Problem 70
Question
Use matrices to solve the system of equations, if possible. Use Gauss-Jordan elimination. $$\left\\{\begin{aligned} 2 x+2 y-z &=2 \\ x-3 y+z &=28 \\ -x+y &=14 \end{aligned}\right.$$
Step-by-Step Solution
Verified Answer
The solutions to the system of equations are \(x=21\), \(y=7\), and \(z=14\).
1Step 1: Write the System of Equations as an Augmented Matrix
The first step is to write the system of equations in matrix form. Thus, the coefficients of the variables x, y, and z form a matrix on the left, and the constants form a matrix on the right. The matrix form is as follows: \[ \begin{bmatrix} 2 & 2 & -1 & 2 \ 1 & -3 & 1 & 28 \ -1 & 1 & 0 & 14 \end{bmatrix}\]
2Step 2: Perform Gauss-Jordan Elimination
Normalize the first row by dividing by 2 and interchange row 2 and 3 to create leading one for second row. Then, add the second row to the first, and add the second row to the third. This results in the matrix \[ \begin{bmatrix} 1 & 0 & -1 & 7 \ 1 & 0 & -0.5 & 14 \ 0 & 1 & 0.5 & 14 \end{bmatrix}\]. Then subtract the second row from the first, and add third row to the second - resulting in the matrix \[ \begin{bmatrix} 0 & 0 & -0.5 & -7 \ 1 & 1 & 0 & 28 \ 0 & 1 & 0.5 & 14 \end{bmatrix}\]. Lastly, multiply first row by -2 and subtract third row from the second - resulting in the matrix \[ \begin{bmatrix} 0 & 0 & 1 & 14 \ 1 & 0 & -0.5 & 14 \ 0 & 1 & 0.5 & 14 \end{bmatrix}\].
3Step 3: Extract the Solutions
The final matrix corresponds to the system of equations \(0x+0y+1z=14\), \(1x+0y-0.5z=14\), and \(0x+1y+0.5z=14\). Solving these gives the solutions \(x=21\), \(y=7\), and \(z=14\).
Key Concepts
System of EquationsAugmented MatrixMatrix OperationsSolution Extraction
System of Equations
A system of equations is a set of multiple equations with multiple variables. The goal is to find the values of these variables that satisfy all the equations at once. In this exercise, we have three equations with three unknowns: \(x\), \(y\), and \(z\). These kinds of problems often appear in mathematics to show scenarios where different conditions must be met at the same time.
- Example Equation 1: \(2x + 2y - z = 2\)
- Example Equation 2: \(x - 3y + z = 28\)
- Example Equation 3: \(-x + y = 14\)
Augmented Matrix
The augmented matrix is a way to represent a system of equations. Instead of writing each equation out, you transform it into a matrix, which is a grid of numbers.
This method combines the coefficients of each variable and the constant terms in separate sections. It looks something like this:\[\begin{bmatrix}2 & 2 & -1 & | & 2 \1 & -3 & 1 & | & 28 \-1 & 1 & 0 & | & 14 \end{bmatrix}\]The vertical line separates the coefficients from the constants. Using this representation allows for more systematic manipulation when solving the equations.
This method combines the coefficients of each variable and the constant terms in separate sections. It looks something like this:\[\begin{bmatrix}2 & 2 & -1 & | & 2 \1 & -3 & 1 & | & 28 \-1 & 1 & 0 & | & 14 \end{bmatrix}\]The vertical line separates the coefficients from the constants. Using this representation allows for more systematic manipulation when solving the equations.
Matrix Operations
Matrix operations are methods we use to manipulate matrices like our augmented matrix. In Gauss-Jordan elimination, our goal is to perform these operations to transform the matrix into what's called "row reduced echelon form."
Here are some common operations:
Here are some common operations:
- Row swapping: Interchanging two rows.
- Scaling: Multiplying all entries of a row by a non-zero constant.
- Row addition: Adding or subtracting entries of one row to/from another row.
Solution Extraction
Solution extraction is the final step in solving the system of equations using the augmented matrix.
Once your matrix is in row reduced echelon form, you can "read off" the solutions. Each row corresponds to an equation, showing directly what a variable equals:
Once your matrix is in row reduced echelon form, you can "read off" the solutions. Each row corresponds to an equation, showing directly what a variable equals:
- \(z = 14\)
- \(x - 0.5z = 14\)
- \(y + 0.5z = 14\)
- \(z = 14\)
- \(x = 21\)
- \(y = 7\)
Other exercises in this chapter
Problem 70
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