Solve the following system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. \[ \left\{\begin{array}{rr} x-2 y+3 z= & 7 \\ 2 x+y+z= & 4 \\ -3 x+4 y-4 z= & -14 \end{array}\right. \] Select the correct choice below and, if necessary, fill in the answer box(es) in your choice. A. The solution is \( \square \) . \( \square \) ). (Simplify your answers.) B. There are infinitely many solutions. The solution can be written as \( \{(x, y, z) \mid x=\square, y=\square, z \) is any real number \( \} \). \( \square \) \( \square \) (Simplify your answers. Type expressions using \( z \) as the variable.) C. There are infinitely many solutions. The solution can be written as \( \{(x, y, z) \mid x= \) \( \square \) , y is any real number, z is any real number\} (Simplify your answer. Type an expression using \( y \) and \( z \) as the variables.) D. The system is inconsistent.
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To solve the given system of equations using matrices and row operations, we start by forming the augmented matrix: \[ \begin{bmatrix} 1 & -2 & 3 & | & 7 \\ 2 & 1 & 1 & | & 4 \\ -3 & 4 & -4 & | & -14 \end{bmatrix} \] Next, we will use row operations to convert this matrix to row echelon form. 1. Start with the first row as it is: \( R_1 \) remains the same. 2. For \( R_2 \), we can make it zero in the first column by updating \( R_2 \) as follows: \( R_2 = R_2 - 2R_1 \): \[ \begin{bmatrix} 1 & -2 & 3 & | & 7 \\ 0 & 5 & -5 & | & -10 \\ -3 & 4 & -4 & | & -14 \end{bmatrix} \] 3. Now for \( R_3 \), make it zero in the first column by: \( R_3 = R_3 + 3R_1 \): \[ \begin{bmatrix} 1 & -2 & 3 & | & 7 \\ 0 & 5 & -5 & | & -10 \\ 0 & -2 & 5 & | & 7 \end{bmatrix} \] 4. Let’s scale \( R_2 \) to simplify: \( R_2 = \frac{1}{5}R_2 \): \[ \begin{bmatrix} 1 & -2 & 3 & | & 7 \\ 0 & 1 & -1 & | & -2 \\ 0 & -2 & 5 & | & 7 \end{bmatrix} \] 5. Now, let's eliminate the second entry of \( R_3 \) using \( R_2 \): \( R_3 = R_3 + 2R_2 \): \[ \begin{bmatrix} 1 & -2 & 3 & | & 7 \\ 0 & 1 & -1 & | & -2 \\ 0 & 0 & 3 & | & 3 \end{bmatrix} \] 6. Finally, simplify \( R_3 \) by dividing \( R_3 \) by 3: \[ \begin{bmatrix} 1 & -2 & 3 & | & 7 \\ 0 & 1 & -1 & | & -2 \\ 0 & 0 & 1 & | & 1 \end{bmatrix} \] From \( R_3 \), we see \( z = 1 \). Now substituting back into \( R_2 \): \( y - 1 = -2 \) gives \( y = -1 \). Now substituting back into \( R_1 \): \[ x - 2(-1) + 3(1) = 7 \implies x + 2 + 3 = 7 \implies x + 5 = 7 \implies x = 2. \] Thus, the solution to the system is: A. The solution is \( (2, -1, 1) \).
