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Magazine Chapter 11: Problem 35
\(17-36\) . Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} x+& z+2 w=\\\ y-2 z &=-3 \\ x+2 y-z &=-2 \\ 2 x+y+3 z-2 w &=0 \end{aligned}\right. $$
Short Answer
Expert verified
The system is consistent with solution: \((x, y, z, w) = (-5, 3, 3, 1)\).
Step by step solution
01
Organize the system into a matrix
Rewrite the given system of equations into an augmented matrix form:\[\begin{bmatrix}1 & 0 & 1 & 2 & \mid & 0 \0 & 1 & -2 & 0 & \mid & -3 \1 & 2 & -1 & 0 & \mid & -2 \2 & 1 & 3 & -2 & \mid & 0 \\end{bmatrix}\]
02
Perform row operations to reach Row-Echelon Form
Apply Gauss-Jordan elimination to transform the matrix into row-echelon form:- Subtract the first row from the third row: \[\begin{bmatrix}1 & 0 & 1 & 2 & \mid & 0 \0 & 1 & -2 & 0 & \mid & -3 \0 & 2 & -2 & -2 & \mid & -2 \2 & 1 & 3 & -2 & \mid & 0 \\end{bmatrix}\]- Subtract 2 times the first row from the fourth row:\[\begin{bmatrix}1 & 0 & 1 & 2 & \mid & 0 \0 & 1 & -2 & 0 & \mid & -3 \0 & 2 & -2 & -2 & \mid & -2 \0 & 1 & 1 & -6 & \mid & 0 \\end{bmatrix}\]
03
Further simplify to find solutions
Continuing with row operations:- Subtract the second row from the fourth row:\[\begin{bmatrix}1 & 0 & 1 & 2 & \mid & 0 \0 & 1 & -2 & 0 & \mid & -3 \0 & 2 & -2 & -2 & \mid & -2 \0 & 0 & 3 & -6 & \mid & 3 \\end{bmatrix}\]- Subtract 2 times the second row from the third row to simplify:\[\begin{bmatrix}1 & 0 & 1 & 2 & \mid & 0 \0 & 1 & -2 & 0 & \mid & -3 \0 & 0 & 2 & -2 & \mid & 4 \0 & 0 & 3 & -6 & \mid & 3 \\end{bmatrix}\]
04
Convert to Reduced Row-Echelon Form
Continue reducing the matrix:- Subtract \(\frac{3}{2}\) times the third row from the fourth row:\[\begin{bmatrix}1 & 0 & 1 & 2 & \mid & 0 \0 & 1 & -2 & 0 & \mid & -3 \0 & 0 & 2 & -2 & \mid & 4 \0 & 0 & 0 & -3 & \mid & -3 \\end{bmatrix}\]
05
Back-substitute to find the solution
Interpret the final matrix to read off solutions and back-substitute:The fourth row gives:\(-3w = -3 \Rightarrow w = 1\)Substitute into the third row:\(2z - 2(1) = 4 \Rightarrow 2z = 6 \Rightarrow z = 3\)Substitute into the second row:\(y - 2(3) = -3 \Rightarrow y = 3\)Substitute into the first row:\(x + 3 + 2(1) = 0 \Rightarrow x = -5\)Thus, the solution to the system is:\[x = -5, \ y = 3, \ z = 3, \ w = 1 \]
06
Conclusion on consistency
Since we have found finite values for \(x\), \(y\), \(z\), and \(w\), the system is consistent and has a unique solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gauss-Jordan elimination
Gauss-Jordan elimination is an essential method for solving linear systems of equations. Imagine you have a set of equations with several unknowns, like a mystery to solve. This technique transforms the mystery into an explicit solution by simplifying the equations step by step. The goal of Gauss-Jordan elimination is to manipulate the coefficients in the system's augmented matrix until a form is reached where each equation corresponds directly to one unknown.
- Start by writing the equations into an augmented matrix, where each row is an equation and each column is a coefficient of a variable.
- Use row operations to simplify the matrix, focusing first on obtaining leading 1鈥檚 (pivot elements) in each row, while zeros are below each pivot.
- Once in row-echelon form, continue simplifying until each row contains only zeros except for a single leading 1.
row-echelon form
The row-echelon form is like reaching the first checkpoint in solving a system of equations. It's a stage in matrix simplification showing strategic placements of elements to get closer to solving for variables. Here's what it involves:
- Each row that contains a non-zero element has a 1, called a leading 1 or pivot, somewhere moving to the right as you go down the rows.
- All elements below a pivot are zeros, creating a sort of staircase effect across the matrix.
- Rows that are entirely zeros are at the bottom of the matrix.
augmented matrix
An augmented matrix is a powerful tool to bring structure to a linear system of equations. Think of it as a table where each row represents an equation, and each column aligns with a variable's coefficient or the constant from the equation's right side. Here's how it's laid out:
- The matrix starts with an arrangement of coefficients of each variable as columns.
- To the right of these columns, separated by a line, are the constants from the equations.
back-substitution
Back-substitution is the final piece in the puzzle of solving a linear system of equations. Once your matrix is in reduced row-echelon form, back-substitution allows you to solve for the unknowns starting from the simplest equation. Here's the process:
- Begin with the last simplified equation. If simpler than others, it should clearly inform you of one variable's value.
- Substitute that known value back into the penultimate equation to solve for another variable.
- Continue this substitution step by step until all variables have known values.
