Questions: Solve using Gauss-Jordan elimination. 2x1 - x2 - 5x3 = 13 x1 - 3x2 = 19 Select the correct choice below and fill in the answer box(es) within your choice. A. The unique solution is x1= , x2= , and x3= 7. B. The system has infinitely many solutions. The solution is x1= , x2= , and x3=t (Simplify your answers. Type expressions using t as the variable.) C. The system has infinitely many solutions. The solution is x1=, x2=s, and x3=t. (Simplify your answer. Type an expression using s and t as the variables.) D. There is no solution.

Solution Steps

To solve the given system of equations using Gauss-Jordan elimination, we first represent the system as an augmented matrix. Then, we perform row operations to transform the matrix into reduced row-echelon form (RREF). From the RREF, we can determine the solution to the system, whether it is a unique solution, infinitely many solutions, or no solution.

To solve the given system of equations using Gauss-Jordan elimination, we will first convert the system into an augmented matrix and then perform row operations to reach the reduced row-echelon form.

The given system of equations is:

\[ \begin{aligned} 2x_1 - x_2 - 5x_3 &= 13 \\ x_1 - 3x_2 &= 19 \end{aligned} \]

The augmented matrix for this system is:

\[ \begin{bmatrix} 2 & -1 & -5 & | & 13 \\ 1 & -3 & 0 & | & 19 \end{bmatrix} \]

We will perform row operations to convert the matrix into reduced row-echelon form.

1. Make the leading coefficient of the first row 1 (if necessary, but here it is already 1 in the second row, so we will swap the rows):

   \[ \begin{bmatrix} 1 & -3 & 0 & | & 19 \\ 2 & -1 & -5 & | & 13 \end{bmatrix} \]

2. Eliminate the first element of the second row by replacing the second row with the second row minus 2 times the first row:

   \[ R_2 = R_2 - 2R_1 \]

   \[ \begin{bmatrix} 1 & -3 & 0 & | & 19 \\ 0 & 5 & -5 & | & -25 \end{bmatrix} \]

3. Make the leading coefficient of the second row 1 by dividing the entire second row by 5:

   \[ R_2 = \frac{1}{5}R_2 \]

   \[ \begin{bmatrix} 1 & -3 & 0 & | & 19 \\ 0 & 1 & -1 & | & -5 \end{bmatrix} \]

4. Eliminate the second element of the first row by replacing the first row with the first row plus 3 times the second row:

   \[ R_1 = R_1 + 3R_2 \]

   \[ \begin{bmatrix} 1 & 0 & -3 & | & 4 \\ 0 & 1 & -1 & | & -5 \end{bmatrix} \]

The matrix is now in reduced row-echelon form, which corresponds to the following system of equations:

\[ \begin{aligned} x_1 - 3x_3 &= 4 \\ x_2 - x_3 &= -5 \end{aligned} \]

From the above equations, we can express \(x_1\) and \(x_2\) in terms of \(x_3\):

\[ \begin{aligned} x_1 &= 4 + 3x_3 \\ x_2 &= -5 + x_3 \end{aligned} \]

Let \(x_3 = t\), where \(t\) is a parameter. Then:

\[ \begin{aligned} x_1 &= 4 + 3t \\ x_2 &= -5 + t \\ x_3 &= t \end{aligned} \]

Final Answer