# Finding reduced row echelon form given only one unique solution

Each equation in the following linear system represents a line in the $xy$-plane:

$$\begin{matrix} a_1x + b_1y = c_1\\ a_2x + b_2y = c_2\\ a_3x + b_3y = c_3\\ \end{matrix}$$

where $a_1$, $a_2$, $a_3$, $b_1$, $b_2$, $b_3$, $c_1$, $c_2$, $c_3$ are constants and for each $i = 1; 2; 3,$ $a_i, b_i$ are not both zero.

Given that the system of linear equations has only one solution, am I correct to say that the three lines intersect at only one unique point?

Also, how would I reduce the augmented matrix of the system to a reduced row echelon form?

• I assume the first equation should be $a_1x+b_1y=c_1$? – 5xum Aug 25 '17 at 8:34
• If the system has only one solution, then yes, the three lines intersect in one point. This is because every point of intersection is one solution to the system, and every solution of the system is one point of intersection. – 5xum Aug 25 '17 at 8:36
• @5xum Noted. Can you advise how to reduce the matrix to a reduced row echelon form? – kevinbobbkoh Aug 25 '17 at 14:49
• Since the system is consistent, you know that the last equation is redundant and the corresponding row of the rref must be zero. For the other rows, think about what the rref matrix of a system of two equations in two unknowns that has a unique solution must look like. – amd Aug 25 '17 at 21:31