Question

Suppose that \( P \) is a \( 4 \times 5 \) matrix such that every solution of the equation \( Px = 0 \) is a scalar multiple of \( [\,2\;5\;4\;3\;1\,]^T \). The rank of \( P \) is \(\underline{\hspace{2cm}}\).

Hint:

If all solutions of \( Ax=0 \) are scalar multiples of a single vector, the nullity of \( A \) is 1.

Answer 1

Correct Answer: 4

Step 1: Interpret the solution space of \( Px = 0 \).
It is given that every solution of \( Px = 0 \) is a scalar multiple of a single non-zero vector. Hence, the null space of \( P \) is one-dimensional.

Step 2: Use the Rank–Nullity Theorem.
For a matrix \( P \) of size \( 4 \times 5 \):
\[ \text{rank}(P) + \text{nullity}(P) = \text{number of columns} = 5 \]

Step 3: Substitute the nullity.
Since the null space is one-dimensional:
\[ \text{nullity}(P) = 1 \]

Step 4: Compute the rank.
\[ \text{rank}(P) = 5 - 1 = 4 \] % Final Answer

Final Answer: \[ \boxed{4} \]