Concept understanding — Consistency by Rank Method
Theorem (Rouché-Capelli). The system AX=B is consistent (has at least one solution) if and only if the rank of the coefficient matrix equals the rank of the augmented matrix: ρ(A)=ρ([A∣B]).
This single test subsumes and explains matrix inversion, Cramer's rule and Gaussian elimination all at once. Row-reduce [A∣B] to echelon form, read off ρ(A) and ρ([A∣B]) from the number of non-zero rows in the left block and in the whole augmented matrix, and apply the standing rule (with n = number of unknowns):
ρ(A)=ρ([A∣B])=n−k<n⇒consistent, infinitely many solutions, forming a k-parameter family (choose k unknowns arbitrarily, solve the rest by back substitution).
ρ(A)=ρ([A∣B])⇒inconsistent, no solution (the echelon form of [A∣B] shows a row like 0=c for some c=0 -- a contradiction).
Worked illustration. For x+y=2,2x+2y=5: [A∣B]=(1212∣∣25)R2→R2−2R1(1010∣∣21). Here ρ(A)=1 but ρ([A∣B])=2 (the bottom row reads 0=1), so the system is inconsistent -- geometrically, two parallel, non-coincident lines.
A system with unknown parameters (e.g. λ,μ appearing among the coefficients or constants) is investigated by row-reducing symbolically until a pivot entry becomes an expression in λ (or a right-hand entry becomes an expression in μ); each case -- that expression zero versus non-zero -- then reads off which of the three outcomes above holds, exactly as the general rule prescribes.
Note
This rank test is the general consistency criterion -- it applies to any m-equation, n-unknown system, square or not, singular or not -- unlike matrix inversion and Cramer's rule, which need a square non-singular coefficient matrix and so can only ever certify the unique-solution case.
Row-reduce each augmented matrix [A∣B] to echelon form and compare ρ(A) with ρ([A∣B]): equal to 3 means a unique solution, equal but less than 3 means infinitely many solutions, unequal means no solution.
(i) unique solution
(ii) infinitely many solutions
(iii) no solution (inconsistent)
(iv) infinitely many solutions
✓Final answer
x=1,y=1,z=1;
x=107−2t,y=2t−101,z=t,t∈R;
inconsistent — no solution;
x=1+2s−2t,y=s,z=t,s,t∈R.
We test consistency by row-reducing [A∣B] to echelon form and comparing ρ(A) with ρ([A∣B]): if the two ranks are equal to the number of unknowns (n=3) the solution is unique, if they are equal but less than 3 there are infinitely many solutions, and if they differ the system is inconsistent.
Step 2. Part (i): read off the ranks and solve. Three non-zero rows, so ρ(A)=ρ([A∣B])=3=n — consistent with a unique solution. Back-substituting: −7z=−7⇒z=1; 3y=3⇒y=1; x−y+2z=2⇒x=2+1−2=1. So x=1,y=1,z=1.
Step 4. Part (ii): read off the ranks and solve. Only two non-zero rows, so ρ(A)=ρ([A∣B])=2<3=n — consistent with infinitely many solutions (one free parameter). Put z=t: from row 2, 10y−5z=−1⇒y=2z−101=2t−101; from row 1, x=1+3y−2z=1+3(2t−101)−2t=107−2t. So x=107−2t,y=2t−101,z=t,t∈R.
Step 6. Part (iii): read off the ranks. The last row reads 0=−2, which is impossible, so ρ(A)=2 (only the first two rows are independent) while ρ([A∣B])=3 (the third row is non-zero purely because of its constant term). Since ρ(A)=ρ([A∣B]), the system is inconsistent — no solution.
(Rows 2 and 3 were already exact multiples of row 1: R2=3R1 and R3=2R1.)
Step 8. Part (iv): read off the ranks and solve. Only one non-zero row, so ρ(A)=ρ([A∣B])=1<3=n — consistent with infinitely many solutions (two free parameters). Put y=s,z=t: from 2x−y+z=2, x=1+2s−2t. So x=1+2s−2t,y=s,z=t,s,t∈R.
✓Final answer
x=1,y=1,z=1;
x=107−2t,y=2t−101,z=t,t∈R;
inconsistent — no solution;
x=1+2s−2t,y=s,z=t,s,t∈R.
Rank method — row-reduce [A∣B] and compare ρ(A) with ρ([A∣B])
Concluding 'no solution' whenever the ranks are unequal without checking which one is bigger — inconsistency needs ρ([A∣B])>ρ(A) specifically.
Stopping at row-echelon form without correctly reading off the free parameter(s) once ρ(A)<n.
Missing that a row like R2=3R1 makes an equation entirely redundant, as in part (iv).