Concept understanding — Elementary Transformations and Rank
An elementary row (column) operation on a matrix is one of three moves: (i) interchange two rows/columns (Ri↔Rj); (ii) multiply a row/column by a non-zero scalar (Ri→λRi); (iii) add to a row/column a non-zero scalar multiple of another row/column (Ri→Ri+λRj). Two matrices related by a sequence of such operations are called equivalent, written A∼B -- an elementary transformation changes the matrix's appearance but never the information (rank, solution set) it encodes.
Row-echelon form. A non-zero matrix E is in row-echelon form if (i) every zero row sits below every non-zero row, (ii) the first non-zero entry of each row (its pivot) lies strictly to the right of the pivot in the row above, and (iii) every entry below a pivot, in its own column, is zero. Any matrix can be driven to this form by repeated pivoting: make the current pivot entry non-zero (swapping rows if needed), then use row operations to zero out everything below it, and move to the next row.
Rank. The rankρ(A) of a matrix A is the order of the largest square sub-matrix of A whose determinant is non-zero (equivalently: the largest r for which some r×r minor is non-zero, while every minor of order r+1 and above vanishes). Basic facts: ρ(A)≥1 once A has a non-zero entry; ρ(In)=n; for an m×n matrix, ρ(A)≤min{m,n}; and a square matrix of order n is invertible exactly when ρ(A)=n.
Theorem (rank via echelon form). The rank of a non-zero matrix equals the number of non-zero rows in any row-echelon form of it -- this is far faster than hunting for the largest non-vanishing minor by hand, especially for a large matrix, since every entry below a pivot is already zero and so contributes nothing extra to a minor. …
For each A we form [A∣I] and use elementary row operations (applied to the whole augmented row) to drive the left block to I; the right block that results is A−1.
Step 1. Part (i): set up [A∣I].A=(25−1−2).
[25−1−21001]
Step 2. Part (i): clear below the pivot.R2→2R2−5R1=2(5,−2,0,1)−5(2,−1,1,0)=(10,−4,0,2)−(10,−5,5,0)=(0,1,−5,2).
[20−111−502]
Step 3. Part (i): clear above the second pivot, then normalise row 1.R1→R1+R2=(2,0,−4,2), then R1→R1/2=(1,0,−2,1).
[1001−2−512]⇒A−1=(−2−512)
(Check: ∣A∣=2(−2)−(−1)(5)=−4+5=1=0, so A is indeed invertible.)
Step 4. Part (ii): set up [A∣I].A=116−10−20−1−3.
116−10−20−1−3100010001
Step 5. Part (ii): clear below the first pivot.R2→R2−R1=(0,1,−1,−1,1,0), R3→R3−6R1=(0,4,−3,−6,0,1).
100−1140−1−31−1−6010001
Step 6. Part (ii): clear below the second pivot.R3→R3−4R2=(0,0,−3+4,−6+4,−4,1)=(0,0,1,−2,−4,1).
100−1100−111−1−201−4001
Step 7. Part (ii): back-substitute to clear above the pivots.R2→R2+R3=(0,1,0,−3,−3,1); then R1→R1+R2new=(1,0,0,1−3,0−3,0+1)=(1,0,0,−2,−3,1) (column 3 of R1 was already 0).