set B= {1+t^2, -2t- t^2, 1 + t + t^2} is a basis for P2. what is the coordinate vector of p(t) = 5 - 13t -2t^2 relative to B.

Answer 1

I got a different answer:

x+xt² -2yt-yt+ z+zt+zt²=5-13t-2t²

t⁰: x+z=5

t¹: -2y+z=-13

t²: x-y+z=-2

Forms the matrix

[1 0 1 5]

[0 -2 1 -13]

[1 -1 1 -2]

Finding the RREF:

[1 0 0 4]

[0 1 0 7]

[0 0 1 1]

<4, 7, 1>

Answer 2

x*(1+t²) + y(-2t-t²) +z(1+t+t²) = x +xt² -2yt² +z +zt + zt²

So, 5 = x + z

-13 = -2y +z

AND

-2 = x -2y + z

Solve for x, y, and z. That gives you the vector.

Subtract the second equation from the third.

11 = x

Substitute this value for x in the first equation:

5 = (11) + z, so z =-6.

Substitute this value for z in the second equation:

-13 = -2y + (-6)

Add 6 to both sides:

-7 = -2y

Divide by -2:

y = 3.5

As a vector coordinate triple: 〈 11, -3.5, -6 〉