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Magazine Chapter 9: Problem 58
Solve using matrices. $$\begin{aligned} &x+y=2 z\\\ &2 x-5 z=4\\\ &x-z=y+8 \end{aligned}$$
Short Answer
Expert verified
The solution is \( x = 7 \), \( y = -3 \), \( z = 2 \).
Step by step solution
01
- Rewrite the System of Equations in Matrix Form
First, express the system of equations as matrix equations. The given system is: \[ x + y = 2z \]\[ 2x - 5z = 4 \]\[ x - z = y + 8 \]. Rewrite the third equation as: \[ x - z - y = 8 \]. This can be written in matrix form as: \[ \begin{bmatrix} 1 & 1 & -2 \ 2 & 0 & -5 \ 1 & -1 & -1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 0 \ 4 \ 8 \end{bmatrix} \].
02
- Set Up the Augmented Matrix
Combine the coefficient matrix and the constants into an augmented matrix:\[ \left[ \begin{array}{ccc|c} 1 & 1 & -2 & 0 \ 2 & 0 & -5 & 4 \ 1 & -1 & -1 & 8 \end{array} \right] \].
03
- Use Row Operations to Achieve Row Echelon Form
Apply row operations to simplify the augmented matrix. Start by making the element in the first row and first column a 1, if it isn't already. Then, use row operations to zero out the elements below this leading 1:1. Subtract 2 times the first row from the second row: \[ R2 = R2 - 2R1 \], giving \[ \left[ \begin{array}{ccc|c} 1 & 1 & -2 & 0 \ 0 & -2 & -1 & 4 \ 1 & -1 & -1 & 8 \end{array} \right] \].2. Subtract the first row from the third row: \[ R3 = R3 - R1 \], giving \[ \left[ \begin{array}{ccc|c} 1 & 1 & -2 & 0 \ 0 & -2 & -1 & 4 \ 0 & -2 & 1 & 8 \end{array} \right] \].
04
- Continue Simplifying
Simplify the second and third rows: 1. Divide the second row by -2: \[ R2 = -\frac{1}{2}R2 \], giving \[ \left[ \begin{array}{ccc|c} 1 & 1 & -2 & 0 \ 0 & 1 & \frac{1}{2} & -2 \ \ 0 & -2 & 1 & 8 \end{array} \right] \].2. Add 2 times the second row to the third row to eliminate the -2 in the second column of the third row:\[ R3 = R3 + 2R2 \], giving \[ \left[ \begin{array}{ccc|c} 1 & 1 & -2 & 0 \ 0 & 1 & \frac{1}{2} & -2 \ \ 0 & 0 & 2 & 4 \end{array} \right] \].
05
- Back-Substitute to Find the Solution
Continue simplifying by converting the leading 2 in the third column of the third row to a 1: 1. Divide the third row by 2:\[ R3 = \frac{1}{2}R3 \], giving\[ \left[ \begin{array}{ccc|c} 1 & 1 & -2 & 0 \ 0 & 1 & \frac{1}{2} & -2 \ \ 0 & 0 & 1 & 2 \end{array} \right] \].Now back-substitute and solve from the bottom row upwards:1. From the third row: \( z = 2 \).2. Substitute \( z = 2 \) into the second row: \[ y + \frac{1}{2}(2) = -2 \]\( y + 1 = -2 \)\[ y = -3 \].3. Substitute \( z = 2 \) and \( y = -3 \) into the first row: \[ x + (-3) = 2(2) \]\[ x - 3 = 4 \]\[ x = 7 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
augmented matrix
When solving a system of linear equations using matrices, one powerful tool we use is the augmented matrix. An augmented matrix combines the coefficient matrix (containing the coefficients of the variables) with the constants from the right-hand side of the equations. This allows us to manage all the information in one place. For instance, in our exercise, the system of equations is represented as an augmented matrix: \[\left[ \begin{array}{ccc|c} 1 & 1 & -2 & 0 \ 2 & 0 & -5 & 4 \ 1 & -1 & -1 & 8 \end{array} \right]\]. This compact format makes it easier to apply row operations and simplifies the path to our solution.
row operations
Row operations are essential for transforming an augmented matrix. They help in achieving the row echelon form and, consequently, facilitate back-substitution. There are three main types of row operations:
- Swapping two rows
- Multiplying a row by a non-zero scalar
- Adding or subtracting a multiple of one row from another row
row echelon form
The row echelon form (REF) is a triangular form obtained through row operations, where each leading entry (non-zero) is 1, and is to the right of the leading entry in the row above it. All entries below a leading entry are zeros. The REF makes it straightforward to use back-substitution to find the solutions to the system of equations. For our exercise, after applying several row operations, we reached the row echelon form of the augmented matrix: \[\left[ \begin{array}{ccc|c} 1 & 1 & -2 & 0 \ 0 & 1 & \frac{1}{2} & -2 \ 0 & 0 & 1 & 2 \end{array} \right]\]. Notice how the leading entries form a staircase pattern, moving from left to right.
back-substitution
Back-substitution is a method used after transforming an augmented matrix to row echelon form. It involves solving the equations from the bottom row upwards. In our solved matrix, the third row gives us that \( z = 2 \). This value of 'z' is then substituted back into the second row equation to solve for 'y', and finally, both 'y' and 'z' values are substituted into the first row to solve for 'x'. Hence, proceeding step-by-step in reverse order, we found:
- \( z = 2 \)
- \( y = -3 \)
- \( x = 7 \)
