Answer :
Answer:
Therefore the solution are
x=-6t-10
y=-3-2t
z=t
Step-by-step explanation:
Given system of the equation is
x-4y -2z=2
2x-11y-10z=13
-x+6y+6z=-8
Transforming matrix of the reduce row echelon form
[tex]\left[\begin{array}{ccc}1&-4&-2\\2&-11&-10\\-1&6&6\end{array}\right \left |\begin{array}{c}2&13&-8\end{array}\right ][/tex]
add -2 times the first row to the second row
[tex]\left[\begin{array}{ccc}1&-4&-2\\0&-3&-6\\-1&6&6\end{array}\right \left |\begin{array}{c}2&9&-8\end{array}\right ][/tex]
add 1 times the row to the third row
[tex]\left[\begin{array}{ccc}1&-4&-2\\0&-3&-6\\0&2&4\end{array}\right \left |\begin{array}{c}2&9&-6\end{array}\right ][/tex]
multiply the second row by [tex]-\frac{1}{3}[/tex]
[tex]\left[\begin{array}{ccc}1&-4&-2\\0&1&2\\-1&6&6\end{array}\right \left |\begin{array}{c}2&-3&-8\end{array}\right ][/tex]
add -2 times the second row to the third row
[tex]\left[\begin{array}{ccc}1&-4&-2\\0&1&2\\0&0&0\end{array}\right \left |\begin{array}{c}2&-3&0\end{array}\right ][/tex]
add 4 times the second row to the first row
[tex]\left[\begin{array}{ccc}1&0&6\\0&1&2\\0&0&0\end{array}\right \left |\begin{array}{c}-10&-3&0\end{array}\right ][/tex]
The reduce form of the system of equation is
x+0.y+6.z=-10......(1)
0.x+y+2z=-3.........(2)
There are infinity of many solution
z= t (say)
Putting the value of z in the equations of (1) and (2)
x +6t=-10
⇒x=-6t-10
and
y+2t=-3
⇒y= -3-2t
Therefore the solution are
x=-6t-10
y=-3-2t
z=t