Problem 35 - Algebra and Trigonometry

Question

\(17-36\) . Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} x+& z+2 w=\\\ y-2 z &=-3 \\ x+2 y-z &=-2 \\ 2 x+y+3 z-2 w &=0 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system is consistent with solution: \((x, y, z, w) = (-5, 3, 3, 1)\).

Step 1: Organize the system into a matrix

Rewrite the given system of equations into an augmented matrix form:\[\begin{bmatrix}1 & 0 & 1 & 2 & \mid & 0 \0 & 1 & -2 & 0 & \mid & -3 \1 & 2 & -1 & 0 & \mid & -2 \2 & 1 & 3 & -2 & \mid & 0 \\end{bmatrix}\]

Step 2: 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:\[\begin{bmatrix}1 & 0 & 1 & 2 & \mid & 0 \0 & 1 & -2 & 0 & \mid & -3 \0 & 2 & -2 & -2 & \mid & -2 \0 & 1 & 1 & -6 & \mid & 0 \\end{bmatrix}\]

Step 3: 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:\[\begin{bmatrix}1 & 0 & 1 & 2 & \mid & 0 \0 & 1 & -2 & 0 & \mid & -3 \0 & 0 & 2 & -2 & \mid & 4 \0 & 0 & 3 & -6 & \mid & 3 \\end{bmatrix}\]

Step 4: Convert to Reduced Row-Echelon Form

Continue reducing the matrix:- Subtract \(\frac{3}{2}\) times the third row from the fourth row:\[\begin{bmatrix}1 & 0 & 1 & 2 & \mid & 0 \0 & 1 & -2 & 0 & \mid & -3 \0 & 0 & 2 & -2 & \mid & 4 \0 & 0 & 0 & -3 & \mid & -3 \\end{bmatrix}\]

Step 5: 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:\[x = -5, \ y = 3, \ z = 3, \ w = 1 \]

Step 6: Conclusion on consistency

Since we have found finite values for \(x\), \(y\), \(z\), and \(w\), the system is consistent and has a unique solution.

Key Concepts

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’s (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.

This step-by-step reduction makes it easier to see the relationships between variables and find their values. This method works systematically to reach a solution, which can often be seen as your system's blueprint.

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.

Achieving this form rearranges the system to where you can start clearly determining values or further narrowing down possibilities. Finally, it sets the stage for the next steps in the matrix solution process, where you aim for further simplification towards the reduced row-echelon form.

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.

This formation allows easy visualization and manipulation of equations through matrix methods, like Gauss-Jordan elimination. It's like turning complex words into numbers and lines, simplifying understanding and calculations. Matrices eliminate the need to juggle numerous numbers and equations, focusing instead on straightforward operations.

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.

This method is like weaving backwards through the pattern of equations, solving one "mystery" at a time to piece together the entire solution. Back-substitution offers closure, ensuring every equation in your system is satisfied by the values found, and confirms the system’s consistency.