Discussion
If
\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 3 & -2 \\ 0 & -6 & 4 \end{pmatrix}, \]
then the rank of the matrix \(AA^{t}\) is :
Answer: B
The rank of \(AA^{t}\) is 2.
For a real matrix, \(\text{rank}(AA^{t}) = \text{rank}(A)\), because \(AA^{t}x = 0\) gives \(x^{t}AA^{t}x = |A^{t}x|^{2} = 0\), so \(AA^{t}\) and \(A^{t}\) have the same null space.
So find the rank of A. The third row \((0, -6, 4)\) is \(-2\) times the second row \((0, 3, -2)\), so the rows are dependent and the rank is at most 2. The first two rows are clearly independent (only the first has a nonzero first entry). Hence rank A = 2, and \(AA^{t}\) has rank 2 as well.
Rank 3 would need \(\det A \ne 0\), but \(\det A = 1 \cdot (3 \cdot 4 - (-2)(-6)) = 0\). Rank 1 or 0 would need the first two rows to be dependent, which they are not.
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