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43 votes
43 votes
Suppose that $P$ is a $4 \times 5$ matrix such that every solution of the equation $\text{Px=0}$ is a scalar multiple of $\begin{bmatrix} 2 & 5 & 4 &3 & 1 \end{bmatrix}^T$. The rank of $P$ is __________
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Best answer
33 votes
33 votes
Every solution to $Px = 0$ is scalar multiple of $\begin{bmatrix} 2 & 5 & 4 &3 & 1 \end{bmatrix}^T$, It means out of 5 column vectors of matrix $P$,  $4$  are linearly independent as we have only one line in NULL Space (along the given vector).

Rank is nothing but the number of linearly independent column vectors in a matrix which is $4$ here.
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15 votes
15 votes
Rank $+$ Nullity = Number of Columns

Here, Nullity is $1$. (Nullity is the dimension of the null space)

Rank : $5\ – 1 = 4$
13 votes
13 votes

This problem is based on Linear Homogeneous equations (System of Linear Equations)

9 votes
9 votes
it is in form PX=0.

P is of order 4X5.  4 EQUATIONS AND 5 UNKNOWNS(5 variables)

n=5.

solution of PX=0 is

k  [2 5 4 3 1] transpose

here we have only 1 arbitary value.(that is only 1 scalar i.e.,K)

no. of arbitary values(scalars)= n-r

1=5-r

r=4
Answer:

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