I've had a little break in solving systems of equations and so I wanted to verify here that my own answer to this problem is 100% correct and done 🙂 So I have the following problem:
Solve the system of equations:
$$3x_1+x_2-4x_3+5x_4=2$$ $$2x_1-3x_2-2x_3+3x_4=5$$
My attempt:
Because this is an underdetermined system we will have infinite amount of solutions. I will solve $x_1$ and $x_2$ explicitly in terms of $x_3$ and $x_4$. By multiplying the first equation by $3$ and adding it to the second we get:
$$11x_1-14x_3+18x_4=11$$
$$x_1=1+\frac{14}{11}x_3-\frac{18}{11}x_4$$
By substituting $x_2$ into the first equation we get:
$$3\left(1+\frac{14}{11}x_3-\frac{18}{11}x_4\right)+x_2-4x_3+5x_4=2$$
From which we get
$$x_2=\frac{2}{11}x_3-\frac{1}{11}x_4-1$$
So the solution is:
$$x_1=1+\frac{14}{11}x_3-\frac{18}{11}x_4$$
$$x_2=\frac{2}{11}x_3-\frac{1}{11}x_4-1,$$
where $x_3,x_4$ are free variables.
Best Answer
Note that
$$ \DeclareMathOperator{rref}{rref} \rref \begin{bmatrix} 3 & 1 & -4 & 5 & 2 \\ 2 & -3 & -2 & 3 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -\frac{14}{11} & \frac{18}{11} & 1 \\ 0 & 1 & -\frac{2}{11} & \frac{1}{11} & -1 \end{bmatrix} $$ This implies \begin{align*} x_1 &= \frac{14}{11}\,x_3-\frac{18}{11}\,x_4+1 \\ x_2 &= \frac{2}{11}\,x_3-\frac{1}{11}\,x_4-1 \end{align*} which confirms your answer.
Do you know how to row-reduce a matrix into its reduced row-echelon form? This is a much more efficient method than substitution.