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liq [111]
2 years ago
5

Solve for homework :)

Mathematics
2 answers:
katrin2010 [14]2 years ago
6 0

Answer:

1) \frac{ {a}^{8} }{ {a}^{6} }  \\  = {a}^{8}  \times  {a}^{ - 6}  =  {a}^{2}  \\ 2) {m}^{2}  {m}^{ - 4}  =  {m}^{ - 2}  \\ thank \: you

Alex73 [517]2 years ago
3 0

Answer:

8. a^2

9. 1/m^2

Step-by-step explanation

8. Subtract the powers

9. Subtract the powers (2-4) and get m^-2, which is 1/m^2

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(A) Set A is linearly independent and spans R^3. Set is a basis for R^3.

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  A= \left(\begin{array}{[c][c][c][c]}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\end{array} \right) , we are to decide which of the given statements is true:

In Matrix A= \left(\begin{array}{[c][c][c][c]}(1) & 0 & 0 & 0\\ 0 & (1) & 0 & 1\\ 0 & 0 & (1) & 1\end{array} \right) , 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. R^3 has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans R^3.

Therefore Set A is linearly independent and spans R^3. Thus it is basis for R^3.

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