Solve the following system of equations using Gaussian elimination or Gauss-Jordan elimination. Use a graphing calculator to check your answer. a+b-c=5 a-b+c=7 3a+b-c=-6 Select the correct choice below and, if necessary fill in the answer boxes to complete your choice. A. There is one solution. The solutior-is (square ,square ,square ) (Type integers or simplified fractions.) B. There are infinitely many solutions of the form (square ,square ,z) , where z is any real number. (Simplify your answers. Type expressions using z as the variable.) C. There is no solution

To solve the system of equations using Gaussian elimination, we first write the augmented matrix for the system:<br /><br />\[<br />\begin{bmatrix}<br />1 & 1 & -1 & | & 5 \\<br />1 & -1 & 1 & | & 7 \\<br />3 & 1 & -1 & | & -6<br />\end{bmatrix}<br />\]<br /><br />Step 1: Use row operations to get a leading 1 in the first row, first column (which is already done).<br /><br />Step 2: Eliminate the first column entries below the leading 1 by subtracting appropriate multiples of the first row from the other rows.<br /><br />- Subtract the first row from the second row:<br /> \[<br /> R_2 = R_2 - R_1 \rightarrow [0, -2, 2, |, 2]<br /> \]<br /><br />- Subtract 3 times the first row from the third row:<br /> \[<br /> R_3 = R_3 - 3R_1 \rightarrow [0, -2, 2, |, -21]<br /> \]<br /><br />The matrix now looks like this:<br /><br />\[<br />\begin{bmatrix}<br />1 & 1 & -1 & | & 5 \\<br />0 & -2 & 2 & | & 2 \\<br />0 & -2 & 2 & | & -21<br />\end{bmatrix}<br />\]<br /><br />Step 3: Use row operations to make the second row, second column entry a leading 1. Divide the second row by -2:<br /><br />\[<br />R_2 = \frac{1}{-2}R_2 \rightarrow [0, 1, -1, |, -1]<br />\]<br /><br />The matrix now looks like this:<br /><br />\[<br />\begin{bmatrix}<br />1 & 1 & -1 & | & 5 \\<br />0 & 1 & -1 & | & -1 \\<br />0 & -2 & 2 & | & -21<br />\end{bmatrix}<br />\]<br /><br />Step 4: Eliminate the second column entries above and below the leading 1 in the second row.<br /><br />- Subtract the second row from the first row:<br /> \[<br /> R_1 = R_1 - R_2 \rightarrow [1, 0, 0, |, 6]<br /> \]<br /><br />- Add 2 times the second row to the third row:<br /> \[<br /> R_3 = R_3 + 2R_2 \rightarrow [0, 0, 0, |, -23]<br /> \]<br /><br />The matrix now looks like this:<br /><br />\[<br />\begin{bmatrix}<br />1 & 0 & 0 & | & 6 \\<br />0 & 1 & -1 & | & -1 \\<br />0 & 0 & 0 & | & -23<br />\end{bmatrix}<br />\]<br /><br />Since the last row corresponds to the equation \(0 = -23\), which is a contradiction, there is no solution to the system of equations.<br /><br />Therefore, the correct choice is:<br /><br />C. There is no solution.