Problem 20 - Precalculus: Mathematical for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} 2 x+y-z &=-8 \\ -x+y+z &=3 \\ -2 x\quad\quad &+4 z=18 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution of the system is \( x = -1 \), \( y = -\frac{5}{2} \), \( z = \frac{7}{2} \).

Step 1: Write the system in augmented matrix form

The given system of equations is written as an augmented matrix: \[\begin{bmatrix}2 & 1 & -1 & | & -8 \-1 & 1 & 1 & | & 3 \-2 & 0 & 4 & | & 18\end{bmatrix}\]

Step 2: Use Row Operations

Perform row operations to simplify the augmented matrix to row-echelon form. First, we'll eliminate the \( x \) coefficient from the second and third rows:1. Add Row 1 to Row 2 to get the new Row 2.2. Add 2 times Row 1 to Row 3 to get the new Row 3.The matrix becomes:\[\begin{bmatrix}2 & 1 & -1 & | & -8 \0 & 2 & 0 & | & -5 \0 & 2 & 2 & | & 2\end{bmatrix}\]

Step 3: Further Simplify and Eliminate Variables

Subtract Row 2 from Row 3 to eliminate the \( y \) term in the third row:\[\begin{bmatrix}2 & 1 & -1 & | & -8 \0 & 2 & 0 & | & -5 \0 & 0 & 2 & | & 7\end{bmatrix}\]Now, apply the inverse operations to isolate terms below main diagonal.

Step 4: Solve for Variables

Back-substitute to solve for the variables:1. From Row 3: \( 2z = 7 \rightarrow z = \frac{7}{2} \).2. Substitute \( z = \frac{7}{2} \) into Row 2: \( 2y = -5 \rightarrow y = -\frac{5}{2} \).3. Substitute \( y = -\frac{5}{2} \) and \( z = \frac{7}{2} \) into Row 1: \( 2x + (-\frac{5}{2}) - \frac{7}{2} = -8 \rightarrow 2x = -8 + 6 \rightarrow x = -1 \).

Step 5: Write the Solution

The solution to the system is:\[x = -1, \quad y = -\frac{5}{2}, \quad z = \frac{7}{2}\]This is the unique solution, indicating the system is consistent.

Key Concepts

Augmented Matrix

To tackle a linear system of equations efficiently, we convert it into an augmented matrix. An augmented matrix is formed by appending the constants of each equation to the respective rows of the coefficients matrix of the variables.Here's how you can form an augmented matrix:

• Write down the coefficients of each variable in each equation.
• Include a vertical line or a different column to separate the constants from the variable coefficients.

For the given system, the augmented matrix looks like this:\[\begin{bmatrix} 2 & 1 & -1 & | & -8 \-1 & 1 & 1 & | & 3 \-2 & 0 & 4 & | & 18 \end{bmatrix}\]By representing the entire system in this compact form, calculations become systematic and computerizable. This method sets the stage for applying row operations to simplify the matrix.

Row-Echelon Form

The next step in solving a linear system using matrices is transforming the augmented matrix into a simpler format called row-echelon form. Row-echelon form consists of having a series of leading ones (1s) proceeding downwards like stair steps across the matrix. Here are key characteristics:

• Any row of all zeros, if present, is at the bottom.
• The leading entry of each non-zero row, called a pivot, is 1.
• Pivots are to the right of pivots in the rows above them.

To achieve this form, perform row operations like adding or subtracting multiples of rows. For this example, we'll eliminate the first variable from the second and third rows: 1. Add row 1 to row 2, making the new second row. 2. Add two times row 1 to row 3, forming the new third row. These operations bring the matrix closer to row-echelon form, simplifying future calculations.

Back Substitution

Once the augmented matrix is in row-echelon form, use back substitution to find the values of the variables. Back substitution involves solving the equations formed by the rows from bottom to top.Performing back substitution:

• Start with the bottom row; solve for the variable within that row.
• Substitute this solution into the row immediately above to find another unknown.
• Continue substituting upwards until all variables are known.

In the example, after simplifying, the last row might provide you with a simple equation like \(2z = 7\). Solve for \(z\) first, then substitute back into the second row to find \(y\), and finally use both \(y\) and \(z\) in the first row to find \(x\).Back substitution thus transforms an abstract matrix into specific solutions, completing the solution of the system.