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UPSC 2018 Maths Optional Paper 1 Q1a — Step-by-Step Solution

10 marks · Section A

Rank of a matrix · Linear Algebra · asked 7× in 13 yrs · Read the full method →

Question

Let $A$ be a $3\times2$ matrix and $B$ a $2\times3$ matrix. Show that $C=A\cdot B$ is a singular matrix.

Technique

Rank of a product is bounded by the ranks of the factors; a square matrix with rank below its order is singular.

Solution

Step 1 — $C$ is a square $3\times3$ matrix

$A$ is $3\times2$ and $B$ is $2\times3$, so the product $C=AB$ is defined and has size $3\times3$. Singularity ($\det C=0$) is therefore a meaningful question.

Step 2 — Rank bound for a product

For any matrices, $\operatorname{rank}(AB)\le\min\{\operatorname{rank}A,\operatorname{rank}B\}$. The reason: every column of $AB$ is a linear combination of the columns of $A$, so the column space of $AB$ is contained in the column space of $A$; hence $\operatorname{rank}(AB)\le\operatorname{rank}A$. (Dually, every row of $AB$ is a combination of the rows of $B$, giving $\operatorname{rank}(AB)\le\operatorname{rank}B$.)

Step 3 — Bound the rank of $A$ and $B$

$A$ has only $2$ columns, so $\operatorname{rank}A\le 2$. Likewise $B$ has only $2$ rows, so $\operatorname{rank}B\le 2$. Therefore

Step 4 — Rank $<$ order $\Rightarrow$ singular

$C$ is $3\times3$ but $\operatorname{rank}C\le2<3$. A square matrix of order $n$ is non‑singular iff its rank equals $n$. Since $\operatorname{rank}C\le2<3$, $C$ cannot have full rank, hence

Answer