# How can I solve this equation system? x + 5y + 2z = 19" "3x + 2y+ z = 13" "2x - y - z = 0 How should I think about this?

Aug 13, 2018

Solution: $x = 2 , y = 3 , z = 1$

#### Explanation:

x+5 y+2 z=19;(1), 3 x+2 y+ z=13; (2)

2 x- y- z=0 ; (3) :. 2 x = y+z . Multiplying equation (1)

by $2$ we get , 2 x + 10 y + 4 z = 38 ;  or

y +z + 10 y + 4 z = 38 ; or 11 y +5 z =38; (4)

Multiplying equation (2) by $2$ we get ,

6 x + 4 y + 2 z = 26 ;  or

3(y +z) + 4 y + 2 z = 26 ; or 7 y +5 z =26; (5)

Subtracting equation (5) from equation (4) we get,

$4 y = 12 \mathmr{and} y = 3 \therefore 5 z = 38 - 11 y$ or

$5 z = 38 - 11 \cdot 3 \mathmr{and} 5 z = 5 \mathmr{and} z = 1$

$\therefore 2 x = y + z \mathmr{and} 2 x = 3 + 1 \mathmr{and} 2 x = 4 \mathmr{and} x = 2$

Solution:$x = 2 , y = 3 , z = 1$ [Ans]

Aug 13, 2018

Thus $x = 2$, $y = 3$ and $z = 1$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}1 & 5 & 2 & | & 19 \\ 3 & 2 & 1 & | & 13 \\ 2 & - 1 & - 1 & | & 0\end{matrix}\right)$

I have written the equations not in the sequence as in the question in order to get $1$ as pivot.

Perform the folowing operations on the rows of the matrix

$R 2 \leftarrow R 2 - 3 R 1$; $R 3 \leftarrow R 3 - 2 R 1$

$A = \left(\begin{matrix}1 & 5 & 2 & | & 19 \\ 0 & - 13 & - 5 & | & - 44 \\ 0 & - 11 & - 5 & | & - 38\end{matrix}\right)$

$R 2 \leftarrow R 2 - R 3$

$A = \left(\begin{matrix}1 & 5 & 2 & | & 19 \\ 0 & - 2 & 0 & | & - 6 \\ 0 & - 11 & - 5 & | & - 38\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{- 2}$

$A = \left(\begin{matrix}1 & 5 & 2 & | & 19 \\ 0 & 1 & 0 & | & 3 \\ 0 & - 11 & - 5 & | & - 38\end{matrix}\right)$

$R 1 \leftarrow R 1 - 5 R 2$; $R 3 \leftarrow R 3 + 11 R 2$

$A = \left(\begin{matrix}1 & 0 & 2 & | & 4 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & - 5 & | & - 5\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{- 5}$

$A = \left(\begin{matrix}1 & 0 & 2 & | & 4 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 1 & | & 1\end{matrix}\right)$

$R 1 \leftarrow R 1 - 2 R 3$

$A = \left(\begin{matrix}1 & 0 & 0 & | & 2 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 1 & | & 1\end{matrix}\right)$

Thus $x = 2$, $y = 3$ and $z = 1$