Question

1. \

$$\begin{cases} -x + 2y - z = -1 \\\\ 2x + y + z = 2 \\\\ x - y - 2z = -13 \\end{cases}$$

Explanation:

Step1: Eliminate \( z \) by adding first and second equations

Add \(-x + 2y - z = -1\) and \(2x + y + z = 2\):
\((-x + 2x) + (2y + y) + (-z + z) = -1 + 2\)
\(x + 3y = 1\) (Equation 4)

Step2: Eliminate \( z \) using second and third equations

Multiply the second equation \(2x + y + z = 2\) by 2: \(4x + 2y + 2z = 4\)
Add to the third equation \(x - y - 2z = -13\):
\((4x + x) + (2y - y) + (2z - 2z) = 4 - 13\)
\(5x + y = -9\) (Equation 5)

Step3: Solve Equations 4 and 5 for \( x \) and \( y \)

From Equation 4: \(x = 1 - 3y\)
Substitute into Equation 5: \(5(1 - 3y) + y = -9\)
\(5 - 15y + y = -9\)
\(5 - 14y = -9\)
\(-14y = -14\)
\(y = 1\)

Step4: Find \( x \) using \( y = 1 \)

Substitute \( y = 1 \) into \( x = 1 - 3y \):
\(x = 1 - 3(1) = -2\)

Step5: Find \( z \) using \( x = -2 \) and \( y = 1 \) in second equation

Substitute into \(2x + y + z = 2\):
\(2(-2) + 1 + z = 2\)
\(-4 + 1 + z = 2\)
\(-3 + z = 2\)
\(z = 5\)

Answer:

\( x = -2 \), \( y = 1 \), \( z = 5 \)