Solve the following system of simultaneous equations for \(x\). \(-2.0 x_{1}+5.0 x_{2}+1.0 x_{3}+3.0 x_{4}+4.0 x_{5}-1.0 x_{6}=0.0\) \(2.0 x_{1}-1.0 x_{2}-5.0 x_{3}-2.0 x_{4}+6.0 x_{5}+4.0 x_{6}=1.0\) \(-1.0 x_{1}+6.0 x_{2}-4.0 x_{3}-5.0 x_{4}+3.0 x_{5}-1.0 x_{6}=-6.0\) \(4.0 x_{1}+3.0 x_{2}-6.0 x_{3}-5.0 x_{4}-2.0 x_{5}-2.0 x_{6}=10.0\) \(-3.0 x_{1}+6.0 x_{2}+4.0 x_{3}+2.0 x_{4}-6.0 x_{5}+4.0 x_{6}=-6.0\) \(2.0 x_{1}+4.0 x_{2}+4.0 x_{3}+4.0 x_{4}+5.0 x_{5}-4.0 x_{6}=-2.0\)

Write the given system of equations as an augmented matrix: \[ \begin{bmatrix} -2 & 5 & 1 & 3 & 4 & -1 & 0 \\ 2 & -1 & -5 & -2 & 6 & 4 & 1 \\ -1 & 6 & -4 & -5 & 3 & -1 & -6 \\ 4 & 3 & -6 & -5 & -2 & -2 & 10 \\ -3 & 6 & 4 & 2 & -6 & 4 & -6 \\ 2 & 4 & 4 & 4 & 5 & -4 & -2 \\ \end{bmatrix} \]

Perform row operations to transform the matrix into reduced row-echelon form. We'll begin by making the element in the top-left corner equal to 1: Swap Row 1 and Row 4: \[ \begin{bmatrix} 4 & 3 & -6 & -5 & -2 & -2 & 10 \\ 2 & -1 & -5 & -2 & 6 & 4 & 1 \\ -1 & 6 & -4 & -5 & 3 & -1 & -6 \\ -2 & 5 & 1 & 3 & 4 & -1 & 0 \\ -3 & 6 & 4 & 2 & -6 & 4 & -6 \\ 2 & 4 & 4 & 4 & 5 & -4 & -2 \\ \end{bmatrix} \] Divide Row 1 by 4: \[ \begin{bmatrix} 1 & \frac{3}{4} & -\frac{3}{2} & -\frac{5}{4} & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ 2 & -1 & -5 & -2 & 6 & 4 & 1 \\ -1 & 6 & -4 & -5 & 3 & -1 & -6 \\ -2 & 5 & 1 & 3 & 4 & -1 & 0 \\ -3 & 6 & 4 & 2 & -6 & 4 & -6 \\ 2 & 4 & 4 & 4 & 5 & -4 & -2 \\ \end{bmatrix} \] Now, use the 1 in the top-left corner to eliminate the numbers below it: Add 2 times Row 1 to Row 2: \[ \begin{bmatrix} 1 & \frac{3}{4} & -\frac{3}{2} & -\frac{5}{4} & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ 0 & 0 & -4 & 0 & 5 & 3 & -4 \\ -1 & 6 & -4 & -5 & 3 & -1 & -6 \\ -2 & 5 & 1 & 3 & 4 & -1 & 0 \\ -3 & 6 & 4 & 2 & -6 & 4 & -6 \\ 2 & 4 & 4 & 4 & 5 & -4 & -2 \\ \end{bmatrix} \] Add 1 times Row 1 to Row 3: \[ \begin{bmatrix} 1 & \frac{3}{4} & -\frac{3}{2} & -\frac{5}{4} & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ 0 & 0 & -4 & 0 & 5 & 3 & -4 \\ 0 & \frac{27}{4} & -\frac{7}{2} & -\frac{9}{4} & 1 & -\frac{1}{2} & -\frac{7}{2} \\ -2 & 5 & 1 & 3 & 4 & -1 & 0 \\ -3 & 6 & 4 & 2 & -6 & 4 & -6 \\ 2 & 4 & 4 & 4 & 5 & -4 & -2 \\ \end{bmatrix} \] Add 2 times Row 1 to Row 4: \[ \begin{bmatrix} 1 & \frac{3}{4} & -\frac{3}{2} & -\frac{5}{4} & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ 0 & 0 & -4 & 0 & 5 & 3 & -4 \\ 0 & \frac{27}{4} & -\frac{7}{2} & -\frac{9}{4} & 1 & -\frac{1}{2} & -\frac{7}{2} \\ 0 & 6 & 0 & 4 & 3 & 0 & 5 \\ -3 & 6 & 4 & 2 & -6 & 4 & -6 \\ 2 & 4 & 4 & 4 & 5 & -4 & -2 \\ \end{bmatrix} \] Add 3 times Row 1 to Row 5: \[ \begin{bmatrix} 1 & \frac{3}{4} & -\frac{3}{2} & -\frac{5}{4} & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ 0 & 0 & -4 & 0 & 5 & 3 & -4 \\ 0 & \frac{27}{4} & -\frac{7}{2} & -\frac{9}{4} & 1 & -\frac{1}{2} & -\frac{7}{2} \\ 0 & 6 & 0 & 4 & 3 & 0 & 5 \\ 0 & \frac{15}{4} & -\frac{5}{2} & \frac{7}{4} & -\frac{9}{2} & \frac{5}{2} & -\frac{17}{2} \\ 2 & 4 & 4 & 4 & 5 & -4 & -2 \\ \end{bmatrix} \] Subtract 2 times Row 1 from Row 6: \[ \begin{bmatrix} 1 & \frac{3}{4} & -\frac{3}{2} & -\frac{5}{4} & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ 0 & 0 & -4 & 0 & 5 & 3 & -4 \\ 0 & \frac{27}{4} & -\frac{7}{2} & -\frac{9}{4} & 1 & -\frac{1}{2} & -\frac{7}{2} \\ 0 & 6 & 0 & 4 & 3 & 0 & 5 \\ 0 & \frac{15}{4} & -\frac{5}{2} & \frac{7}{4} & -\frac{9}{2} & \frac{5}{2} & -\frac{17}{2} \\ 0 & \frac{13}{4} & 5 & \frac{11}{4} & \frac{9}{2} & -\frac{11}{2} & -\frac{5}{2} \\ \end{bmatrix} \] Next, we will attempt to make the element in the second row and second column equal to 1: We can't divide the second row by any scalar to achieve this. However, we can multiply the second row by 2 and then add it to the third row. Add 2 times Row 2 to Row 3: \[ \begin{bmatrix} 1 & \frac{3}{4} & -\frac{3}{2} & -\frac{5}{4} & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ 0 & 0 & -4 & 0 & 5 & 3 & -4 \\ 0 & \frac{27}{4} & -8 & -\frac{9}{4} & 11 & \frac{5}{2} & -\frac{15}{2} \\ 0 & 6 & 0 & 4 & 3 & 0 & 5 \\ 0 & \frac{15}{4} & -\frac{5}{2} & \frac{7}{4} & -\frac{9}{2} & \frac{5}{2} & -\frac{17}{2} \\ 0 & \frac{13}{4} & 5 & \frac{11}{4} & \frac{9}{2} & -\frac{11}{2} & -\frac{5}{2} \\ \end{bmatrix} \] This is where we encounter a problem. The third row has non-zero elements in the third and higher columns, but its second column has a 0. This means the system of equations does not have a unique solution. We can't use Gaussian elimination to find the solution to this system of equations.