Questions: Solve the system of equations using the row echelon method. -21x - 2y + z = -13 12x + y = 10 -24x - 2y + z = -16 Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is . (Type an exact answer in simplified form. If there are infinitely many solutions, type an expression involving z for each coordinate where z re B. The solution is the empty set.

Solve the system of equations using the row echelon method. -21x - 2y + z = -13 12x + y = 10 -24x - 2y + z = -16 Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is . (Type an exact answer in simplified form. If there are infinitely many solutions, type an expression involving z for each coordinate where z re B. The solution is the empty set.
Transcript text: Solve the system of equations using the row echelon method. \[ \begin{aligned} -21 x-2 y+z & =-13 \\ 12 x+y & =10 \\ -24 x-2 y+z & =-16 \end{aligned} \] Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is $\{\square$ $\square$ $\square$ )\}. (Type an exact answer in simplified form. If there are infinitely many solutions, type an expression involving z for each coordinate where z re B. The solution is the empty set.
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Solution

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Solution Steps

To solve the system of equations using the row echelon method, we need to perform Gaussian elimination. This involves transforming the system's augmented matrix into an upper triangular form, from which we can then perform back-substitution to find the solutions for \(x\), \(y\), and \(z\).

Step 1: Formulate the Augmented Matrix

The given system of equations can be represented as an augmented matrix: \[ \begin{bmatrix} -21 & -2 & 1 & | & -13 \\ 12 & 1 & 0 & | & 10 \\ -24 & -2 & 1 & | & -16 \end{bmatrix} \]

Step 2: Row Echelon Form

Using Gaussian elimination, we transform the augmented matrix into row echelon form: \[ \begin{bmatrix} 1 & 0.0952 & -0.0476 & | & 0.6190 \\ 0 & 1 & -4 & | & -18 \\ 0 & 0 & 1 & | & 4 \end{bmatrix} \]

Step 3: Back Substitution

From the row echelon form, we can express the variables:

  1. From the third row: \( z = 4 \)
  2. From the second row: \( y - 4z = -18 \) leads to \( y - 16 = -18 \) or \( y = -2 \)
  3. From the first row: \( x + 0.0952y - 0.0476z = 0.6190 \) leads to \( x + 0.0952(-2) - 0.0476(4) = 0.6190 \) or \( x - 0.1904 - 0.1904 = 0.6190 \) which simplifies to \( x = 1 \)

Final Answer

The solution set is: \[ \boxed{(x, y, z) = (1, -2, 4)} \]

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