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.
Elementary matrices and the Gauss-Jordan method. An elementary matrix is what you get by applying exactly one elementary row operation to the identity matrix In. Its defining property: pre-multiplying any matrix A by that elementary matrix has exactly the same effect as applying the row operation directly to A. Chaining k row operations that reduce a non-singular A all the way to In is therefore the same as pre-multiplying by a product of elementary matrices Ek⋯E2E1, so Ek⋯E2E1A=In⇒A−1=Ek⋯E2E1.
This is exactly what the Gauss-Jordan method automates: augment A with In to form [A∣In], then apply the same row operations to both halves until the left block becomes In -- at that point the right block has become A−1:
[A∣In]row ops[In∣A−1].
Tip
Gauss-Jordan needs no cofactors or determinants at all -- for a large matrix it is usually faster than computing A−1=∣A∣1adjA. If, partway through, an entire row on the left becomes zero, A is singular and no inverse exists.
Start with the highest-order minor a matrix's shape allows; if it vanishes, drop to the next lower order until a nonzero minor turns up — that order is the rank.
(i) the only 2×2 minor is 0, but a 1×1 entry is nonzero, so ρ=1.
(ii)–(v) a nonzero minor is found already at the largest possible order.
✓Final answer
ρ(A)=1;
ρ(A)=2;
ρ(A)=2;
ρ(A)=3;
ρ(A)=3.
For each matrix we test the highest-order minor its shape permits (min(m,n) for an m×n matrix). If that minor is 0, we drop to the next lower order and keep testing until a nonzero minor appears; that order is the rank ρ(A).
Step 1. Part (i): test the only 2×2 minor.A=(2−1−42) is itself 2×2, so the highest possible order is 2.
∣A∣=2(2)−(−4)(−1)=4−4=0.
The order-2 minor vanishes, so we drop to order 1.
Step 2. Part (i): test an order-1 minor. The entry a11=2=0, so a nonzero 1×1 minor exists. Hence ρ(A)=1.
Step 3. Part (ii): test an order-2 minor.A=−1433−7−4 is 3×2, so the highest possible order is min(3,2)=2. Take the minor formed by rows 1,2:
−143−7=(−1)(−7)−(3)(4)=7−12=−5=0.
A nonzero order-2 minor exists, so ρ(A)=2 (no need to check the other row pairs).
Step 4. Part (iii): test an order-2 minor.A=(13−2−6−1−301) is 2×4, so the highest possible order is 2. Take the minor formed by columns 1,4:
1301=1(1)−0(3)=1=0.
A nonzero order-2 minor exists, so ρ(A)=2.
Step 5. Part (iv): test the order-3 minor (the whole determinant).A=125−2413−6−1 is 3×3, so the highest possible order is 3. Expanding along row 1: