GATE CSE 2021 Set 2 | Question: 24

Question

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 __________

Comments

Detailed Video Solution – Here

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What I understand is that rank(A)= # pivots in the echelon form of a matrix.

As, the given matrix is of 5 unknown variable but with 4 equations, so,  $\exists$ 1 free variable.

So, the # pivot is (5 – 1)= 4. The rank is 4.

Q. How this # LI or LD eigenvector is relevant to this question?

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see simple apply rank nullity theorem :

rank + nullity = n(no of Dim)

okay, now as to question there's matrix 4*5 mean m*n as to that n is 5 also remember no.of column also represent dimension.
then in question there's has been mentioned that there only of scalar multiple which you can call only one vector space .
remember :

no.of vector space = no.of Null space(Nullity)

okay so we get 1 nullity.
then now apply theorem:-
we get,
r + 1 = 5
r = 5 - 1
r=4

THIS IS ANOTHER WAY OF SOLVING THIS QUESTION HOPE IT HELPED YOU!!

Answer 1

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.

Comments

Since every other solution can be expressed in the form of a given vector u, so we can write $X=k⋅u. $ (where k is scalar)

This means u form the basis of null(P) and null space has dimension 1. Hence $nullity(P)=1$,

$Rank = 5(no of column vectors)- 1(nullity)=4 $

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When we say that Ax=0 has some non trivial solution then that means c1v1+c2v2+...+cnVn=0 and here at least one of the ci is not zero as solution is non trivial. So A will be a linearly dependent matrix.

@Anuj2000

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Concept of the question-

“Every solution is a scalar multiple of $[2 \ 5\  4\  3\  1]^T$”

It means, $[2 \ 5\  4\  3\  1]^T$” is the only linearly independent solution.

There is a difference between Linearly Independent Solution and Linearly Independent Column.

# of LI Solution = # free variables = Nullity = 1 (here)

While, #LI Columns = # pivot elements = Rank

Hence, Rank = $n – \ nullity = 5-1 = 4$ (Ans).

 

Answer 2

Rank $+$ Nullity = Number of Columns

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

Rank : $5\ – 1 = 4$

Comments

Very well explained. But I have one suggestion - along with “Nullity is the dimension of the null space”, you can also mention the fact that “Nullity of a matrix A is no. of possible column vectors X satisfying the equation AX = 0”. It will be more intuitive for some people.

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how nullity is one

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@Anuj,

Nullity is the dimension of the null space

in other words,

Nullity of a matrix A is no. of possible column vectors X satisfying the equation AX = 0

In the given question, we are interested in finding the nullity of P. So we need to find out the number of possible column vectors X satisfying the equation PX = 0. In the question, it is also given that

every solution of the equation PX=0 is a scalar multiple of [25431]$^T$

this means that X is of the form k [25431]$^T$ where k is a scalar. Clearly, the Null Space of P contains only 1 vector. Hence Nullity is 1.

You can also refer to this video for better understanding.

Answer 3

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

Comments

simple and nice explanation

Answer 4

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