Question

x+y+z=9

2x-3y+4z=13

3x+4y+5z=40

Answer 1

x + y + z = 9

2x - 3y + 4z = 13

3x + 4y + 5z = 40   to find x, y, z we use Row echelon form

$\begin{bmatrix} 1 & 1 & 1 && 9 \\ 2 & -3 & 4 && 13 \\ 3 & 4 & 5 && 40 \\ \end{bmatrix}$ = $\begin{vmatrix} R_2 = 2R_1 - R_2 \\ R_3 = 2R_3-3R_2 \\ \end{vmatrix}$ =

$\begin{bmatrix} 1 & 1 & 1 && 9 \\ 0 & 5 & -2 && 5 \\ 0 & 17 & -2 && 41 \\ \end{bmatrix}$ = $\begin{vmatrix} R_3 = 5R_3-17R_2 \\ \end{vmatrix}$ =

$\begin{bmatrix} 1 & 1 & 1 && 9 \\ 0 & 5 & -2 && 5 \\ 0 & 0 & 24 && 120 \\ \end{bmatrix}$ = $\begin{bmatrix} 1 & 1 & 1 && 9 \\ 0 & 1 & -0.4 && 1 \\ 0 & 0 & 1 && 5 \\ \end{bmatrix}$ => z = 5;

y - 0.4z = 1 => y - 0.4*5 = 1 => y =  3;

x + y + z = 9 => x + 3 + 5 = 9 => x = 1;

Answer: [1, 3, 5]