Question:medium

Let \(A\) be an \(m\times n\) real matrix with non-zero entries. Which of the following statements is/are true?

Show Hint

For an \(m\times n\) matrix, compare \(m\) and \(n\) using rank-nullity. A homogeneous system with more unknowns than equations always has a non-trivial solution.
Updated On: Jun 4, 2026
  • If \(m<n\), then there exists a non-zero \(n\times 1\) real matrix \(X\) such that \(AX=0\), where \(0\) is the \(m\times 1\) zero matrix
  • If \(m>n\), then \(A^TA\) is a singular matrix
  • If \(n=1\) and \(m>1\), then \(AA^T\) has nullity \(m-1\)
  • If there exists an \(n\times m\) matrix \(B\) such that \(AB=I_m\) and an \(n\times m\) matrix \(C\) such that \(CA=I_n\), then \(n=m\)
Show Solution

The Correct Option is A, C, D

Solution and Explanation

Step 1: Tackle (A).
If $m<n$, the system $AX=0$ has more unknowns than equations, so a non-zero solution must exist. (A) holds.

Step 2: Tackle (B).
When $m>n$, the $n$ columns can be independent, making $A^TA$ non-singular. So $A^TA$ need not be singular and (B) fails.

Step 3: Tackle (C).
With $n=1$, $A$ is a non-zero column, so $\mathrm{rank}(A)=1$ and $\mathrm{rank}(AA^T)=1$. By rank-nullity on the $m\times m$ matrix $AA^T$, the nullity is $m-1$. (C) holds.

Step 4: Tackle (D).
$AB=I_m$ forces $\mathrm{rank}(A)=m$, so $m\le n$; $CA=I_n$ forces $\mathrm{rank}(A)=n$, so $n\le m$. Together $m=n$. (D) holds.

Step 5: Collect.
\[ \boxed{(A),(C),(D)} \]
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