0
$\begingroup$

How to solve system linear of equations? I'm so confused I don't know how to find the result. $$ \begin{array}{rcr} 2p - 2q - r + 3s &=& 4 \\ p - q +2s &=& 1 \\ -2p +2q -4s &=& -2 \end{array} $$ Thank you.

$\endgroup$
4
$\begingroup$

We can use Gauss-Jordan elimination. The operations are \begin{align*} \left[\begin{array}{rrrr|r} 2 & -2 & -1 & 3 & 4 \\ 1 & -1 & 0 & 2 & 1 \\ -2 & 2 & 0 & -4 & -2 \end{array}\right] \xrightarrow{(1/2)\cdot R_{1}\to R_{1}}\left[\begin{array}{rrrr|r} 1 & -1 & -\frac{1}{2} & \frac{3}{2} & 2 \\ 1 & -1 & 0 & 2 & 1 \\ -2 & 2 & 0 & -4 & -2 \end{array}\right] \\ \xrightarrow{ \begin{array}{rcr} R_{2}-R_{1} &\to& R_{2} \\ R_{3}+2\cdot R_{1} &\to& R_{3} \\ \end{array}}\left[\begin{array}{rrrr|r} 1 & -1 & -\frac{1}{2} & \frac{3}{2} & 2 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} & -1 \\ 0 & 0 & -1 & -1 & 2 \end{array}\right] \\ \xrightarrow{2\cdot R_{2}\to R_{2}}\left[\begin{array}{rrrr|r} 1 & -1 & -\frac{1}{2} & \frac{3}{2} & 2 \\ 0 & 0 & 1 & 1 & -2 \\ 0 & 0 & -1 & -1 & 2 \end{array}\right] \\ \xrightarrow{ \begin{array}{rcrcr} R_{1}+1/2\cdot R_{2} &\to& R_{1} \\ R_{3}+R_{2} &\to& R_{3} \\ \end{array}}\left[\begin{array}{rrrr|r} 1 & -1 & 0 & 2 & 1 \\ 0 & 0 & 1 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{align*} These steps show that $$ \operatorname{rref}\left[\begin{array}{rrrr|r} 2 & -2 & -1 & 3 & 4 \\ 1 & -1 & 0 & 2 & 1 \\ -2 & 2 & 0 & -4 & -2 \end{array}\right]=\left[\begin{array}{rrrr|r} 1 & -1 & 0 & 2 & 1 \\ 0 & 0 & 1 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] $$ The variables $\{p, r\}$ correspond pivot columns and are called "dependent" variables. The other variables $\{q, s\}$ are called "free" variables. To find the general solution to the system, we write each dependent variable in terms of the free variables. This gives $$ \left[\begin{array}{r} p \\ q \\ r \\ s \end{array}\right] = \left[\begin{array}{r} q - 2 \, s + 1 \\ q \\ -s - 2 \\ s \end{array}\right] = \left[\begin{array}{r} 1 \\ 0 \\ -2 \\ 0 \end{array}\right]+q\left[\begin{array}{r} 1 \\ 1 \\ 0 \\ 0 \end{array}\right]+s\left[\begin{array}{r} -2 \\ 0 \\ -1 \\ 1 \end{array}\right] $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.