Solve the system of equations using Gaussian elimination or Gauss-Jordan elimination. Use a graphing calculator to check your answer. 2x-11y-3z=3 x+2y+z=-1 6x-9y-z=1 Select the correct choice below and, if necessary.fill in the answer box(es) to complete your choice. A. There is one solution. The solution is (square cdot square cdot square ) (Type integers or simplified fractions.) B. There are infinitely many solutions. The solutions are of the form (square ,square ,z) , where z is any real number. square (Simplify your answers. Use integers or fractions for any numbers in the expression.) C. There is no solution.

To solve the system of equations using Gaussian elimination or Gauss-Jordan elimination, we first write the augmented matrix for the system:<br /><br />\[<br />\begin{pmatrix}<br />2 & -11 & -3 & | & 3 \\<br />1 & 2 & 1 & | & -1 \\<br />6 & -9 & -1 & | & 1<br />\end{pmatrix}<br />\]<br /><br />We will perform row operations to transform this matrix into reduced row echelon form.<br /><br />Step 1: Make the leading coefficient of the first row 1 (if it is not already). In this case, it is already 1 in the second row, so we can swap the first and second rows:<br /><br />\[<br />\begin{pmatrix}<br />1 & 2 & 1 & | & -1 \\<br />2 & -11 & -3 & | & 3 \\<br />6 & -9 & -1 & | & 1<br />\end{pmatrix}<br />\]<br /><br />Step 2: Eliminate the x-term from the second and third rows by subtracting appropriate multiples of the first row:<br /><br />Row 2 = Row 2 - 2 * Row 1:<br />\[<br />\begin{pmatrix}<br />1 & 2 & 1 & | & -1 \\<br />0 & -15 & -5 & | & 5 \\<br />6 & -9 & -1 & | & 1<br />\end{pmatrix}<br />\]<br /><br />Row 3 = Row 3 - 6 * Row 1:<br />\[<br />\begin{pmatrix}<br />1 & 2 & 1 & | & -1 \\<br />0 & -15 & -5 & | & 5 \\<br />0 & -21 & -7 & | & 7<br />\end{pmatrix}<br />\]<br /><br />Step 3: Make the leading coefficient of the second row 1 by dividing the entire row by -15:<br /><br />\[<br />\begin{pmatrix}<br />1 & 2 & 1 & | & -1 \\<br />0 & 1 & \frac{1}{3} & | & -\frac{1}{3} \\<br />0 & -21 & -7 & | & 7<br />\end{pmatrix}<br />\]<br /><br />Step 4: Eliminate the y-term from the first and third rows by adding appropriate multiples of the second row:<br /><br />Row 1 = Row 1 - 2 * Row 2:<br />\[<br />\begin{pmatrix}<br />1 & 0 & \frac{1}{3} & | & -\frac{1}{3} \\<br />0 & 1 & \frac{1}{3} & | & -\frac{1}{3} \\<br />0 & -21 & -7 & | & 7<br />\end{pmatrix}<br />\]<br /><br />Row 3 = Row 3 + 21 * Row 2:<br />\[<br />\begin{pmatrix}<br />1 & 0 & \frac{1}{3} & | & -\frac{1}{3} \\<br />0 & 1 & \frac{1}{3} & | & -\frac{1}{3} \\<br />0 & 0 & 0 & | & 0<br />\end{pmatrix}<br />\]<br /><br />Step 5: Make the leading coefficient of the third row 1 (if necessary) and eliminate the z-term from the first and second rows. However, since the third row is all zeros, we can skip this step.<br /><br />The resulting matrix corresponds to the following system of equations:<br /><br />\[<br />\begin{cases}<br />x + \frac{1}{3}z = -\frac{1}{3} \\<br />y + \frac{1}{3}z = -\frac{1}{3} \\<br />0 = 0<br />\end{cases}<br />\]<br /><br />This indicates that there are infinitely many solutions. We can express \(x\) and \(y\) in terms of \(z\):<br /><br />\[<br />x = -\frac{1}{3} - \frac{1}{3}z<br />\]<br />\[<br />y = -\frac{1}{3} - \frac{1}{3}z<br />\]<br /><br />Thus, the solutions are of the form \((-\frac{1}{3} - \frac{1}{3}z, -\frac{1}{3} - \frac{1}{3}z, z)\), where \(z\) is any real number.<br /><br />Therefore, the correct choice is:<br /><br />B. There are infinitely many solutions. The solutions are of the form \((- \frac{1}{3} - \frac{1}{3}z, - \frac{1}{3} - \frac{1}{3}z, z)\), where \(z\) is any real number.