How solve the following system of linear equations?
The question goes as the following $$x-3y+6z=21$$ $$3x+2y-5z=-30$$ $$2x-5y+2z=-6$$
for this one I am really confused and don't even know how to solve it.
I tried to do the elimination method but i came with fractions.
I woud show the process but I'm currently using my phone.
$\endgroup$ 33 Answers
$\begingroup$$$\left\{\begin{array}{lcr}(A)\;\phantom{1}x-3y+6z&=&21\\ (B)\;3x+2y-5z&=&-30\\ (C)\;2x-5y+2z&=&-6\end{array}\right.$$ is equivalent (through $(B)\mapsto(B-3A)$ and $(C)\mapsto (C-2A)$) to$$\left\{\begin{array}{lcr}(A)\;\phantom{1}x-3y+6z&=&21\\ (B)\;\phantom{3x+}11y-23z&=&-93\\ (C)\;\phantom{2x-\,\,\,}y-10z&=&-48\end{array}\right.$$ that is equivalent (through $(B)\mapsto(B-11C)$) to $$\left\{\begin{array}{lcr}(A)\;\phantom{1}x-3y+6z&=&21\\ (B)\;\phantom{3x+11y-}87z&=&435\\ (C)\;\phantom{2x-\,\,\,}y-10z&=&-48\end{array}\right.$$ Now use $(B)$ to find the value of $z$, plug it into $(C)$ to find the value of $y$, plug them both into $(A)$ to find the value of $x$. This is known as the Gaussian elimination method. A good alternative for $3\times 3$ systems is Cramer's rule. They both lead to: $$ (x,y,z)=\color{red}{(-3,2,5)} $$
$\endgroup$ 1 $\begingroup$multiplying the first equation by -3 and adding to the seconde one we get $$11y-23z=-93$$ (I) multiplying the first by -2 and adding to the third we get $$y-10z=-48$$ (II) can you proceed? multiplying (II) by -11 and adding this to (I) we obtain $$87z=435$$ thus $$z=5$$
$\endgroup$ $\begingroup$You can row reduce the matrix with no fractions to
$$\left[\begin{array}{ccc|c} 1&0&0&-3\\ 0&1&0&2\\ 0&0&1&5\\ \end{array}\right]$$
giving $x=-3, y=2,z=5$
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