Answered

IDNLearn.com: Your reliable source for finding expert answers. Join our community to receive timely and reliable responses to your questions from knowledgeable professionals.

Use Gaussian elimination to find the complete solution to the following system of equations, or show that none exists.

[tex]\[
\begin{cases}
5x_1 + 12x_2 + 7x_3 = 11 \\
2x_1 + 5x_2 + 5x_3 = -2 \\
x_1 + 2x_2 - 3x_3 = 6
\end{cases}
\][/tex]

Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

A. The unique solution is [tex]\( x_1 = \ \square, x_2 = \ \square, x_3 = \ \square \)[/tex].

B. The system has infinitely many solutions. The solution is [tex]\( x_1 = \ \square, x_2 = \ \square, x_3 = t \)[/tex].
(Simplify your answer. Type an expression using [tex]\( t \)[/tex] as the variable.)

C. The system has infinitely many solutions. The solution is [tex]\( x_1 = \ \square, x_2 = s, x_3 = t \)[/tex].
(Simplify your answer. Type an expression using [tex]\( s \)[/tex] and [tex]\( t \)[/tex] as the variables.)

D. There is no solution.

Sagot :

GinnyAnsweridn
To solve the given system of equations using Gaussian elimination, we start by writing the augmented matrix for the system:

[tex]\[ \begin{pmatrix} 5 & 12 & 7 & | & 11 \\ 2 & 5 & 5 & | & -2 \\ 1 & 2 & -3 & | & 6 \end{pmatrix} \][/tex]

We will perform Gaussian elimination to try to reach the reduced row echelon form (RREF).

### Step 1: Form Row Echelon Form

1. First pivot: Use the (1,1) position as the first pivot to eliminate the entries in the first column below it.
[tex]\[ R2 = R2 - \left(\frac{2}{5}\right) R1 \quad \text{and} \quad R3 = R3 - \left(\frac{1}{5}\right) R1 \][/tex]
This gives:
[tex]\[ \begin{pmatrix} 5 & 12 & 7 & | & 11 \\ 0 & \left(5 - \frac{24}{5}\right) & \left(5 - \frac{14}{5}\right) & | & \left(-2 - \frac{22}{5}\right) \\ 0 & \left(2 - \frac{12}{5}\right) & \left(-3 - \frac{7}{5}\right) & | & \left(6 - \frac{11}{5}\right) \end{pmatrix} \][/tex]
Simplify the rows:
[tex]\[ R2: \quad 0, \quad \frac{1}{5}, \quad \frac{11}{5}, \quad \left| \quad \frac{-32}{5} \quad \right. \][/tex]
[tex]\[ R3: \quad 0, \quad \frac{-2}{5}, \quad - \frac{22}{5}, \quad \left| \quad \frac{19}{5} \quad \right. \][/tex]
Simplified matrix:
[tex]\[ \begin{pmatrix} 5 & 12 & 7 & | & 11 \\ 0 & \left(\frac{1}{5} \cdot 5\right) & \left(\frac{11}{5}\cdot 5\right) & | & \left(\frac{-32}{5}\cdot 5\right) \\ 0 & \left(\frac{-2}{5} \cdot 5\right) & \left(\frac{-22}{5} \cdot 5\right) & | & \left(\frac{19}{5} \cdot 5\right) \end{pmatrix} \][/tex]

[tex]\[ \begin{pmatrix} 5 & 12 & 7 & | & 11 \\ 0 & 1 & 11 & | & -6 \\ 0 & -2 & -22 & | & 19 \end{pmatrix} \][/tex]

### Step 2: Eliminate below the second row's pivot.

2. Second pivot: Use the (2,2) position as the pivot to eliminate the entries in the second column below it.
[tex]\[ R3 = R3 + 2R2 \][/tex]
Simplified matrix:
[tex]\[ \begin{pmatrix} 5 & 12 & 7 & | & 11 \\ 0 & 1 & 11 & | & -6 \\ 0 & 0 & 0 & | & 7 \end{pmatrix} \][/tex]

### Step 3: Interpretation.

Notice that after reaching the Row Echelon Form, our system has a resulting matrix row:

[tex]\[ 0 \quad 0 \quad 0 \quad \mid \quad 7 \][/tex]

Such a row indicates inconsistency. It translates to the equation:
[tex]\[ 0 = 7 \][/tex]

This contradicts any possible solution because such an equation is never true.

### Conclusion
Given this inconsistency, we conclude that the system of equations has no solution. Thus, we select:

[tex]\[ \boxed{\text{D. There is no solution.}} \][/tex]
Your engagement is important to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and come back for more insightful information.