Question:
Let \(A\) be an \(m\times n\) real matrix with non-zero entries. Which of the following statements is/are true?
Let \(A\) be an \(m\times n\) real matrix with non-zero entries. Which of the following statements is/are true?
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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 \(n=m\)
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The Correct Option is A, C, D
Solution and Explanation
Step 1: Check option (A).
Here, \(A\) is an \(m\times n\) matrix and \(m<n\).
The homogeneous system is
\[ AX=0 \] This system has \(n\) unknowns and \(m\) equations.
Since
\[ n>m, \] there are more unknowns than equations. Therefore, the system must have a non-trivial solution.
So, there exists a non-zero \(n\times 1\) real matrix \(X\) such that
\[ AX=0 \] Hence, option (A) is true.
Step 2: Check option (B).
If \(m>n\), then \(A\) is a tall matrix.
But \(A^TA\) is an \(n\times n\) matrix.
The matrix \(A^TA\) is singular only when the columns of \(A\) are linearly dependent.
However, when \(m>n\), it is possible that the \(n\) columns of \(A\) are linearly independent.
In that case,
\[ A^TA \] is non-singular.
Therefore, option (B) is not always true.
Hence, option (B) is false.
Step 3: Check option (C).
If
\[ n=1, \] then \(A\) is an \(m\times 1\) column matrix.
Since all entries of \(A\) are non-zero, \(A\neq 0\).
Thus,
\[ \operatorname{rank}(A)=1 \] Now,
\[ AA^T \] is an \(m\times m\) matrix.
Also,
\[ \operatorname{rank}(AA^T)=\operatorname{rank}(A)=1 \] Using the rank-nullity theorem for \(AA^T\),
\[ \text{nullity}(AA^T)=m-\operatorname{rank}(AA^T) \] \[ =m-1 \] Hence, option (C) is true.
Step 4: Check option (D).
Given
\[ AB=I_m \] This means \(A\) has a right inverse. Hence,
\[ \operatorname{rank}(A)=m \] So,
\[ m\leq n \] Also, given
\[ CA=I_n \] This means \(A\) has a left inverse. Hence,
\[ \operatorname{rank}(A)=n \] So,
\[ n\leq m \] Combining both inequalities,
\[ m\leq n \] and
\[ n\leq m \] Therefore,
\[ m=n \] Hence, option (D) is true.
Step 5: Final conclusion.
The true statements are
\[ \boxed{(A),\,(C)\text{ and }(D)} \]
Here, \(A\) is an \(m\times n\) matrix and \(m<n\).
The homogeneous system is
\[ AX=0 \] This system has \(n\) unknowns and \(m\) equations.
Since
\[ n>m, \] there are more unknowns than equations. Therefore, the system must have a non-trivial solution.
So, there exists a non-zero \(n\times 1\) real matrix \(X\) such that
\[ AX=0 \] Hence, option (A) is true.
Step 2: Check option (B).
If \(m>n\), then \(A\) is a tall matrix.
But \(A^TA\) is an \(n\times n\) matrix.
The matrix \(A^TA\) is singular only when the columns of \(A\) are linearly dependent.
However, when \(m>n\), it is possible that the \(n\) columns of \(A\) are linearly independent.
In that case,
\[ A^TA \] is non-singular.
Therefore, option (B) is not always true.
Hence, option (B) is false.
Step 3: Check option (C).
If
\[ n=1, \] then \(A\) is an \(m\times 1\) column matrix.
Since all entries of \(A\) are non-zero, \(A\neq 0\).
Thus,
\[ \operatorname{rank}(A)=1 \] Now,
\[ AA^T \] is an \(m\times m\) matrix.
Also,
\[ \operatorname{rank}(AA^T)=\operatorname{rank}(A)=1 \] Using the rank-nullity theorem for \(AA^T\),
\[ \text{nullity}(AA^T)=m-\operatorname{rank}(AA^T) \] \[ =m-1 \] Hence, option (C) is true.
Step 4: Check option (D).
Given
\[ AB=I_m \] This means \(A\) has a right inverse. Hence,
\[ \operatorname{rank}(A)=m \] So,
\[ m\leq n \] Also, given
\[ CA=I_n \] This means \(A\) has a left inverse. Hence,
\[ \operatorname{rank}(A)=n \] So,
\[ n\leq m \] Combining both inequalities,
\[ m\leq n \] and
\[ n\leq m \] Therefore,
\[ m=n \] Hence, option (D) is true.
Step 5: Final conclusion.
The true statements are
\[ \boxed{(A),\,(C)\text{ and }(D)} \]
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