Translate the matrix into equations
Let A have rows (2,1) and (1,−1), with unknown vector (x,y) and right-hand side (7,2). Then Ax=b means 2x+y=7 and x−y=2. Each row is a constraint on the same unknowns.
MIT linear algebra materials introduce elimination as a way to work with matrices. Track equivalent systems rather than treating row operations as unmotivated arithmetic.
Work through elimination
Replace row two with row two minus half of row one. The augmented rows become (2,1|7) and (0,−1.5|−1.5). Thus y=1 and substitution gives x=3.
The right-hand side must change too: 2−7/2=−1.5. Leaving it at 2 would define a different equation. Record one operation per line, such as R₂←R₂−0.5R₁, so errors can be traced.
Recognise dependent and inconsistent equations
For augmented rows (2,1|7) and (4,2|14), subtracting twice row one from row two produces (0,0|0). There is one independent constraint and infinitely many solutions.
Change the second right-hand side to 15 and the operation produces (0,0|1). This asserts 0=1 and there is no solution. A unique inverse-matrix answer cannot be forced in either case.
Check the original system
Substituting (3,1) into the original equations gives 7 and 2 as required. Use the original equations for checking so that an earlier copying error is not silently reused.
Self-check: change the original right-hand side to (8,1). The same elimination yields y=2 and x=3. Bring your own working and explain why each operation preserves solutions; keep assessed submissions independent.
References
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