Please help me with the answer plz

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

1. Which inequality is true?<br>A.6 π > 18<br>B.6 π < 18<br>C.6 π = 18<br>D.18 > 6 π<br>

lilavasa [31]

Answer:

a

Step-by-step explanation:

a is the answer ..........

7 0

3 years ago

Which function is equivalent to y = x2 - 6x + 10?

valentina_108 [34]

Option D
I hope that is right

3 0

3 years ago

1. Howard's CD House sells new CDs for $15 each and used CDs for $6 each. Bre'anna bought 51 CDs and spent a

Yanka [14]

The numbers of new CDs is 30 and the old CDs is 21.

To solve this problem, we have to write a system of equations in which we would know the numbers of each types of CD she bought.

<h3>System of Equations</h3>

This is used to solve a series or complex of word problems in which we can represent in a mathematical statements;

Let n represents numbers of new CDs
Let u represents numbers of old CDs

The equations can be represented as

$n + u = 51 ...equation(i)\\15n + 6u = 576 ... equation(ii)$

Let's take equation (i)

Making n the subject of formula

$n + u = 51\\n = 51 - u...equation(iii)$

we can put equation (iii) into equation (ii)

$15n + 6u =576\\n = 51 - u\\15(51 - u) + 6u = 576\\765-15u + 6u = 576\\collect like terms\\765-576=15u - 6u\\189=9u\\9u/9 = 189/9\\u = 21$

Let's put the value of u into equation (i)

$n + u = 51\\u = 21\\n + 21 = 51\\n = 51 - 21\\n = 30$

From the calculations above, the numbers of new CDs is 30 and the old CDs is 21.

Learn more on system of equations here;

brainly.com/question/13729904

4 0

3 years ago

Write an equation for a parabola in which the set of all points in the plane are equidistant from the focus and line. F(–2, 0);

UNO [17]

The line is called the directrix. Here we have a vertical directrix, so a parabola sideways from usual.

Geometry is best done with squared distances. The squared distance from an arbitrary point (x,y) to the vertical line x=2 is $(x-2)^2.$

We equate that to the squared distance of (x,y) to the focus (-2,0):

$(x-2)^2 = (x - -2)^2 + (y - 0)^2$

$x^2 -4x + 4=x^2 +4x +4 + y^2$

$-8x = y^2$

We could call that done. A more standard form might be

$x =- \dfrac 1 8 \ y^2$

8 0

3 years ago