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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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14 Answers

2 votes
2 votes

If we focus on the question it is stating that the solution is scalar multiple of \begin {bmatrix}2\\5\\4\\3\\1\end{bmatrix}We are basically saying the solution can be generated only using this. Therefore Number of LI solution is equal to 1 that translates the Matrix P has only 1 Free Variable and Number of Pivot Variables/ Number of LI Columns in P are equal to 4 which means the Rank is equal to 4.

If we clearly understand and dont’t confuse between LI Solution and LI Columns this can be solved easily.

1 votes
1 votes
Question : Given  P is a matrix of 4 x 5 matrix and  solution for homogenous system i.e. Px = 0 has solution( set of x) is scalar multiple of [ 2 5 4 3 1]^T:

(in simple word they are saying that solution with is “x” is multiple of  [ 2 5 4 3 1]^T i.e.
x = k [ 2 5 4 3 1]^T where k is scaler .)

Solution 1: here number of equation are 4 and number of variables are 5

and there is only one free variable and all the other are Linearly Independent => nullity is 1

we know rank + nullity = number of free variables
rank + 1 = 5 => rank = 4.
Solution 2 : x = k [ 2 5 4 3 1]^T means that number of Linear Independent solution is 1  which is multiple of  [ 2 5 4 3 1]^T  => nullity = 1 => rank = 5 -1 = 4
0 votes
0 votes
This is a homogenous equation and having a non-trivial solution which means it’s solution set is obtained from solving using parametric method.

In question it’s given that matrix is having 4 rows and 5 cols and since all the solutions are  scalar multiple of 1-column vector means there will be only one free variable in the P matrix.

Since, there is only 1 free variable we are left with 4 columns in P with Pivots which will be counted as a rank bcz rank of matrix = no. of cols with pivots.
Answer:

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