Rank of a matrix

At a Glance

• Frequency: 7 sub-parts across 7 of 13 years (2013, 2014, 2015, 2018, 2019, 2021, 2023)
• Priority tier: T2
• Marks (count): 8 (1), 10 (4), 15 (2)
• Average solve time: ~7 min
• Difficulty mix: easy 5, medium 2
• Section: A | Dominant type: computation

Why This Chapter Matters

Rank appears in 7 of the last 13 years — exclusively in Section A — making it one of the most consistent Paper 1 Linear Algebra atoms. Six of the seven questions are straightforward row-reduction problems; only the 2018 question (show that a $3\times2$ times a $2\times3$ matrix is singular) uses a rank inequality as a proof. If you can reduce a matrix to echelon form reliably under exam conditions and verify your answer by exhibiting a non-vanishing minor, you will handle every question in the corpus.

Minimum Theory

Rank. The rank of a matrix $A$ is the number of non-zero rows in its row echelon form (equivalently, the dimension of the row space, column space, or image). A square $n\times n$ matrix is non-singular (invertible, $\det A\ne0$) iff $\operatorname{rank}A=n$.

Row echelon form (REF). A matrix is in REF when (i) every row of zeros is below all non-zero rows, and (ii) the leading non-zero entry (pivot) of each non-zero row is strictly to the right of the pivot in the row above. Reduced REF (RREF) additionally requires each pivot to be 1, with zeros above and below it.

Elementary row operations. Three operations do not change the rank:

1. Swap two rows.
2. Multiply a row by a non-zero scalar.
3. Add a multiple of one row to another.

Finding rank by reduction. Apply elementary row operations to produce REF; count the non-zero rows.

Verifying rank. Rank $\ge r$ iff some $r\times r$ submatrix has non-zero determinant. To confirm rank $=r$, show an $r\times r$ minor is non-zero and every $(r+1)\times(r+1)$ minor is zero (the row reduction already guarantees this).

Rank inequality for products. For any matrices: $\operatorname{rank}(AB)\le\min\{\operatorname{rank}A,\,\operatorname{rank}B\}.$ Proof: every column of $AB$ is a linear combination of the columns of $A$, so $\operatorname{col}(AB)\subseteq\operatorname{col}(A)$ and $\operatorname{rank}(AB)\le\operatorname{rank}A$. Dually (via rows) $\operatorname{rank}(AB)\le\operatorname{rank}B$.

Question Archetypes

Two patterns cover every rank question in the corpus.

rank-by-reduction (6 question(s); 2013, 2014, 2015, 2019, 2021, 2023)

Recognition Cues

• “Find the rank of $A$” or “reduce to (row-reduced) echelon form; find rank.”
• Given a $4\times4$, $5\times5$, or $4\times5$ matrix with integer entries.

Solution Template

1. Look for obvious row dependencies (e.g., $R_3=R_1+R_2$) to zero a row cheaply.
2. Use the row with a leading $1$ (or the simplest non-zero entry) as the first pivot; this avoids fractions.
3. Eliminate below each pivot column by column.
4. Swap zero rows to the bottom; count the non-zero rows.
5. Optional verification: exhibit a non-vanishing $r\times r$ minor where $r$ is the claimed rank.

Worked Example(s)

2014 Paper 1, 2014-P1-Q1b (10 marks, compulsory)

Find the rank of $A=\begin{bmatrix}0&1&-3&-1\\ 0&0&1&1\\ 3&1&0&2\\ 1&1&-2&0\end{bmatrix}$.

$R_1\leftrightarrow R_4$ (bring a leading 1 to row 1): $\begin{bmatrix}1&1&-2&0\\ 0&0&1&1\\ 3&1&0&2\\ 0&1&-3&-1\end{bmatrix}.$

$R_3\to R_3-3R_1$: $\begin{bmatrix}1&1&-2&0\\ 0&0&1&1\\ 0&-2&6&2\\ 0&1&-3&-1\end{bmatrix}.$

$R_2\leftrightarrow R_4$ (bring a non-zero entry into the column-2 pivot): $\begin{bmatrix}1&1&-2&0\\ 0&1&-3&-1\\ 0&-2&6&2\\ 0&0&1&1\end{bmatrix}.$

$R_3\to R_3+2R_2$: $\begin{bmatrix}1&1&-2&0\\ 0&1&-3&-1\\ 0&0&0&0\\ 0&0&1&1\end{bmatrix}.$

$R_3\leftrightarrow R_4$: $\begin{bmatrix}1&1&-2&0\\ 0&1&-3&-1\\ 0&0&1&1\\ 0&0&0&0\end{bmatrix}.\quad(\text{REF})$

Three non-zero rows: $\boxed{\operatorname{rank}(A)=3.}$

2019 Paper 1, 2019-P1-Q3c-i (15 marks)

Find the rank of $A=\begin{pmatrix}5&7&2&1\\1&1&-8&1\\2&3&5&0\\3&4&-3&1\end{pmatrix}$.

Pivot on $R_2$ (leading 1 is cheapest). Rearrange to put $R_2$ first:

$R_1-5R_2,\;R_3-2R_2,\;R_4-3R_2$ (using $R_2=(1,1,-8,1)$): $\begin{bmatrix}1&1&-8&1\\ 0&2&42&-4\\ 0&1&21&-2\\ 0&1&21&-2\end{bmatrix}.$

Note $R_3=\tfrac{1}{2}R_2$ and $R_4=R_3$. Eliminate: $R_2-2R_3$ and $R_4-R_3$: $\begin{bmatrix}1&1&-8&1\\ 0&1&21&-2\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix}.\quad(\text{REF})$

Two non-zero rows: $\boxed{\operatorname{rank}(A)=2.}$

Verification. The $2\times2$ minor at rows $\{1,2\}$, columns $\{1,2\}$ is $\det\begin{pmatrix}1&1\\1&1\end{pmatrix}=0$… pick columns $\{1,3\}$: $\det\begin{pmatrix}1&-8\\1&21\end{pmatrix}=29\ne0$. So rank $\ge2$ ✓. Four zero rows in REF confirms rank $\le2$.

2023 Paper 1, 2023-P1-Q4a (15 marks)

Find the rank of $A=\begin{bmatrix}1&2&-1&0\\-1&3&0&-4\\2&1&3&-2\\1&1&1&-1\end{bmatrix}$ by reducing to RREF.

$R_2+R_1,\;R_3-2R_1,\;R_4-R_1$: $\begin{bmatrix}1&2&-1&0\\ 0&5&-1&-4\\ 0&-3&5&-2\\ 0&-1&2&-1\end{bmatrix}.$

$R_3+\tfrac{3}{5}R_2,\;R_4+\tfrac{1}{5}R_2$: $\begin{bmatrix}1&2&-1&0\\ 0&5&-1&-4\\ 0&0&22/5&-22/5\\ 0&0&9/5&-9/5\end{bmatrix}.$

$R_4-\tfrac{9}{22}R_3$ zeros out row 4 entirely (since $9/5-(9/22)(22/5)=0$): $\begin{bmatrix}1&2&-1&0\\ 0&5&-1&-4\\ 0&0&22/5&-22/5\\ 0&0&0&0\end{bmatrix}.\quad(\text{REF})$

Three pivots: $\boxed{\operatorname{rank}(A)=3.}$

Common Traps

• Always pivot on the row with a leading $1$ (not the first row mechanically). Pivoting on $R_2$ in the 2019 example avoids all fractions.
• After row-reducing, count non-zero rows — not columns, not pivots by column position.
• For a sanity check: exhibit a non-vanishing minor of size equal to the rank. A $2\times2$ determinant takes 10 seconds.
• Rows $R_3$ and $R_4$ in the 2019 example are identical after one sweep — a pattern easy to miss if you rush.

rank-inequality-proof (1 question(s); 2018)

Recognition Cues

• “Show that $AB$ is singular” where $A$ is $m\times k$ and $B$ is $k\times n$ with $k<\min(m,n)$.
• The word “singular” for a square matrix means rank $<$ order.

Solution Template

1. State: $\operatorname{rank}(AB)\le\min\{\operatorname{rank}A,\operatorname{rank}B\}$.
2. Justify the bound: columns of $AB$ lie in the column space of $A$; thus rank$(AB)\le$ rank$(A)\le k$.
3. Since $k<n$ (order of the square matrix $AB$), rank$(AB)<n$ $\Rightarrow$ $AB$ is singular.

Worked Example(s)

2018 Paper 1, 2018-P1-Q1a (10 marks, compulsory)

Let $A$ be $3\times2$ and $B$ be $2\times3$. Show $C=AB$ is singular.

$C=AB$ is $3\times3$ (the product is defined and square). For any matrices, $\operatorname{rank}(AB)\le\min\{\operatorname{rank}A,\,\operatorname{rank}B\}.$ Why: every column of $AB$ is a linear combination of the columns of $A$, so $\operatorname{col}(AB)\subseteq\operatorname{col}(A)$ and $\operatorname{rank}(AB)\le\operatorname{rank}A$. But $A$ has only 2 columns, so $\operatorname{rank}A\le2$. Therefore $\operatorname{rank}(C)\le 2 < 3 = \operatorname{order}(C).$ A $3\times3$ matrix with rank $<3$ is singular: $\boxed{\;\det(C)=0,\quad C=AB\text{ is singular.}\;}$

Common Traps

• State the rank-of-product inequality and justify it (column-space containment). Merely asserting it loses marks.
• The key numbers: $A$ is $3\times2$, so rank$(A)\le2$; this is the tight bound that forces singularity of the $3\times3$ product.

Marks-Aware Writing

8-mark questions (2013-Q2b-ii): Show the row-reduction steps explicitly — at least 2–3 labelled steps — then state the rank. If you spot a dependency by inspection ($R_3=R_1+R_2$), state it; it counts as a valid first step.

10-mark questions (2014-Q1b, 2015-Q1b, 2018-Q1a, 2021-Q4a-i): Full row-reduction to echelon form with all steps shown; state the rank from the count of non-zero rows. For the inequality proof, write the theorem, justify it, and conclude.

15-mark questions (2019-Q3c-i, 2023-Q4a): Produce RREF (not just REF) and verify by exhibiting the linear-dependence relations among the original rows. Also exhibit a non-vanishing minor of the claimed rank.

Practice Set