CSIR NET MATHEMATICS | Linear Algebra | Rank Nullity Theorem / Dimension Theorem DEC 2011

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CSIR NET MATHEMATICS | Linear Algebra | Rank Nullity Theorem DEC 2011
Linear Algebra, Rank Nullity Theorem, Dimension Theorem
Example (marks 3)
Let A be a 54 matrix with real entries such that the space of all solution of the linear system AX^t=[1 2 3 4 5]^t is given by , where {[1+2s, 2+3s, 3+4s, 4+5s]^t :sR}. Then the rank of A is equal to
4 (2) 3 (3) 2 (4) 1
CSIR NET Answer Key December 2011,
CSIR NET Answer Key June 2019, Mathematical Science,
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Welcome, friends. In this video,
we are going to discuss a problem of linear algebra from CS&E December 2011.
Example. Let A be a five cross four matrix with the real entries such that.
The space of all solution of the linear system.
Ax transpose is equal to one two three four five transpose, is given by.
The set one plus two s, comma two plus three s, comma three plus four s.
Com four plus five s transpose, s belongs to R.
Then the rank of A is equal to. We have four options. First option four.
Second option three, third option two, and the fourth option one.
Now, in this problem, we have to find the rank using the solution of.
Non-homogeneous system of linear equation AX transpose
is equal to one two three four five transpose.
And we have a relation between the solution space and the rank of A.
Which is called Rank Nullity Theorem, and the statement is.
If A is a matrix of order m cross n, then.
रैंक ऑफ़ ए प्लस नॉलिटी ऑफ़ ए इक्वल टू
वेर, एन इज़ द नंबर ऑफ़ कॉलम्स इन ए, एंड नॉलिटी ऑफ़ ए
Is the dimension of solution space
of homogeneous system A X transpose equal to zero?
Now, in the given example,
we have given matrix is of order five cross four, therefore.
Number of columns in A is four, therefore n equal to four.
Now the solution space is one plus two s, two plus three s.
Three plus four s and four plus five s.
If we rewrite, in the two vector form, one two three four plus.
Two s, three s, four s, five s.
So this solution has two part. One is a constant vector.
Second vector is a multiple of s.
And the second vector, which is a multiple of s.
Is a solution of homogeneous system ax transpose is equal to zero.
Therefore.
It will generate the null space of A, and.
The nullity is nothing but the dimension of null space of A.
This whole space generated by the single vector two, three, four file.
All the other vectors are the multiple of this vector. Therefore.
The dimension of null space is one because it's generated by this single vector,
and nullity of A is equal to one.
By the rank nullity theorem, rank of A plus nullity of A equal to n.
Therefore, rank of A is equal to n minus nullity of A,
where n equal to four, nullity equal to one.
So four minus one equal to three.
So rank of a is equal to three, which is option number two, and it's correct answer.
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