Problem 23 - Precalculus Mathematics for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{array}{rr} x+2 y-z= & -2 \\ x & +z=0 \\ 2 x-y-z= & -3 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{3}{2}\), \(y = -\frac{5}{2}\), \(z = -\frac{3}{2}\).

Step 1: Write the Augmented Matrix

First, we'll represent the system of equations as an augmented matrix. The system is:\[\begin{align*}x + 2y - z &= -2 \x + 0y + z &= 0 \2x - y - z &= -3\end{align*}\]This can be written as the augmented matrix:\[\begin{bmatrix}1 & 2 & -1 & | & -2 \1 & 0 & 1 & | & 0 \2 & -1 & -1 & | & -3\end{bmatrix}\]

Step 2: Eliminate First Column Below Pivot

Use the first row to eliminate the first column for the second and third rows. Subtract the first row from the second row, and subtract twice the first row from the third row.New matrix:\[\begin{bmatrix}1 & 2 & -1 & | & -2 \0 & -2 & 2 & | & 2 \0 & -5 & 1 & | & 1\end{bmatrix}\]

Step 3: Eliminate Second Column Below Pivot

Now use the second row to eliminate the second column of the third row by adding \(-\frac{5}{2}\) times the second row to the third row.New matrix:\[\begin{bmatrix}1 & 2 & -1 & | & -2 \0 & -2 & 2 & | & 2 \0 & 0 & -4 & | & 6\end{bmatrix}\]

Step 4: Back Substitution

Start with the last row, which reads \(-4z = 6\). Solve for \(z\). Then continue upwards to solve for \(y\) and \(x\).- From \(-4z = 6\): \(z = -\frac{3}{2}\)- Substitute \(z\) in the second row: \(-2y + 2(-\frac{3}{2}) = 2 \rightarrow -2y - 3 = 2 \rightarrow -2y = 5 \rightarrow y = -\frac{5}{2}\)- Substitute \(y\) and \(z\) into the first row: \(x + 2(-\frac{5}{2}) - (-\frac{3}{2}) = -2 \rightarrow x - 5 + \frac{3}{2} = -2 \rightarrow x - \frac{7}{2} = -2 \rightarrow x = \frac{3}{2}\)

Step 5: Verify Solution

Plug the values \(x = \frac{3}{2}\), \(y = -\frac{5}{2}\), and \(z = -\frac{3}{2}\) back into the original equations to confirm they satisfy all equations. This is important to ensure that no arithmetic errors were made in the simplification steps.

Key Concepts

Gauss-Jordan Elimination

Gauss-Jordan elimination is an essential method for solving systems of linear equations. It transforms the system into a simpler form by manipulating the rows of its augmented matrix. Unlike Gaussian elimination, which only aims for an upper triangular form, Gauss-Jordan goes further by reducing the matrix to its row-reduced echelon form (RREF). This means:

• Leading coefficients (also called pivots) are 1.
• Each pivot is the only non-zero entry in its column.

The ultimate goal is to simplify the system so each variable can be read directly from the matrix, offering a clear path to the solution. This process requires successive row operations, including:

• Swapping two rows.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting a multiple of one row to another.

Through these steps, the augmented matrix morphs into a series of simple linear equations making back substitution straightforward.

Augmented Matrix

An augmented matrix is a convenient tool for representing a system of linear equations. It combines both the coefficients of the variables and the constants from the right-hand side of the equations into a single matrix form. For example, for the system:

• \(x + 2y - z = -2\)
• \(x + 0y + z = 0\)
• \(2x - y - z = -3\)

the augmented matrix is written as:\[\begin{bmatrix}1 & 2 & -1 & | & -2 \1 & 0 & 1 & | & 0 \2 & -1 & -1 & | & -3\end{bmatrix}\]The vertical line separates the coefficients of the variables from the constants. This matrix form is advantageous because it provides a compact and structured way to work with the equations simultaneously, facilitating operations like row reduction used in Gauss-Jordan elimination.

Back Substitution

Back substitution is the final step after the matrix has been transformed, typically using Gaussian elimination, into an upper triangular form. In this stage, solutions for the variables are determined starting from the last row and working upwards systematically. This process involves:

• Solving for the last variable directly from its respective equation.
• Substituting this value into the preceding equations, one by one, to solve for the other variables.

Using the given example, once the matrix is simplified to reveal the equation \(-4z = 6\), we solve for \(z\). With \(z = -\frac{3}{2}\), this value is then used in the next equation to find \(y\), and eventually, both values help determine \(x\). Back substitution effectively converts the abstract matrix manipulations into concrete values for each variable, completing the solution.