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Yanka [14]
3 years ago
12

Answered

Mathematics
1 answer:
Nutka1998 [239]3 years ago
8 0
True


The question is not a problem with this article
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Use Gauss-Jordan elimination to solve the following linear system.
marysya [2.9K]
Just put the coefients in to a matrix

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

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

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

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

B is answer
7 0
3 years ago
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