All sides of AABC are tangent to circle P.
1 answer:
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Answer:
(A) Set A is linearly independent and spans . Set is a basis for
.
Step-by-Step Explanation
<u>Definition (Linear Independence)</u>
A set of vectors is said to be linearly independent if at least one of the vectors can be written as a linear combination of the others. The identity matrix is linearly independent.
<u>Definition (Span of a Set of Vectors)</u>
The Span of a set of vectors is the set of all linear combinations of the vectors.
<u>Definition (A Basis of a Subspace).</u>
A subset B of a vector space V is called a basis if: (1)B is linearly independent, and; (2) B is a spanning set of V.
Given the set of vectors , we are to decide which of the given statements is true:
In Matrix , the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column.
has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans
.
Therefore Set A is linearly independent and spans . Thus it is basis for
.
Answer:
Option C
Step-by-step explanation:
We are given a coefficient matrix along and not the solution matrix
Since solution matrix is not given we cannot check for infinity solutions.
But we can check whether coefficient matrix is 0 or not
If coefficient matrix is zero, the system is inconsistent and hence no solution.
Option A)
|A|=
since II row is a multiple of I row
Hence no solution or infinite
OPtion B
|B|=
Hence no solution or infinite
Option C
Hence there will be a unique solution
Option D
=0
(since I row is -5 times III row)
Hence there will be no or infinite solution
Option C is the correct answer
Answer:
Step-by-step explanation:
Please see the picture below.
1. Given the function f(x) = |x|, applying a vertical stretch with scale factor , we have the transformed function:
2. Applying a translation of 3 units to the right, we have:
3. Finally applying a translation down of 1 unit, we have:
