Write the equations in matrix,
[tex] \left[\begin{array}{ccc}5&-1&1\\1&2&-1\\2&3&-3\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right] [/tex]
Using row transformation,
R₂ <---> R₃
[tex] \left[\begin{array}{ccc}5&-1&1\\2&3&-3\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right] [/tex]
Using,
R₂ ---> R₂ - 2R₃
[tex] \left[\begin{array}{ccc}5&-1&1\\0&-1&-1\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\-5\\5\end{array}\right] [/tex]
Using,
R₂ --- > (-1)R₂
[tex] \left[\begin{array}{ccc}5&-1&1\\0&1&1\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right] [/tex]
Using row transformation,
R₂ <----> R₃
[tex] \left[\begin{array}{ccc}5&-1&1\\1&2&-1\\0&1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right] [/tex]
Using,
R₂ ---> R₂ - R₁/5
[tex] \left[\begin{array}{ccc}5&-1&1\\0&11/5&-6/5\\0&1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\21/5\\5\end{array}\right] [/tex]
Using,
R₃ ---> R₃ - 5R₂/11
[tex] \left[\begin{array}{ccc}5&-1&1\\0&11/5&-6/5\\0&0&17/11\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\21/5\\34/11\end{array}\right] [/tex]
∴ 5x-y+z = 4 ====(i)
11y-6z = 21 === (ii)
17z=34 === (iii)
from iii,
z=2.
Plug z=2 in ii to get y,
∴y=3.
Plug y and z values in i to get x,
∴x=1
Therefore the solution to the system of equations is (1,3,2)
