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Magazine Chapter 12: Problem 66
Solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. $$ \left\\{\begin{array}{rr} x+y+z+w= & 4 \\ -x+2 y+z= & 0 \\ 2 x+3 y+z-w= & 6 \\ -2 x+y-2 z+2 w= & -1 \end{array}\right. $$
Short Answer
Expert verified
The solution to the system is \(x = 2\), \(y = 1\), \(z = 0\), \(w = 1\).
Step by step solution
01
Write the Augmented Matrix
Start by creating the augmented matrix from the given system of equations. \[ A = \begin{bmatrix} 1 & 1 & 1 & 1 & | & 4 \ -1 & 2 & 1 & 0 & | & 0 \ 2 & 3 & 1 & -1 & | & 6 \ -2 & 1 & -2 & 2 & | & -1 \end{bmatrix} \]
02
Perform Row Operations to Get a Leading 1 in the First Row
We already have a leading 1 in the first row. Next, use row operations to make other entries in the first column 0. \[ R2 = R2 + R1 \rightarrow \begin{bmatrix} 1 & 1 & 1 & 1 & | & 4 \ 0 & 3 & 2 & 1 & | & 4 \ 2 & 3 & 1 & -1 & | & 6 \ -2 & 1 & -2 & 2 & | & -1 \end{bmatrix} \]\[ R3 = R3 - 2R1 \rightarrow \begin{bmatrix} 1 & 1 & 1 & 1 & | & 4 \ 0 & 3 & 2 & 1 & | & 4 \ 0 & 1 & -1 & -3 & | & -2 \ -2 & 1 & -2 & 2 & | & -1 \end{bmatrix} \]\[ R4 = R4 + 2R1 \rightarrow \begin{bmatrix} 1 & 1 & 1 & 1 & | & 4 \ 0 & 3 & 2 & 1 & | & 4 \ 0 & 1 & -1 & -3 & | & -2 \ 0 & 3 & 0 & 4 & | & 7 \end{bmatrix} \]
03
Make Leading 1 in Second Row and Zero Out Other Entries of Second Column
Next, we make the second row's leading coefficient 1 and zero out the second column's other entries. \[ R2 = \frac{1}{3}R2 \rightarrow \begin{bmatrix} 1 & 1 & 1 & 1 & | & 4 \ 0 & 1 & \frac{2}{3} & \frac{1}{3} & | & \frac{4}{3} \ 0 & 1 & -1 & -3 & | & -2 \ 0 & 3 & 0 & 4 & | & 7 \end{bmatrix} \]\[ R3 = R3 - R2 \rightarrow \begin{bmatrix} 1 & 1 & 1 & 1 & | & 4 \ 0 & 1 & \frac{2}{3} & \frac{1}{3} & | & \frac{4}{3} \ 0 & 0 & -\frac{5}{3} & -\frac{10}{3} & | & -\frac{10}{3} \ 0 & 3 & 0 & 4 & | & 7 \end{bmatrix} \]\[ R4 = R4 - 3R2 \rightarrow \begin{bmatrix} 1 & 1 & 1 & 1 & | & 4 \ 0 & 1 & \frac{2}{3} & \frac{1}{3} & | & \frac{4}{3} \ 0 & 0 & -\frac{5}{3} & -\frac{10}{3} & | & -\frac{10}{3} \ 0 & 0 & -2 & 3 & | & 3 \end{bmatrix} \]
04
Make Leading 1 in Third Row and Zero Out Other Entries of Third Column
Adjust the third row to make the leading coefficient a 1 and zero out the third column's other entries. \[ R3 = -\frac{3}{5}R3 \rightarrow \begin{bmatrix} 1 & 1 & 1 & 1 & | & 4 \ 0 & 1 & \frac{2}{3} & \frac{1}{3} & | & \frac{4}{3} \ 0 & 0 & 1 & 2 & | & 2 \ 0 & 0 & -2 & 3 & | & 3 \end{bmatrix} \]\[ R4 = R4 + 2R3 \rightarrow \begin{bmatrix} 1 & 1 & 1 & 1 & | & 4 \ 0 & 1 & \frac{2}{3} & \frac{1}{3} & | & \frac{4}{3} \ 0 & 0 & 1 & 2 & | & 2 \ 0 & 0 & 0 & 7 & | & 7 \end{bmatrix} \]
05
Make Leading 1 in Fourth Row
Finally, adjust the fourth row to finish transforming our matrix. \[ R4 = \frac{1}{7}R4 \rightarrow \begin{bmatrix} 1 & 1 & 1 & 1 & | & 4 \ 0 & 1 & \frac{2}{3} & \frac{1}{3} & | & \frac{4}{3} \ 0 & 0 & 1 & 2 & | & 2 \ 0 & 0 & 0 & 1 & | & 1 \end{bmatrix} \]
06
Perform Back Substitution to Find the Solution
Now perform back substitution to find the solution. Starting from the last row, substitute backwards. \[ w = 1 \]\[ z + 2w = 2 \rightarrow z + 2 \cdot 1 = 2 \rightarrow z = 0 \]\[ y + \frac{2}{3}z + \frac{1}{3}w = \frac{4}{3} \rightarrow y + \frac{2}{3}(0) + \frac{1}{3}(1) = \frac{4}{3} \rightarrow y + \frac{1}{3} = \frac{4}{3} \rightarrow y = 1 \]\[ x + y + z + w = 4 \rightarrow x + 1 + 0 + 1 = 4 \rightarrow x + 2 = 4 \rightarrow x = 2 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solving Systems of Equations
