Answer to Question #278371 in Linear Algebra for mim

Question

4x-3y+z=-8

-2x+y-3z=-4

x-y+2z=3

Answer 1

Question is incomplete.

Let us solve for $x,y , z$

$4x-3y+z=-8\\ -2x+y-3z=-4\\ x-y+2z=3$

Step 1: Swap Row 1 and Row 3 . The result is:

$\begin{array}{ccc} x -y+2 z =3...(1) \\ -2 x+y +2 z =-4...(2) \\ 4 x-3 y+z =-8...(3) \end{array}$

Step 2: Multiply first equation by 2 and add the result to the second equation. The result is:

$\begin{aligned} x-y+2 z =3 \\ -y-y+2 z =2 \\ 4 x-3 y+z =-8 \end{aligned}$

Step 3: Multiply first equation by -4 and add the result to the third equation. The result is:

$\begin{aligned} x-y+2 z =3 \\ -y+2 z =2 \\ y -7 z=-20 \end{aligned}$

Step 4: Multiply second equation by 1 and add the result to the third equation. The result is:

$\begin{aligned} x-y+2 z =3 \\ -y+2 z =2 \\ -5 z=-18 \end{aligned}$

Step 5: solve for z.

$\begin{aligned} -5 z =-18 \\ z =\frac{18}{5} \end{aligned}$

Step 6: solve for y using second equation:

$\begin{aligned} -y+2 z =2 \\ -y+2 \cdot \frac{18}{5} =2 \\ y =\frac{26}{5} \end{aligned}$

Step 7: solve for x by substituting $y=\frac{26}{5}$ and $z=\frac{18}{5}$ into the first equation.

$x-\frac{26}{5}+2\times\frac{18}{5}=3\\ x-\frac{26}{5}+\frac{36}{5}=3\\ x+2=3\\ x=1$

Hence, $x=1, y=\frac{26}{5}, z =\frac{18}{5}$