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castortr0y [4]
4 years ago
5

Express 80 as a product of prime factors using index notation

Mathematics
1 answer:
hammer [34]4 years ago
4 0
80 = 2 x 2 x 2 x 2 x 5
80 = 2 ^4 x 5
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70 degrees.

Step-by-step explanation:

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Step-by-step explanation:

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T(v) = Av represents the linear transformation T. Find a basis for the kernel of T and the range of T.
lawyer [7]

Answer:

Ker(T) has basis \emptyset

Range(T) has basis \left\{\left(\begin{array}{c}2&3\end{array}\right), \left(\begin{array}{c}1&4\end{array}\right) \right\}.

Step-by-step explanation:

If the matrix you wanted to represent  is A=\left(\begin{array}{cc}2&1\\3&4\end{array}\right), then:

ker(T)=\left\{{\bf x}\in \mathbb{R}^{2}: A{\bf x}=0\right\}.

So, to find ker(T) you must solve the homogenous equation

A{\bf x}=\left(\begin{array}{cc}2&1\\3&4\end{array}\right) \left(\begin{array}{c} x&y \end{array}\right)=\left(\begin{array}{c}0&0\end{array}\right)

Using Gauss elimination we obtain the simpler equivalent system

\left(\begin{array}{cc} -6&3\\0&5\end{array}\right) \left(\begin{array}{c} x&y\end{array}\right)=\left(\begin{array}{c}0&0\end{array}\right)

Then, we have that

x=0,y=0.

We have that ker(T)=\{\left(\begin{array}{c}0&0\end{array}\right)\}. On this case we say that the basis is the empty set \emptyset.

The range of T is the set of vectors of the form

\left(\begin{array}{c} \alpha &\beta\end{array}\right)=\left(\begin{array}{cc}2&1\\3&4\end{array}\right)\left(\begin{array}{c}x&y\end{array}\right)=x\left(\begin{array}{c}2&3\end{array}\right)+y\left(\begin{array}{c}1&4\end{array}\right)

So,

Range(T)=\langle \left(\begin{array}{c}2&3\end{array}\right), \left(\begin{array}{c}1&4\end{array}\right) \rangle. Where the angle brakets denotes the span.

5 0
4 years ago
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