Answered

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.



Answer :

GinnyAnswer
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]

Other Questions