Use Gauss-Jordan elimination to solve the following linear system.
x – 6y – 3z = 4
–2x – 3z = –8
–2x + 2y – 3z = –14

A. (–3,–2,4)
B. (–2,–3,4)
C. (5,3,–5)
D. (–5,–3,3)

Respuesta :

just put the coefients in to a matrix

1x-6y-3z=4
-2x+0y-3z=-8
-2x+2y-3z=-14

[tex] \left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\-2&2&-3|-14\end{array}\right] [/tex]
mulstiply 2nd row by -1 and add to 3rd
[tex] \left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&2&0|-6\end{array}\right][/tex]
divde last row by 2
[tex] \left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&1&0|-3\end{array}\right][/tex]
multiply 2rd row by 6 and add to top one
[tex] \left[\begin{array}{ccc}1&0&-3|-14\\-2&0&-3|-8\\0&1&0|-3\end{array}\right][/tex]
multiply 1st row by -1 and add to 2nd
[tex] \left[\begin{array}{ccc}1&0&-3|-14\\-3&0&0|6\\0&1&0|-3\end{array}\right][/tex]
divide 2nd row by -3
[tex] \left[\begin{array}{ccc}1&0&-3|-14\\1&0&0|-2\\0&1&0|-3\end{array}\right][/tex]
mulstiply 2nd row by -1 and add to 1st row
[tex] \left[\begin{array}{ccc}0&0&-3|-12\\1&0&0|-2\\0&1&0|-3\end{array}\right][/tex]
divide 1st row by -3
[tex] \left[\begin{array}{ccc}0&0&1|4\\1&0&0|-2\\0&1&0|-3\end{array}\right][/tex]

rerange
[tex]\left[\begin{array}{ccc}1&0&0|-2\\0&1&0|-3\\0&0&1| 4\end{array}\right][/tex]

x=-2
y=-3
z=4
(x,y,z)
(-2,-3,4)

B is answer
333ipb

Answer:

(-2,-3,4)

Step-by-step explanation:

Yes. : )