Solving systems of equations involves finding the values of variables that satisfy all equations simultaneously. Here, we solve a system of linear equations using matrices and row operations. Linear equations can be represented in matrix form, which makes it easier to handle multiple variables and equations. The solution can be found using techniques such as elimination, substitution, or matrix operations like Gaussian elimination.
Augmented Matrix
An augmented matrix is an essential tool when solving systems of linear equations. It combines the coefficients and constants from each equation into a single matrix. For example, from the given equations: \(x + y + z + w = 4\), \(-x + 2y + z = 0\), \(2x + 3y + z - w = 6\), \(-2x + y - 2z + 2w = -1\), we create an augmented matrix:
\
This matrix is useful as it allows us to perform systematic row operations to solve the system by converting it to a simpler (usually row-echelon) form.
\
\[ \begin{bmatrix} 1 & 1 & 1 & 1 & | & 4\ -1 & 2 & 1 & 0 & | & 0\ 2 & 3 & 1 & -1 & | & 6\ -2 & 1 & -2 & 2 & | & -1 \end{bmatrix} \]This matrix is useful as it allows us to perform systematic row operations to solve the system by converting it to a simpler (usually row-echelon) form.
Matrix Row Operations
Matrix row operations are the steps we use to simplify an augmented matrix. The goal is to transform the matrix into row-echelon form or reduced row-echelon form. There are three main types of row operations:
\
By applying these operations, we systematically zero out elements below lead coefficients, ultimately solving the equations. For example, in Step 2, we transform the second row by adding the first row: \(R2 = R2 + R1\), which adjusts the second row's entries.
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- Row swapping (interchanging two rows): \(R_i \leftrightarrow R_j\)
- Row multiplication (scaling a row by a non-zero number): \(kR_i\)
- Row addition/subtraction (adding or subtracting a multiple of one row to another): \(R_i = R_i + kR_j\)
By applying these operations, we systematically zero out elements below lead coefficients, ultimately solving the equations. For example, in Step 2, we transform the second row by adding the first row: \(R2 = R2 + R1\), which adjusts the second row's entries.
Back Substitution
Back substitution is the final step after transforming the augmented matrix into row-echelon form. It involves solving for variables starting from the last row and moving upwards. This method ensures that each variable is solved in terms of already determined variables. For instance, after transforming the matrix to: \[ \begin{bmatrix} 1 & 1 & 1 & 1 & | & 4\ 0 & 1 & \frac{2}{3} & \frac{1}{3} & | & \frac{4}{3}\ 0 & 0 & 1 & 2 & | & 2\ 0 & 0 & 0 & 1 & | & 1 \end{bmatrix} \] we solve from the bottom:
\
This sequential approach leads to the complete solution: \(x = 2\), \(y = 1\), \(z = 0\), \(w = 1\).
\
- Find \(w\): \(w = 1\)
- Substitute \(w\) to find \(z\): \(z + 2w = 2 \rightarrow z = 0\)
- Use \(z\) and \(w\) to find \(y\): \(y + \frac{1}{3}w = \frac{4}{3} \rightarrow y = 1\)
- Finally, solve for \(x\): \(x + 1 + 0 + 1 = 4 \rightarrow x = 2\)
This sequential approach leads to the complete solution: \(x = 2\), \(y = 1\), \(z = 0\), \(w = 1\).
