Revise Systems of linear equations for IB Mathematics: Analysis and Approaches (1.8)

Before touching the algebra, picture the problem. In two dimensions a linear equation $ax+by=c$ is a straight line, and two such lines either cross once (a unique solution), run parallel and never meet (no solution), or lie exactly on top of each other (infinitely many solutions). Moving up to three dimensions, a linear equation $ax+by+cz=d$ is no longer a line but a flat plane stretching out forever. The solution of a system of three such equations is the set of points that lie on all three planes at once.

There are only three things three planes can do. They can pinch together at a single point, they can share a whole common line, or they can fail to share any common point at all. These three pictures are exactly the three algebraic cases you will meet, so learning to read one from the other is the heart of this topic.

Row reduction is a disciplined way of eliminating variables one at a time until the system becomes a 'staircase' that you can solve from the bottom up. The tools are the three elementary row operations, and the crucial fact is that each of them leaves the solution set completely unchanged — you are rewriting the same system, never a different one.

The three elementary row operations are:

1. Swap two equations: $R_i \leftrightarrow R_j$.
2. Multiply an equation by a non-zero constant: $R_i \to k R_i$, $k\neq 0$.
3. Add a multiple of one equation to another: $R_i \to R_i + k R_j$. A 3×3 system can be stacked into an augmented matrix $\left[\begin{array}{ccc|c} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3 & b_3 & c_3 & d_3 \end{array}\right]$ and reduced until zeros sit below the leading entries.

The strategy is always the same: use the first equation to remove the first variable from the other two, then use the resulting second equation to remove the second variable from the third. That leaves a single equation in one unknown, which you solve and then feed back up the staircase — this last part is called back-substitution.

Sometimes elimination does not leave a tidy value for each variable. If a row collapses to a contradiction such as $0=5$, no values can satisfy the system: it is inconsistent and has no solution. If a row collapses to an identity such as $0=0$, that equation is telling you nothing new: one variable is free and there are infinitely many solutions, which you must then write in parametric form. The two look similar on the page — a row of zeros on the left — so the deciding factor is the number on the right.

On Paper 1 the marks are analytic: each is tied to a specific line of working. A method mark (M) rewards a correct approach even if the arithmetic later slips; an accuracy mark (A) rewards a correct result and is usually dependent on the method mark being earned. Follow-through (FT) means a correct step performed on your own earlier (wrong) value still scores, ISW ('ignore subsequent working') means a correct answer is not un-marked by a later fumble, and equivalent correct forms are accepted. All of that protection exists only if your method is on the page. Study how each of the six marks below is earned by a specific line.

Everything here is one procedure — eliminate systematically, then read the reduced system — applied to three outcomes. The same row-reduction habit reappears when you meet vectors and the intersection of planes later in the course, where the parametric solution you wrote for a dependent system becomes the vector equation of the line of intersection. Master the routine now — label, eliminate the same variable from two pairs, back-substitute, then interpret the geometry — and those later topics become variations on a method you already own.