Question

type the correct answer in each box. use numerals instead of words. what is the solution to this matrix equation? \\(\

$$\begin{bmatrix}1 & 2 & -1 \\\\ 1 & 3 & -1 \\\\ 2 & -1 & 2\\end{bmatrix}$$

\

$$\begin{bmatrix}x \\\\ y \\\\ z\\end{bmatrix}$$

= \

$$\begin{bmatrix}17 \\\\ 35 \\\\ 80\\end{bmatrix}$$

\\) \\(x = \square\\) \\(y = \square\\) \\(z = \square\\)

Explanation:

Step1: Convert to linear system

From matrix multiplication, we get:

$$\begin{cases} x + 2y - z = 17 \quad (1) \\ x + 3y - z = 35 \quad (2) \\ 2x - y + 2z = 80 \quad (3) \end{cases}$$

Step2: Eliminate $x,z$ to find $y$

Subtract equation (1) from (2):
$$(x + 3y - z) - (x + 2y - z) = 35 - 17$$
$$y = 18$$

Step3: Simplify equations with $y=18$

Substitute $y=18$ into (1):
$$x + 2(18) - z = 17 \implies x - z = 17 - 36 \implies x - z = -19 \quad (4)$$
Substitute $y=18$ into (3):
$$2x - 18 + 2z = 80 \implies 2x + 2z = 98 \implies x + z = 49 \quad (5)$$

Step4: Solve for $x$ using (4)+(5)

Add equations (4) and (5):
$$(x - z) + (x + z) = -19 + 49$$
$$2x = 30 \implies x = 15$$

Step5: Solve for $z$ using (5)

Substitute $x=15$ into (5):
$$15 + z = 49 \implies z = 49 - 15 = 34$$

Answer:

$x = 15$
$y = 18$
$z = 34$