or subtracting multiples of one row from another to eliminate coefficients in certain positions.
Efficient use of these operations transforms an augmented matrix from its original state to row-echelon form efficiently, helping to solve for the unknown variables quickly and accurately.
