Polynomial Curve Fitting, Matrices

**kaylakutie** · February 4th 2010, 08:20 PM

I've done over 50 problems for a Linear Algebra class tonight and I'm sooo burnt out. I'm giving up on these ones.. if you can help me, that would be wonderful. Otherwise, I'm turning in what I have. Strangely enough, it's the odd problems that I already have solutions to that I don't understand. Got the even ones already.

11) In the "Polynomial Curve Fitting" section:
The graph of a cubic polynomial function has horizontal tangents at (1, -2) and (-1,2). Find an equation for the cubic and sketch its graph.
Somehow the answer is p(x) = -3x + x^3. Just want to know the steps.

29) Use a system of equations to write the partial fraction decomposition of the rational expression. Then solve the system using matrices.

$ \frac{4x^2}{(x+1)^2(x-1)} = \frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^2} $

And the final answer should be:

$ \frac{1}{1-x}+\frac{3}{1+x}-\frac{2}{(x+1)^2} $

47) Consider the matrix..
$ A=\begin{bmatrix} 1 &k &2 \\ -3 &4 &1 \\ \end{bmatrix} $
If A is the augmented matrix of a system of linear equations, find the value(s) of k such that the system is consistent.
(Answer is all real k not equal to -4/3. Just want to know how they got this so I understand it.

58) True or false: Every matrix has a unique reduced row-echelon form.

Thank you in advance. I appreciate it.

**math2009** · February 5th 2010, 04:54 AM

ref attachment

**HallsofIvy** · February 6th 2010, 05:13 AM

Originally Posted by kaylakutie

I've done over 50 problems for a Linear Algebra class tonight and I'm sooo burnt out. I'm giving up on these ones.. if you can help me, that would be wonderful. Otherwise, I'm turning in what I have. Strangely enough, it's the odd problems that I already have solutions to that I don't understand. Got the even ones already.

11) In the "Polynomial Curve Fitting" section:
The graph of a cubic polynomial function has horizontal tangents at (1, -2) and (-1,2). Find an equation for the cubic and sketch its graph.
Somehow the answer is p(x) = -3x + x^3. Just want to know the steps.

Any cubic polynomial can be written in the form $f(x)= ax^3+ bx^2+ cx+ d$ and then $f'(x)= 3ax^2+ 2bx+ c$ .

Saying that it has a horizontal tangent at (1, -2) tells you two things: its value at x= 1 is $f(1)= a(1)^3+ b(1)^2+ c(1)+ d= a+ b+ c+ d= -2$ and its derivative there is $f'(1)= 3a(1)^2+ 2b(1)+ c= 0$ . Do the same at x= -1 to get four equations for a, b, c, and d.

[quote]29) Use a system of equations to write the partial fraction decomposition of the rational expression. Then solve the system using matrices.

$ \frac{4x^2}{(x+1)^2(x-1)} = \frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^2} $
Multiply both sides of the equation by $(x+1)^2(x-1)$ to get
$4x^2= A(x+1)^2+ B(x-1)(x+1)+ C(x-1)= Ax^2+ 2Ax+ A+ Bx^2- B+ Cx- C$
$4x^2= (A+ B)x^2+ (2A+ C)x+ (A- B+ C)$
Equating coefficients, A+ B= 4, 2A+ C= 0, and A- B+ C= 0.
Those correspond to the matrix equation
$\begin{bmatrix}1 & 1 & 0 \\ 2& 0 & 1 \\ 1 & -1 & 1\end{bmatrix}\begin{bmatrix}A \\ B \\ C\end{bmatrix}= \begin{bmatrix}4 \\ 0 \\ 0\end{bmatrix}$

And the final answer should be:

$ \frac{1}{1-x}+\frac{3}{1+x}-\frac{2}{(x+1)^2} $

47) Consider the matrix..
$ A=\begin{bmatrix} 1 &k &2 \\ -3 &4 &1 \\ \end{bmatrix} $
If A is the augmented matrix of a system of linear equations, find the value(s) of k such that the system is consistent.
(Answer is all real k not equal to -4/3. Just want to know how they got this so I understand it.

Row reduce the matrix just as you would to solve it. Since there are only two rows, that is simple: Add 3 times the first row to the second to get
$\begin{bmatrix} 1 & k & 2 \\0 & 4+ 3k & 7\end{bmatrix}$
That last row corresponds to (4+3k)y= 7. To solve that you must divide by 4+ 3k which you cannot do if 4+ 3k= 0.

58) True or false: Every matrix has a unique reduced row-echelon form.

True, of course. You can find the reduced row-echelon form by following a specific formula which, if done correctly, will always give the same result for the same matrix.

Thank you in advance. I appreciate it.

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