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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

A philanthropist pledges to donate 10% of a fund each year. If the fund initially has $450,000 how much will

Ann [662]

After 7 years fund balance will be $2152336.05 if the philanthropist pledges to donate 10% of a fund each year.

<h3>What is an exponential function?</h3>

It is defined as the function that rapidly increases and the value of the exponential function is always a positive. It denotes with exponent. $\rm y = a^x$

where a is a constant and a>1

We have:

A philanthropist pledges to donate 10% of a fund each year.

We can find the amount left after 7 years as follows:

A = 450000(1 - 0.1)⁷

10% of a fund each year.

Rate r = 10% = 0.1

A = 450000(0.9)⁷

A = $2152336.05

Thus, after 7 years fund balance will be $2152336.05 if the philanthropist pledges to donate 10% of a fund each year.

Learn more about the exponential function here:

brainly.com/question/11487261

#SPJ1

6 0

2 years ago

Solve for w p=1.2w/h^2

andreyandreev [35.5K]

Solve for w:
p = (1.2 w)/h^2
(1.2 w)/h^2 = (6 w)/(5 h^2):
p = (6 w)/(5 h^2)
p = (6 w)/(5 h^2) is equivalent to (6 w)/(5 h^2) = p:
(6 w)/(5 h^2) = p
Multiply both sides by (5 h^2)/6:
Answer: w = (5 h^2 p)/6

6 0

3 years ago

Evaluate when x=-4 and y=2 2xy+ysquared

Bas_tet [7]

The answer would be -12

8 0

3 years ago

What is the answer to A and B?

Alex_Xolod [135]

For A) u must start at -3 on y lines. Then move points up 1 and 3 to the right. Makes point and continue to make lines
For B) start at 5 on y lines, and remember -x is same thing is -1. And negative means go down! So move 1 down and 1 to the right and make point then move 1 down and make 1 right and make second points and continue. Draw a lines.

The solution is when u see two lines from two equations crosses to each other. There’s solution shown.

3 0

3 years ago