Problem 26 - Precalculus Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$\left\\{\begin{aligned} 2 y+z &=3 \\ 5 x+4 y+3 z &=-1 \\ x-3 y &=-2 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The system has a unique solution: \( (x, y, z) = \left( \frac{10}{19}, \frac{24}{19}, \frac{49}{19} \right) \).

Step 1: Write the system of equations in matrix form

Express the given system of equations as an augmented matrix. The equations are:\[\begin{aligned} 0x + 2y + z &= 3, \ 5x + 4y + 3z &= -1, \ x - 3y + 0z &= -2. \end{aligned}\] The augmented matrix is:\[\left[\begin{array}{ccc|c} 0 & 2 & 1 & 3 \ 5 & 4 & 3 & -1 \ 1 & -3 & 0 & -2 \\end{array}\right].\]

Step 2: Use row operations to obtain an upper triangular form

Perform row operations to simplify the matrix into an upper triangular form. 1. Swap rows 1 and 3 to get a non-zero leading coefficient in the first row. The matrix becomes: \[\begin{array}{ccc|c}1 & -3 & 0 & -2 \5 & 4 & 3 & -1 \0 & 2 & 1 & 3 \\end{array}\] 2. Subtract 5 times the first row from the second row to eliminate the first element in the second row.The result is:\[\begin{array}{ccc|c}1 & -3 & 0 & -2 \0 & 19 & 3 & 9 \0 & 2 & 1 & 3 \\end{array}\]

Step 3: Continue row operations for further simplification

Perform more row operations to achieve reduced row-echelon form.1. Subtract \(\left(\frac{2}{19}\right)\) times the second row from the third row to eliminate the second-row leading element in the third row.The matrix simplifies to:\[\begin{array}{ccc|c}1 & -3 & 0 & -2 \0 & 19 & 3 & 9 \0 & 0 & \frac{19}{19} & \frac{49}{19} \\end{array}\] 2. Scale the third row by its coefficient to make the diagonal leading coefficient 1.The matrix becomes:\[\begin{array}{ccc|c}1 & -3 & 0 & -2 \0 & 19 & 3 & 9 \0 & 0 & 1 & \frac{49}{19} \\end{array}\]

Step 4: Perform back substitution to find the solutions

Use back substitution to find values for each variable starting from the last row.1. Retrieve from Row 3: \( z = \frac{49}{19} \).2. Substitute \( z \) into Row 2 and solve for \( y \): \[ 0 \cdot x + 19y + 3 \cdot \frac{49}{19} = 9 \Rightarrow 19y + \frac{147}{19} = 9 \Rightarrow 19y = 9 - \frac{147}{19} \Rightarrow y = \frac{24}{19}. \]3. Substitute \( y \) and \( z \) into Row 1 and solve for \( x \): \[ x - 3 \cdot \frac{24}{19} + 0 \cdot \frac{49}{19} = -2 \Rightarrow x - \frac{72}{19} = -2 \Rightarrow x = -2 + \frac{72}{19} = \frac{10}{19}. \]

Step 5: Write the solution

The solution to the system of equations is the set of values found for \( x \), \( y \), and \( z \):\[( x, y, z ) = \left( \frac{10}{19}, \frac{24}{19}, \frac{49}{19} \right).\] The system is consistent and has a single unique solution.

Key Concepts

Augmented Matrix

In solving linear systems of equations, an important tool we use is the augmented matrix. An augmented matrix is essentially a compact way to represent a system of linear equations. What you do is take the coefficients of each variable from the equations and line them up in a matrix form. Plus, you tack on an extra column for the constants from the equations. This means, for a system like:

• \(0x + 2y + z = 3\),
• \(5x + 4y + 3z = -1\), and
• \(x - 3y = -2\)

The augmented matrix becomes:\[\begin{array}{ccc|c}0 & 2 & 1 & 3 \5 & 4 & 3 & -1 \1 & -3 & 0 & -2\\end{array}\]This matrix is now a much more manageable form to work with, allowing you to visually organize your computations as you work through the solution.

Row Operations

The magic of solving systems using the augmented matrix lies in row operations. These operations include:

• Switching two rows
• Multiplying a row by a nonzero scalar
• Adding or subtracting the product of a row by a scalar to another row

These operations are aimed at simplifying the matrix into an upper triangular form, which makes solving for each variable more straightforward. You start off by looking for a leading 1 in the top-left corner and try to turn the matrix into an upper triangular shape.

In our example, swapping rows to get \(1\) at the top-left corner and performing row operations like subtracting five times the first row from the second row helped in reducing the problem. As you can see, as the augmented matrix changes, it gets closer to revealing each variable's value. With these operations, each step is justified as they lead us closer to the final solution without changing the system's integrity.

Back Substitution

After performing enough row operations, your matrix should nicely sit in what math fans call an 'upper triangular form'. This is when you have zeroes all below the main diagonal in your matrix. Once you're there, it's time for back substitution.

In back substitution, you start from the bottom-most non-zero row and work upwards solving for each variable in terms of the ones below it. In our system:

• For the third row, you directly find \( z = \frac{49}{19} \).
• Next, substitute \( z \) in the second row to find \( y = \frac{24}{19} \).
• Finally, use the values of \( y \) and \( z \) in the first row to solve for \( x = \frac{10}{19} \).

Back substitution turns the reduced matrix into a full solution, translating the simplified results back into the context of the original equations, showing us the unique set of solutions for \( x, y, \) and \( z \).