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
Answer:
D
Step-by-step explanation:
our basic Pythagorean identity is cos²(x) + sin²(x) = 1
we can derive the 2 other using the listed above.
1. (cos²(x) + sin²(x))/cos²(x) = 1/cos²(x)
1 + tan²(x) = sec²(x)
2.(cos²(x) + sin²(x))/sin²(x) = 1/sin²(x)
cot²(x) + 1 = csc²(x)
A. sin^2 theta -1= cos^2 theta
this is false
cos²(x) + sin²(x) = 1
isolating cos²(x)
cos²(x) = 1-sin²(x), not equal to sin²(x)-1
B. Sec^2 theta-tan^2 theta= -1
1 + tan²(x) = sec²(x)
sec²(x)-tan(x) = 1, not -1
false
C. -cos^2 theta-1= sin^2
cos²(x) + sin²(x) = 1
sin²(x) = 1-cos²(x), our 1 is positive not negative, so false
D. Cot^2 theta - csc^2 theta=-1
cot²(x) + 1 = csc²(x)
isolating 1
1 = csc²(x) - cot²(x)
multiplying both sides by -1
-1 = cot²(x) - csc²(x)
TRUE
The answer should be 38/7
Answer:
Slope = Rise over Run
Rise = 2
Run = 3
Slope = 2/3
Let me know if this helps!
