Question: If the subspace of all solutions of
Ax = 0
has a basis consisting of vectors and if A is a matrix, what is the rank of A.
Note: The rank of A can only be determined if the dimension of the matrix A is given, and the number of vectors is known. Here in this question, neither the dimension, nor the number of vectors is given.
Assume: The number of vectors is 3, and the dimension is 5 × 8.
Answer:
The rank of the matrix A is 5.
Step-by-step explanation:
In the standard basis of the linear transformation:
f : R^8 → R^5, x↦Ax
the matrix A is a representation.
and the dimension of kernel of A, written as dim(kerA) is 3.
By the rank-nullity theorem, rank of matrix A is equal to the subtraction of the dimension of the kernel of A from the dimension of R^8.
That is:
rank(A) = dim(R^8) - dim(kerA)
= 8 - 3
= 5
3.7/2 = 1.85
2 goes into 3 1 time.
you have a remainder of 1
That one gets a 7 after it
now you have 17
2 goes into 17 8 times
remainder of 1
bring down the 0 that's there, but is not shown because it's not important in standard form or decimals.
You get 10 and that's 5
so you have 1.85
1.85$
Answer:
(f o g)(4) = 45
Step-by-step explanation:
f(x)=4x+1
g(x)=x²-5
(f o g)(4)=?
(f o g)(4) = f(g(4))
Calculating g(4):
x=4→g(4)=4²-5
g(4)=16-5
g(4)=11
Replacing g(4)=11
(f o g)(4) = f(g(4))
(f o g)(4) = f(11)
Calculating f(11)
x=11→f(11)=4(11)+1
f(11)=44+1
f(11)=45
Replacing f(11)=45:
(f o g)(4) = f(11)
(f o g)(4) = 45
