Use Gauss-Jordan elimination to solve the following system of equations. If there is a solution, write your answer in the format $(w, x, y, z)$. \[ \left\{\begin{aligned} 3 w-2 y & =0 \\ w-x+4 z & =-15 \\ -3 w-4 y-z & =21 \\ x+y & =-2 \end{aligned}\right. \] Answer Keyboard Shortcuts Indicate the number of solutions to this system. State the solution, if one exists, and if there are infinitely many solutions, express the solution set in terms of one of the variables. Selecting an option will display any text boxes needed to complete your answer. No Solution One Solution Infinitely Many Solutions

Step 1 ：First, we write the system of equations in augmented matrix form. The augmented matrix is: \[\begin{bmatrix} 3 & 0 & -2 & 0 & 0 \\ 1 & -1 & 0 & 4 & -15 \\ -3 & 0 & -4 & -1 & 21 \\ 0 & 1 & 1 & 0 & -2 \end{bmatrix}\]

Step 2 ：Next, we swap row 1 and row 2 to get the leading 1 in the top left corner. The new matrix is: \[\begin{bmatrix} 1 & -1 & 0 & 4 & -15 \\ 3 & 0 & -2 & 0 & 0 \\ -3 & 0 & -4 & -1 & 21 \\ 0 & 1 & 1 & 0 & -2 \end{bmatrix}\]

Step 3 ：Then, we add row 1 to row 3 and subtract 3 times row 1 from row 2 to eliminate the coefficients of w in the second and third equations. The new matrix is: \[\begin{bmatrix} 1 & -1 & 0 & 4 & -15 \\ 0 & 3 & -2 & -12 & 45 \\ 0 & -1 & -4 & 3 & 6 \\ 0 & 1 & 1 & 0 & -2 \end{bmatrix}\]

Step 4 ：Next, we add row 2 to row 3 and row 4 to row 2 to eliminate the coefficients of x in the second and third equations. The new matrix is: \[\begin{bmatrix} 1 & 0 & 0 & 4 & -17 \\ 0 & 4 & -1 & -12 & 43 \\ 0 & 0 & -6 & -9 & 51 \\ 0 & 1 & 1 & 0 & -2 \end{bmatrix}\]

Step 5 ：Then, we divide row 2 by 4 and row 3 by -6 to make the leading coefficients of x and y to be 1. The new matrix is: \[\begin{bmatrix} 1 & 0 & 0 & 4 & -17 \\ 0 & 1 & -0.25 & -3 & 10.75 \\ 0 & 0 & 1 & 1.5 & -8.5 \\ 0 & 1 & 1 & 0 & -2 \end{bmatrix}\]

Step 6 ：Next, we subtract row 2 from row 4 to eliminate the coefficient of x in the fourth equation. The new matrix is: \[\begin{bmatrix} 1 & 0 & 0 & 4 & -17 \\ 0 & 1 & -0.25 & -3 & 10.75 \\ 0 & 0 & 1 & 1.5 & -8.5 \\ 0 & 0 & 1.25 & 3 & -12.75 \end{bmatrix}\]

Step 7 ：Then, we subtract row 3 from row 4 to eliminate the coefficient of y in the fourth equation. The new matrix is: \[\begin{bmatrix} 1 & 0 & 0 & 4 & -17 \\ 0 & 1 & 0 & -4.5 & 13 \\ 0 & 0 & 1 & 1.5 & -8.5 \\ 0 & 0 & 0 & 1.5 & -4.25 \end{bmatrix}\]

Step 8 ：Finally, we divide row 4 by 1.5 to make the leading coefficient of z to be 1. The new matrix is: \[\begin{bmatrix} 1 & 0 & 0 & 4 & -17 \\ 0 & 1 & 0 & -3 & 8.67 \\ 0 & 0 & 1 & 1 & -5.67 \\ 0 & 0 & 0 & 1 & -2.83 \end{bmatrix}\]

Step 9 ：From the final matrix, we can read off the solutions: \(w = -17 - 4z\), \(x = 8.67 + 3z\), \(y = -5.67 - z\), and \(z = -2.83\). Substituting \(z = -2.83\) into the other equations, we get \(w = 4.32\), \(x = 0.17\), and \(y = -2.84\).

Step 10 ：So, the solution to the system of equations is \(\boxed{(w, x, y, z) = (4.32, 0.17, -2.84, -2.83)}\).