You have:
$$T = \begin{bmatrix}a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3}\end{bmatrix}$$
Multiply that by the vector:
$$\begin{bmatrix}x \\ y \\ z\end{bmatrix}$$
You get:
$$\begin{bmatrix}a_{1,1}x+a_{1,2}y+a_{1,3}z \\ a_{2,1}x+a_{2,2}y+a_{2,3}z \\ a_{3,1}x+a_{3,2}y+a_{3,3}z\end{bmatrix} = \begin{bmatrix}x-2z \\ 2x-y+3z \\ 4x+y+8z\end{bmatrix}$$
Then, just equate coefficients to get:
$$T = \begin{bmatrix}1 & 0 & -2 \\ 2 & -1 & 3 \\ 4 & 1 & 8\end{bmatrix}$$
To show that this linear transformation is an isomorphism you must show that it preserves vector addition and scalar multiplication and is invertible. Here, T maps (x, y, z) to (x- 2z, 2x- y+ 3z, 4x+ y+ 8z). It maps (x1+ x2, y1+ y2, z1+ z2) to (x1+ x2- 2(z1+ z2), 2(x1+ x2)- (y1+ y2)+ 3(z1+ z2), 4(x1+ x2)+ (y1+ y2)+ 8(z1+ z2))= (x1+ x2- 2z1- 2z2, 2x1+ 2x2- y1- y2+ 3z1+ 3z2, 4x1+ 4x1+ y1+ y1+ 8z1+ 8z2)= (x1- 2z1, 2x1- y1+ 3z1, 4x1+ y1+ 8z1)+ (x2- 2z2, 2x2- y2+ 3z2, 4x2+ y2+ 8z2)= T(x1, y1, z1)+ T(x2, y2, z2). Yes, T preserves vector addition.
T maps a(x, y, z)= (ax, ay, az) to (ax- 2az, 2ax- ay+ 3az, 4ax+ ay+ 8az)= (a(x- 2z), a(2x- y+ 3z), a(4x+ y+ 8z))= a(x- 2z, 2x- y+ 3z, 4x+ y+ 8z)= aT(x, y, z). Yes, T preserves scalar multiplication.
Finally, if T(x, y, z)= (x- 2z, 2x- y+ 3z, 4x+ y+ 8z)= (p, q, r) then x- 2z= p, 2x- y+ 3z= q, and 4x+ y+ 9z= r. Adding the second and third equations eliminates y: 5x+ 7z= q+ r. Add -5 times x- 2z= p to that; 5x+ 7z- 5x+ 10z= q+ r- 5p or 17z= q+ r- p so z= (q+ r- p)/17. Then x- 2z= x- (q+ r- p)/17= q+ r, x= (q+ r- p)/17+ q+ r= (q+ r- p+ 17q+ 17r)/17= (18q+ 18r- p)/17. And then 4x+ y+ 8z= (72q+ 72r- 4p)/17+ y+ (8q+ 8r- 8p)/17= (80q+ 80r- 12p)/17+ y= r so y= r- (80q+ 80r- 12p)/17= (17r- 80q- 80r+ 12p)/17= (12p- 63r- 80q)/17. That show that this transformation is invertible.