Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, express x, y, and z in terms of the parameter t.)
3x+3y+12z=9
x+y+4z=3
2x+5y+20z=15
-x+2y+8z=6
(x, y, z) = ( _____ )


Answer :

The solution for the given system of equation is (x,y,z) is = (0 , (3-4t) , t) .

In the question ,

it is given that the system of equations are ,

3x + 3y + 12z = 9

x + y + 4z = 3

2x + 5y + 20z = 15

-x + 2y + 8z = 6

we have to solve the equations for x, y , z.

writing the given equations in augmented matrix form ,

we get ,

[tex]\left[\begin{array}{ccc}3&3&12\\1&1&4\\2&5&20\\-1&2&8\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}9\\3\\15\\6\end{array}\right][/tex]

Applying the operation , R₁/3 ⇒ R₁

[tex]\left[\begin{array}{ccc}1&1&4\\1&1&4\\2&5&20\\-1&2&8\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}3\\3\\15\\6\end{array}\right][/tex]

Applying the row operation , R₂ - R₁ ⇒ R₂  .

[tex]\left[\begin{array}{ccc}1&1&4\\0&0&0\\2&5&20\\-1&2&8\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}3\\0\\15\\6\end{array}\right][/tex]

After applying more row operations , in the above matrix ,

to make the lower triangle matrix 0 ,

we get ,

[tex]\left[\begin{array}{ccc}1&0&0\\0&1&4\\0&0&0\\0&0&0\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}0\\3\\0\\0\end{array}\right][/tex]

The above result can be expressed as x = 0,

and y + 4z = 3 .

So , the system of equations have infinite solutions .

So , the solution x, y, and z expressed in terms of the parameter t is

(0 , (3-4t) , t)  .

Therefore , the solution is (x,y,z) is = (0 , (3-4t) , t) .

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