Questions: Solve the system of equations using matrices. Use the Gaussian elimination method with back-substitution. x+y-z= -5 2x-y+z= 5 -x+3y-2z= 1 Use the Gaussian elimination method to obtain the matrix in row-echelon form. Choose the correct answer below. B. [1 1 -1 -5; 0 1 -1 -5; 0 0 1 16] C. [1 0 0 -5; -1 1 0 -5; -1 1 1 16] D. [1 1 -5 -1; 0 1 -5 -1; 0 0 1 16]

Solve the system of equations using matrices. Use the Gaussian elimination method with back-substitution.

x+y-z= -5
2x-y+z= 5
-x+3y-2z= 1

Use the Gaussian elimination method to obtain the matrix in row-echelon form. Choose the correct answer below.

B. [1 1 -1 -5; 0 1 -1 -5; 0 0 1 16]
C. [1 0 0 -5; -1 1 0 -5; -1 1 1 16]
D. [1 1 -5 -1; 0 1 -5 -1; 0 0 1 16]

Transcript text: Solve the system of equations using matrices. Use the Gaussian elimination method with back-substitution. \[ \left\{\begin{array}{rr} x+y-z= & -5 \\ 2 x-y+z= & 5 \\ -x+3 y-2 z= & 1 \end{array}\right. \] Use the Gaussian elimination method to obtain the matrix in row-echelon form. Choose the correct answer below. B. $\left[\begin{array}{rrr|r}1 & 1 & -1 & -5 \\ 0 & 1 & -1 & -5 \\ 0 & 0 & 1 & 16\end{array}\right]$ C. $\left[\begin{array}{rrr|r}1 & 0 & 0 & -5 \\ -1 & 1 & 0 & -5 \\ -1 & 1 & 1 & 16\end{array}\right]$ D. $\left[\begin{array}{rrr|r}1 & 1 & -5 & -1 \\ 0 & 1 & -5 & -1 \\ 0 & 0 & 1 & 16\end{array}\right]$

Solution

Solution Steps

To solve the system of equations using the Gaussian elimination method, we first represent the system as an augmented matrix. Then, we perform row operations to transform the matrix into row-echelon form. Finally, we use back-substitution to find the solution to the system. The correct row-echelon form of the matrix will be one of the given options.

Step 1: Formulate the Augmented Matrix

The given system of equations is: \[ \begin{align_} x + y - z &= -5 \\ 2x - y + z &= 5 \\ -x + 3y - 2z &= 1 \end{align_} \] We can represent this system as an augmented matrix: \[ A = \begin{bmatrix} 1 & 1 & -1 & | & -5 \\ 2 & -1 & 1 & | & 5 \\ -1 & 3 & -2 & | & 1 \end{bmatrix} \]

Step 2: Perform Gaussian Elimination

We perform row operations to convert the matrix into row-echelon form. The resulting matrix after applying the Gaussian elimination method is: \[ A = \begin{bmatrix} 1 & 1 & -1 & | & -5 \\ 0 & 1 & -1 & | & -5 \\ 0 & 0 & 1 & | & 16 \end{bmatrix} \]

Step 3: Back-Substitution

From the row-echelon form, we can express the equations as: