L is 0 because it cancels itself out
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
C
Y=X+9
9 is called y intercept that's where pt starts
The number is front of X is called slope if no number it's 1 so it's 1/1 one up and one right up positive right positive start from 9
since by the provided tickmarks WE ≅ ER, and QE ≅ ET, then both triangles are congruent by LL theorem, Leg Leg.
anyhow, since those legs are congruent, their hypotenuses must also be congruents, and we could use the HL theorem as well.
The composite is 4 .
I hope this helps and please mark as branilyest I need 4 more to level up in rank.
Turtle14526