Discussion
If P is a \(3 \times 3\) matrix of rank three and :
\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}, \]
then the rank of PA is :
Answer: B
The rank of PA is 2.
A \(3 \times 3\) matrix of rank three is invertible, and multiplying by an invertible matrix does not change rank. PA and A have the same null space, since \(PAx = 0\) gives \(Ax = P^{-1}0 = 0\). So rank PA = rank A.
For A, the rows are in arithmetic progression: row 3 minus row 2 equals row 2 minus row 1, both \((3, 3, 3)\). So row 1 - 2 row 2 + row 3 = 0 and \(\det A = 0\), while the first two rows are independent. Hence rank A = 2 and rank PA = 2.
3 would need A invertible. 1 would need all rows proportional. 'Depends upon matrix P' is the trap: it would be right if P could be singular, but rank three rules that out.
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