# Simple matrix/linear algebra question, help

1. Apr 28, 2015

### ilyas.h

1. The problem statement, all variables and given/known data
Here is the question, i know how to do part (i) but I do not understand part (ii):

3. The attempt at a solution

here's the solution from the marking scheme:

i understand how they formed the matrix from their working out (i can se the pattern), but I do not understand the actual working out. Thanks to anyone who could explain it to me.

why does L(1) = t? why does L(t) = 1 + (-2)t? etc

2. Apr 28, 2015

### Staff: Mentor

If f is a function in P3, then L(f) = f' + t * f(-2). In other words, to get the image of a function f, add the derivative of the function and t times f(-2). So L(1) = 0 + t = t. Can you figure out why L(t) = 1 - 2t?

3. Apr 28, 2015

### ilyas.h

Still doesn't make sense.

L(1) = 1' + 1*(-2)t = d(1)/dt + (-2)t =

0 + (-2)t = (-2)t

=/= t.

what is f(-2) equal to anyway?

Last edited: Apr 28, 2015
4. Apr 28, 2015

### SammyS

Staff Emeritus
Indeed !

What is ƒ(-2) equal to?

What is ƒ ?

You were considering finding L(1) .

That makes ƒ(x) = 1, a constant function, so it maps everything to 1. It even maps -2 to 1.

5. Apr 28, 2015

### ilyas.h

I think i understand.

L(1) means that we are considering the function f(x) = 1 (a straight line through y=1):

L(f) = f' + f(-2)t

L(1) = d(1)/dt + f(-2)t

f(-2) is equal to 1 in this case, so:

d(1)/dt + f(-2)t = 0 +t = t

However, if we consider L(t):

L(t) = t' + f(-2)t
=d(t)/dt + f(-2)t
= 1 + f(-2)t

in this case, we are considering f(x) = t, another constant function (y=t throughout), so f(-2) = t:

L(t) = 1 + f(-2)t = 1 + t*t

= 1 + t^2

this does not equal 1+(-2)t as indicated in the markscheme.

EDIT: nvm, i understand now.

if f(t) = t then:

f(-2) = -2.

Plug it all in and it all works.

thanks,

Last edited: Apr 28, 2015
6. Apr 28, 2015

### ilyas.h

can you look at my edit? thanks.

7. Apr 28, 2015

### SammyS

Staff Emeritus
Yup.

(I quickly deleted a post I had made in reply. Now that you edited that post, all is well.)