Using a calculator, work out:<br> V2704 x 22

Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

Applying the coordinates 2,6 and -2 to the basis E we obtain

$2+6t-2t^{2}$

That was the original result of T[e1(t)]

8 0

3 years ago

????? HELP PLEASEEE!!

lys-0071 [83]

What is the question?

5 0

2 years ago

What is the domain of <br> F(x)=√x ?<br><br> (PLEASE HELP ASAP)

leva [86]

Option C. all real numbers greater than 0 is the correct answer

Further explanation:

Domain is the set of all inputs on which the function is defined or real.

The given function is:

$f(x)=\sqrt{x}$

As the function involves a square root, we have to avoid negative numbers.

All positive real numbers will produce defined output so the positive real numbers will be included in the domain.
If we look at negative real numbers, we will be taking square root of a negative number which will not result in a real number

So the domain will only include positive real numbers.

Hence, Option C. all real numbers greater than 0 is the correct answer.

Keywords: Domain, Domain of radical functions

Learn more about domain at:

#LearnwithBrainly

5 0

3 years ago

-2(2n-2)>18-3n how do you solve this inequality

Svet_ta [14]

-2 * 2n -2 equals -4n + 4 > 18n -3n

8 0

3 years ago

Find the perimeter of the following rectangle.

Ede4ka [16]

So the formula to workout the area of a rectangle is height X width

so 11/3 x 5/8

which is 55/24 ft if they want a mixed number then... 2 $\frac{7}{24\\}$

hope this helps

6 0

2 years ago

Using a calculator, work out: V2704 x 22 - 149