Problem 43 - A Graphical Approach to Precalculus with Limits

Question

Use row operations on an augmented matrix to solve each system of equations. Round to nearest thousandth when appropriate. $$\begin{aligned} x+y-z &=6 \\ 2 x-y+z &=-9 \\ x-2 y+3 z &=1 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = -1\), \(y = 23\), \(z = 16\).

Step 1: Write the Augmented Matrix

Convert the given system of equations into an augmented matrix. The system of equations\[\begin{align*}x + y - z &= 6 \2x - y + z &= -9 \x - 2y + 3z &= 1\end{align*}\]becomes\[\begin{bmatrix}1 & 1 & -1 & | & 6 \2 & -1 & 1 & | & -9 \1 & -2 & 3 & | & 1\end{bmatrix}.\]

Step 2: Obtain a Leading 1 in the First Row, First Column

The first row already has a leading 1 in the first column, so you don't need to make any changes to the first row. Keep the matrix as it is:\[\begin{bmatrix}1 & 1 & -1 & | & 6 \2 & -1 & 1 & | & -9 \1 & -2 & 3 & | & 1\end{bmatrix}.\]

Step 3: Eliminate First Column Entries Below the Leading 1

Subtract 2 times the first row from the second row to eliminate the 2 in the second row:\[R_2 = R_2 - 2R_1\].The new second row becomes:\[0 & -3 & 3 & | & -21\].Similarly, subtract the first row from the third row to eliminate the 1 in the third row:\[R_3 = R_3 - R_1\].The new third row becomes:\[0 & -3 & 4 & | & -5\].The matrix is now:\[\begin{bmatrix}1 & 1 & -1 & | & 6 \0 & -3 & 3 & | & -21 \0 & -3 & 4 & | & -5\end{bmatrix}.\]

Step 4: Obtain a Leading 1 in the Second Row, Second Column

Divide the second row by -3 to turn the leading entry into 1:\[R_2 = \frac{1}{-3}R_2\],resulting in:\[0 & 1 & -1 & | & 7\].The matrix becomes:\[\begin{bmatrix}1 & 1 & -1 & | & 6 \0 & 1 & -1 & | & 7 \0 & -3 & 4 & | & -5\end{bmatrix}.\]

Step 5: Eliminate the Second Column Entry in the Third Row

Add 3 times the second row to the third row to eliminate the -3 in the third row:\[R_3 = R_3 + 3R_2\].The new third row becomes:\[0 & 0 & 1 & | & 16\].Now the matrix is:\[\begin{bmatrix}1 & 1 & -1 & | & 6 \0 & 1 & -1 & | & 7 \0 & 0 & 1 & | & 16\end{bmatrix}.\]

Step 6: Back Substitution to Solve the System

Starting with the third row, which gives:\[z = 16\].Substitute \(z = 16\) into the second row:\[y - 16 = 7\], therefore,\[y = 23\].Substitute \(y = 23, z = 16\) into the first row:\[x + 23 - 16 = 6\], therefore,\[x = -1\].The solution for the system is \(x = -1\), \(y = 23\), \(z = 16\).

Key Concepts

Row Operations

Row operations are techniques we use to manipulate rows in an augmented matrix to make solving a system of linear equations easier. These operations do not change the solutions of the system.
There are three main types of row operations we can apply:

• Swapping two rows: Sometimes we swap rows to get a better position for a leading entry.
• Multiplying a row by a non-zero scalar: This is useful to turn entries into ones or zeroes.
• Adding or subtracting multiples of one row to another: This is critical in eliminating terms and working towards a reduced form.

In our exercise, we used these operations to manipulate the augmented matrix until it was in a form that allowed for straightforward back substitution.

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. The objective is to find values for these unknowns that satisfy all equations in the system simultaneously.
For example, in our original exercise, we have three equations with three variables:

• \(x + y - z = 6\)
• \(2x - y + z = -9\)
• \(x - 2y + 3z = 1\)

Solving these equations involves finding the values of \(x, y,\) and \(z\) that work for all three equations. By converting the system into an augmented matrix and applying row operations, we reduce it to a much simpler form to solve.

Back Substitution

Back substitution is a method used to find the solutions of a system of equations once it has been transformed into an upper triangular or row-echelon form using row operations.
With the system in this form, as was the case in our exercise, solving becomes straightforward:

• Start with the bottommost equation (from the reduced matrix) to solve for one variable since it typically contains only one unknown.
• Use the solution of that variable and substitute back into the equation above to solve for another variable.
• Continue substituting known values back into higher equations until all variables are found.

In our case, we started with \(z = 16\) from the third row, used it to find \(y = 23\) from the second row, and finally substituted these into the first row to get \(x = -1\).

Leading Entry

The leading entry in a row is the first non-zero number from the left in that row. It plays a crucial role in performing row operations effectively.
Here's why leading entries are important:

• They guide which row operations to perform, particularly when eliminating other entries in the matrix.
• Having a leading entry of 1 makes calculations easier, which is why it's a common practice to turn these elements into ones where possible.
• They help in checking the progress of transforming the matrix towards an upper triangular or reduced row-echelon form.

In our exercise, we focused on obtaining leading 1s by dividing the entire row by the leading number, as seen in the second row where we turned a -3 into 1 by dividing by -3. This strategy ensures smoother operations as we continue simplifying the matrix.