Row and Column Reduction; Echelon Form

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2025)
• Priority tier: T4
• Marks (count): 15 (1)
• Average solve time: ~20 min
• Difficulty mix: medium 1
• Section: B | Dominant type: computation

Why This Chapter Matters

Row reduction to echelon form is the algorithmic backbone of linear algebra: it computes rank, solves $Ax = b$, and finds the null space all in one sweep. UPSC 2025 posed a 15-mark Section B question asking candidates to reduce a matrix to echelon form and then use it to determine rank or solve a system — rewarding careful, displayed work. Although the atom has appeared only once, it underpins nearly every other linear-algebra topic, making fluency with the algorithm essential.

Minimum Theory

Elementary row operations (EROs):

1. $R_i \leftrightarrow R_j$ — swap two rows.
2. $R_i \to c R_i$, $c \neq 0$ — scale a row by a nonzero constant.
3. $R_i \to R_i + c R_j$, $i \neq j$ — add a scalar multiple of one row to another.

Each ERO is invertible; hence they do not change the row space or the rank.

Row Echelon Form (REF): A matrix is in REF if

• Every zero row is below all nonzero rows.
• The leading (first nonzero) entry — the pivot — of each nonzero row lies strictly to the right of the pivot of the row above it.

Reduced Row Echelon Form (RREF): REF plus

• Every pivot equals 1.
• Every other entry in a pivot column is 0.

Gaussian elimination converts any matrix to REF; Gauss–Jordan elimination continues to RREF.

Rank: $\text{rank}(A)$ = number of nonzero rows in any REF of $A$ = number of pivots.

Rank–Nullity Theorem: For $A$ an $m\times n$ matrix, $\text{rank}(A) + \text{nullity}(A) = n.$

Solving $Ax = b$: Augment to $[A \mid b]$ and row-reduce.

• Consistent iff no pivot appears in the last column (i.e., $\text{rank}(A) = \text{rank}([A\mid b])$).
• Unique solution iff additionally $\text{rank}(A) = n$ (no free variables).

Column reduction: Analogous operations on columns preserve the column space structure and may be combined with row reduction (PAQ decomposition) to find the rank normal form.

Normal form under row and column operations: Every $m \times n$ matrix of rank $r$ is equivalent (under EROs and ECOs) to $\begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}.$

Question Archetypes

reduce-and-rank (1 question(s); 2025)

Recognition Cues

• A concrete matrix (often $3\times 3$ or $4\times 4$ with integer entries) is given.
• Asked to “reduce to echelon form” — may say REF or RREF.
• Rank is asked, or the reduction is the first step of a larger system-solving problem.
• 15 marks implies full displayed work: every row operation labelled.

Solution Template

1. Write the matrix (or augmented matrix $[A\mid b]$).
2. Use $R_i \to R_i - \frac{a_{ij}}{a_{jj}} R_j$ to create zeros below each pivot column; label each operation.
3. (For RREF) Scale each pivot row to make the leading entry 1; then eliminate upward.
4. Count the nonzero rows — this is the rank.
5. If solving $Ax=b$: identify free variables; back-substitute to write the general solution.

Worked Example

2025 Paper 1, 2025-P1-Q3b (15 marks)

Reduce the following matrix to row echelon form and hence determine its rank. Also find the complete solution of the system $Ax = b$ where $b = (1, 2, 3, 0)^T$.

$A = \begin{pmatrix} 1 & 2 & 1 & 3 \\ 2 & 4 & 3 & 7 \\ 3 & 6 & 4 & 10 \\ 1 & 2 & 2 & 4 \end{pmatrix}$

Step 1. Form the augmented matrix $[A \mid b]$:

$\left(\begin{array}{cccc|c} 1 & 2 & 1 & 3 & 1 \\ 2 & 4 & 3 & 7 & 2 \\ 3 & 6 & 4 & 10 & 3 \\ 1 & 2 & 2 & 4 & 0 \end{array}\right)$

Step 2. $R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - 3R_1$, $R_4 \to R_4 - R_1$:

$\left(\begin{array}{cccc|c} 1 & 2 & 1 & 3 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & -1 \end{array}\right)$

Step 3. $R_3 \to R_3 - R_2$, $R_4 \to R_4 - R_2$:

$\left(\begin{array}{cccc|c} 1 & 2 & 1 & 3 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 \end{array}\right)$

Step 4. $R_4 \to R_4 - 0 \cdot R_3$ (already reduced). The last row reads $0 = -1$, which is a contradiction, so the system is inconsistent for $b = (1,2,3,0)^T$.

The matrix $A$ itself (ignore the last column) has two nonzero rows in echelon form:

$\text{REF of } A = \begin{pmatrix} 1 & 2 & 1 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

$\boxed{\text{rank}(A) = 2}$

Step 5. Null space (for completeness, with $b=0$): free variables are $x_2 = s$ and $x_4 = t$.

From row 2: $x_3 = -t$. From row 1: $x_1 = -2s - x_3 - 3t = -2s + t - 3t = -2s - 2t$.

General solution of $Ax = 0$:

$x = s\begin{pmatrix}-2\\1\\0\\0\end{pmatrix} + t\begin{pmatrix}-2\\0\\-1\\1\end{pmatrix}, \quad s,t\in\mathbb{R}.$

$\boxed{\text{nullity}(A) = 2 \text{ (confirming rank} + \text{nullity} = 4 = n\text{)}}$

Common Traps

• Forgetting to apply the same row operation to the augmented column $b$ — gives a wrong consistency verdict.
• Stopping at REF when RREF is required, or vice versa — read the question carefully.
• Mis-labelling free vs. pivot variables when columns 2 and 4 are non-pivot: always check which columns contain pivots after full reduction.
• Arithmetic sign errors when $R_i \to R_i - cR_j$ with negative entries — write out the multiplier explicitly before subtracting.

Marks-Aware Writing

At 15 marks (Section B), UPSC expects full displayed work. Allocate roughly: 4 marks for setting up and performing all row operations with labels; 2 marks for identifying the REF/RREF clearly; 2 marks for stating the rank; 4 marks for the consistency check and back-substitution; 3 marks for writing the general solution or null-space basis. Do not skip steps or bundle multiple operations — each operation on its own line earns partial credit.

Practice Set

Only one historical question on this atom (shown above).