Lesson 38: Solve \[$3 \times 3\$\] SystemsAfter Steps In The Process Of Simplifying The Given Matrix To Row-echelon Form, Fill In The Missing Numbers In The Steps Shown.$\[ \begin{aligned} &\left[\begin{array}{rrr|r} 1 & 2 & -2 & -8 \\ 2 &
Introduction
In this lesson, we will learn how to solve 3x3 systems of linear equations using the method of row-echelon form. We will start with a given matrix and simplify it to row-echelon form, and then fill in the missing numbers in the steps shown.
Step 1: Write the Augmented Matrix
The first step in solving a 3x3 system of linear equations is to write the augmented matrix. The augmented matrix is a matrix that includes the coefficients of the variables and the constant terms.
${
\begin{aligned}
&\left[\begin{array}{rrr|r}
1 & 2 & -2 & -8 \\
2 & 3 & -1 & -5 \\
4 & 5 & -3 & -12 \\
\end{array}\right]
}$
$
Step 2: Perform Row Operations
The next step is to perform row operations to simplify the matrix to row-echelon form. We will start by making the first column have zeros below the first row.
Step 2.1: Multiply Row 1 by -2 and Add to Row 2
To make the first column have zeros below the first row, we will multiply row 1 by -2 and add it to row 2.
${
\begin{aligned}
&\left[\begin{array}{rrr|r}
1 & 2 & -2 & -8 \\
0 & -1 & 3 & 11 \\
4 & 5 & -3 & -12 \\
\end{array}\right]
}$
$
Step 2.2: Multiply Row 1 by -4 and Add to Row 3
Next, we will multiply row 1 by -4 and add it to row 3.
${
\begin{aligned}
&\left[\begin{array}{rrr|r}
1 & 2 & -2 & -8 \\
0 & -1 & 3 & 11 \\
0 & -3 & 5 & 0 \\
\end{array}\right]
}$
$
Step 3: Continue Performing Row Operations
We will continue performing row operations to simplify the matrix to row-echelon form.
Step 3.1: Multiply Row 2 by -1
To make the second column have zeros below the second row, we will multiply row 2 by -1.
${
\begin{aligned}
&\left[\begin{array}{rrr|r}
1 & 2 & -2 & -8 \\
0 & 1 & -3 & -11 \\
0 & -3 & 5 & 0 \\
\end{array}\right]
}$
$
Step 3.2: Add 3 Times Row 2 to Row 3
Next, we will add 3 times row 2 to row 3.
${
\begin{aligned}
&\left[\begin{array}{rrr|r}
1 & 2 & -2 & -8 \\
0 & 1 & -3 & -11 \\
0 & 0 & -4 & 33 \\
\end{array}\right]
}$
$
Step 4: Write the System of Linear Equations
Now that we have simplified the matrix to row-echelon form, we can write the system of linear equations.
${
\begin{aligned}
x + 2y - 2z &= -8 \\
y - 3z &= -11 \\
-4z &= 33 \\
\end{aligned}
}$
$
Step 5: Solve the System of Linear Equations
We can now solve the system of linear equations using back-substitution.
Step 5.1: Solve for z
We will start by solving for z.
${
\begin{aligned}
-4z &= 33 \\
z &= -\frac{33}{4} \\
\end{aligned}
}$
$
Step 5.2: Solve for y
Next, we will solve for y.
${
\begin{aligned}
y - 3z &= -11 \\
y + \frac{99}{4} &= -11 \\
y &= -11 - \frac{99}{4} \\
y &= -\frac{44}{4} - \frac{99}{4} \\
y &= -\frac{143}{4} \\
\end{aligned}
}$
$
Step 5.3: Solve for x
Finally, we will solve for x.
${
\begin{aligned}
x + 2y - 2z &= -8 \\
x + 2\left(-\frac{143}{4}\right) - 2\left(-\frac{33}{4}\right) &= -8 \\
x - \frac{286}{4} + \frac{66}{4} &= -8 \\
x - \frac{220}{4} &= -8 \\
x &= -8 + \frac{220}{4} \\
x &= -8 + 55 \\
x &= 47 \\
\end{aligned}
}$
$
Conclusion
Introduction
In this lesson, we learned how to solve 3x3 systems of linear equations using the method of row-echelon form. We started with a given matrix and simplified it to row-echelon form, and then filled in the missing numbers in the steps shown. We also wrote the system of linear equations and solved it using back-substitution. In this Q&A section, we will answer some common questions related to solving 3x3 systems of linear equations.
Q: What is the purpose of row-echelon form?
A: The purpose of row-echelon form is to simplify the matrix to a form where it is easier to solve the system of linear equations. Row-echelon form is a way of organizing the matrix so that the variables can be solved for one at a time.
Q: How do I know when to stop performing row operations?
A: You know when to stop performing row operations when the matrix is in row-echelon form. Row-echelon form is characterized by having zeros below the leading entry in each row.
Q: What is the leading entry in a row?
A: The leading entry in a row is the first non-zero entry in the row. It is the entry that is used to eliminate the other entries in the row.
Q: How do I solve for the variables using back-substitution?
A: To solve for the variables using back-substitution, you start with the last row of the matrix and work your way up to the first row. You solve for the last variable first, and then use that value to solve for the next variable, and so on.
Q: What if I have a system of linear equations with no solution?
A: If you have a system of linear equations with no solution, it means that the equations are inconsistent. This can happen if the matrix is not in row-echelon form, or if the system of linear equations is inconsistent.
Q: What if I have a system of linear equations with infinitely many solutions?
A: If you have a system of linear equations with infinitely many solutions, it means that the equations are dependent. This can happen if the matrix is not in row-echelon form, or if the system of linear equations is dependent.
Q: How do I determine if a system of linear equations has a unique solution, no solution, or infinitely many solutions?
A: To determine if a system of linear equations has a unique solution, no solution, or infinitely many solutions, you need to examine the matrix and the system of linear equations. If the matrix is in row-echelon form and the system of linear equations is consistent, then the system has a unique solution. If the matrix is not in row-echelon form or the system of linear equations is inconsistent, then the system has no solution. If the matrix is in row-echelon form and the system of linear equations is dependent, then the system has infinitely many solutions.
Conclusion
In this Q&A section, we answered some common questions related to solving 3x3 systems of linear equations. We discussed the purpose of row-echelon form, how to know when to stop performing row operations, and how to solve for the variables using back-substitution. We also discussed what to do if the system of linear equations has no solution or infinitely many solutions.