Problem 25 - Intermediate Algebra

Question

Solve the system $$ \left(\begin{array}{r} x-3 y-2 z+w=-3 \\ -2 x+7 y+z-2 w=-1 \\ 3 x-7 y-3 z+3 w=-5 \\ 5 x+y+4 z-2 w=18 \end{array}\right) $$

Step-by-Step Solution

Answer

Solution: \( x = \frac{14}{9}, y = \frac{-49}{9}, z = \frac{14}{9}, w = 4 \).

Step 1: Write the System of Equations

We start with the given system of linear equations:1. \( x - 3y - 2z + w = -3 \)2. \( -2x + 7y + z - 2w = -1 \)3. \( 3x - 7y - 3z + 3w = -5 \)4. \( 5x + y + 4z - 2w = 18 \)

Step 2: Form the Augmented Matrix

Write the coefficients of the variables and the constants into an augmented matrix:\[\begin{bmatrix}1 & -3 & -2 & 1 & | & -3 \-2 & 7 & 1 & -2 & | & -1 \3 & -7 & -3 & 3 & | & -5 \5 & 1 & 4 & -2 & | & 18\end{bmatrix}\]

Step 3: Row Reduction to Row Echelon Form

Apply row operations to convert the matrix to row echelon form. Multiply the first row by 2 and add it to the second row to eliminate the \(x\) in the second row. Also, adjust other rows for simplification.\[\begin{bmatrix}1 & -3 & -2 & 1 & | & -3 \0 & 1 & -3 & 0 & | & -7 \3 & -7 & -3 & 3 & | & -5 \5 & 1 & 4 & -2 & | & 18\end{bmatrix}\]Continue to make similar adjustments to the third and fourth rows.

Step 4: Further Row Reduction

Continue reducing by making the leading coefficients except diagonals zero, using previous rows. After necessary operations, check for a diagonal of 1s in leading positions:\[\begin{bmatrix}1 & -3 & -2 & 1 & | & -3 \0 & 1 & -3 & 0 & | & -7 \0 & 0 & -9 & 0 & | & -14 \0 & 0 & 0 & 1 & | & 4\end{bmatrix}\]

Step 5: Back Substitution

Now substitute back from the bottom row to solve for each variable. Begin by solving for \(w\): - From the fourth row: \(w = 4\)- From the third row: \(-9z = -14 \rightarrow z = \frac{14}{9}\)- From the second row: \(y - 3(\frac{14}{9}) = -7 \rightarrow y = \frac{-49}{9}\)- From the first row: \(x - 3(\frac{-49}{9}) - 2(\frac{14}{9}) + 4 = -3 \rightarrow x = \frac{14}{9} \)

Step 6: Solution of the System

After performing back substitution, we find the solution of the system as:\[ x = \frac{14}{9},\, y = \frac{-49}{9},\, z = \frac{14}{9},\, w = 4 \]

Key Concepts

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear equations. It is essential for solving systems of equations, transformations in space, and many applications in engineering and physics. In the context of systems of equations, linear algebra allows us to use matrices and other powerful techniques to find solutions more efficiently than simple substitution methods.

• Vectors and Matrices: Vectors are quantities defined by both magnitude and direction, and matrices are grids of numbers used to represent linear transformations and systems of equations.
• Systems of Equations: These are collections of equations that share variables, and linear algebra offers ways to solve for those unknowns using structured techniques.
• Applications: Linear algebra applies to fields such as computer graphics, statistics, machine learning, and more.

Breaking down large sets of linear equations into matrix forms showcases the beauty of linear algebra in organizing and solving mathematical problems.

Augmented Matrix

An augmented matrix is a crucial tool in linear algebra for simplifying systems of equations. It combines the coefficients of variables and the constants from each equation into a single matrix form.
Consider this system of equations:

• Equation 1: \(x - 3y - 2z + w = -3\)
• Equation 2: \(-2x + 7y + z - 2w = -1\)
• Equation 3: \(3x - 7y - 3z + 3w = -5\)
• Equation 4: \(5x + y + 4z - 2w = 18\)

From these, the augmented matrix is formed as follows:\[\begin{bmatrix}1 & -3 & -2 & 1 & | & -3 \-2 & 7 & 1 & -2 & | & -1 \3 & -7 & -3 & 3 & | & -5 \5 & 1 & 4 & -2 & | & 18\end{bmatrix}\]Here, each row represents an equation, and each column corresponds to one of the variables. The vertical line separates the coefficients from the constants, helping visualize the balance between the two sides of each equation.

Row Reduction

Row reduction is an effective technique for converting an augmented matrix into a form that is easier to interpret and solve. This process involves applying elementary row operations to restructure the matrix into what is known as row-echelon form.
Elementary row operations include:

• Swapping two rows
• Multiplying a row by a non-zero constant
• Adding or subtracting a multiple of one row from another row

The goal of row reduction is to systematically eliminate variables by creating leading coefficients (typically 1s) and zeros below the leading ones. This form simplifies solving the system since each row corresponds to an equation with fewer unknowns.
After performing various row operations, the matrix for our example transforms to:\[\begin{bmatrix}1 & -3 & -2 & 1 & | & -3 \0 & 1 & -3 & 0 & | & -7 \0 & 0 & -9 & 0 & | & -14 \0 & 0 & 0 & 1 & | & 4\end{bmatrix}\]Each transformation step should make it progressively easier to solve the equations by eliminating one variable at a time.

Back Substitution

Once we have achieved a row-echelon form, the next step in solving a system of linear equations is back substitution. This method involves starting with the bottom-most equation and solving for one variable at a time.

• Begin at the last row: Solve for the straightforward equation involving the last variable.
• Substitute back: Use the known value from the last equation to solve the preceding rows, working your way up.
• Continue substituting: Repeat this process to derive each previous unknown in each step.

In our scenario, we begin with the fourth row, which in its simplified form is straightforward enough to solve directly. For example:- From the fourth row: \(w = 4\)
- Substitute \(w = 4\) into the prior rows to uncover the value of \(z\), and continue the process: - From the third row: \(-9z = -14 \rightarrow z = \frac{14}{9}\)
- From the second row: \(y - 3\left(\frac{14}{9}\right) = -7 \rightarrow y = \frac{-49}{9}\)
- Finally, substitute all known variables in the first row: \(x - 3\left(\frac{-49}{9}\right) - 2\left(\frac{14}{9}\right) + 4 = -3 \rightarrow x = \frac{14}{9}\)
This results in the complete solution: \(x = \frac{14}{9}, y = \frac{-49}{9}, z = \frac{14}{9}, w = 4\). The stepwise nature of back substitution allows for systematic problem-solving, ensuring no variable remains unaccounted for.