# Thread: Help on a Linear Algebra Problem

1. ## Help on a Linear Algebra Problem

Hi All,

I got this problem and it seems like it should be straight forward, but I get the wrong answer. I listed below my reasoning. What is wrong with what I am doing? Any help would be greatly appreciated.

Let B={$\displaystyle 2x^3-5x^2+3x+7,x^2-4x+9,1/2x+5,3$}

Find <$\displaystyle 3x^3+6x^2-8x+27$> by method of comparing coefficients.
Here is what I did.
<$\displaystyle 3x^3+6x^2-8x+2$> =C1<$\displaystyle 2x^3-5x^2+3x+7$> +C2<$\displaystyle x^2-4x+9$> +C3<$\displaystyle 1/2x+5$>+C4*3

2C1=3, C1=3/2
C2-5=6 C2= 11
1/2C3-4+3= -8
1/2C3=-7
C3=-14
3C4+9+7+5=27
3C4=6
C4=2
<3/2,11, -14,2>

2. Originally Posted by egshih
Hi All,

I got this problem and it seems like it should be straight forward, but I get the wrong answer. I listed below my reasoning. What is wrong with what I am doing? Any help would be greatly appreciated.

Let B={$\displaystyle 2x^3-5x^2+3x+7,x^2-4x+9,1/2x+5,3$}

Find <$\displaystyle 3x^3+6x^2-8x+27$> by method of comparing coefficients.
Here is what I did.
<$\displaystyle 3x^3+6x^2-8x+2$> =C1<$\displaystyle 2x^3-5x^2+3x+7$> +C2<$\displaystyle x^2-4x+9$> +C3<$\displaystyle 1/2x+5$>+C4*3

2C1=3, C1=3/2
C2-5=6 C2= 11
1/2C3-4+3= -8
1/2C3=-7
C3=-14
3C4+9+7+5=27
3C4=6
C4=2
<3/2,11, -14,2>
You should be using the value of $\displaystyle C1$ in your computation of $\displaystyle C2$ as $\displaystyle b_1$ has a second degree term, ...

You could try it this way:

$\displaystyle p(x)=3x^3+6x^2-8x+27$

Then as the leading coefficient of $\displaystyle b_1$ is $\displaystyle 2$ we have:

$\displaystyle p(x)=\frac{3}{2}b_1(x)+\frac{27}{2}x^2-8x+12$

similarly as the leading coefficient of $\displaystyle b_2$ is $\displaystyle 1$ we have:

$\displaystyle p(x)=\frac{3}{2}b_1(x)+\frac{27}{2}b_2(x)+46x-\frac{219}{2}$

etc.

CB

3. Originally Posted by egshih
Hi All,

I got this problem and it seems like it should be straight forward, but I get the wrong answer. I listed below my reasoning. What is wrong with what I am doing? Any help would be greatly appreciated.

Let B={$\displaystyle 2x^3-5x^2+3x+7,x^2-4x+9,1/2x+5,3$}

Find <$\displaystyle 3x^3+6x^2-8x+27$> by method of comparing coefficients.
Here is what I did.
<$\displaystyle 3x^3+6x^2-8x+2$> =C1<$\displaystyle 2x^3-5x^2+3x+7$> +C2<$\displaystyle x^2-4x+9$> +C3<$\displaystyle 1/2x+5$>+C4*3
Better if you go ahead and do the combination on the right:
$\displaystyle 3x^3+6x^2-8x+2= C1x^3+(-5C1+C2)x^2+ (3C1- 4C2$$\displaystyle + (1/2)C3)x+ (7C1+9C2+ 5C3+ 3C4)$

2C1=3, C1=3/2
Yes, that is correct.

C2-5=6 C2= 11
-5C1+ C2= -5(3/2)+ C2= C2- 15/2= 11. C2= 11+ 15/2= 37/2.
You need -5C1, not just -5.

1/2C3-4+3= -8
1/2C3=-7
C3=-14
3C4+9+7+5=27
3C4=6
C4=2
<3/2,11, -14,2>