Linear Algebra Problem

  • Thread starter Mathman23
  • Start date
  • #26
254
0
Hi

I try again.


Is its correct now?

[tex]\left[ \begin{array}{c} a \\ b \end{array} \right] = \left[ \begin{array}{*{40} cc} \frac{ x \cdot x}{x \cdot x -N \overline{x}^2} & \frac{-\overline{x }N}{x \cdot x - N \overline{x} ^2} \\ \frac{-\overline{x} N}{x \cdot x - N \overline{x}^2} & \frac{N}{x \cdot x - N \overline{x}^2} \end{array} \right] \left[ \begin{array}{c} \overline{y} \\ \frac{x \cdot y}{N} \end{array} \right] = \left[ \begin{array}{c} \frac{( x \cdot x) \overline{y}}{x \cdot x -N \overline{x}^2} + \frac{-\overline{x }N (x \cdot y)}{(x \cdot x)N - N^2 \overline{x} ^2} \\ \frac{-\overline{x }N(\overline{y})}{x \cdot x - N \overline{x} ^2} + \frac{N (x \cdot y)}{N(x \cdot x) - N^2 \overline{x}^2} \end{array} \right] [/tex]

Sincerely and Best Regards

Fred

OlderDan said:
You can reduce the upper left term in your inverse by dividing numeratior and denominator by N. Other than that it looks fine.
[tex]A^{-1} = \frac{N}{x \cdot x - N\overline{x}^2}} \left[ {\begin{array}{*{20}c} {\frac{{ x \cdot x }}{N}} & -{\overline x } \\ -{\overline x } & 1\\\end{array}} \right] = \left[ \begin{array}{*{40} cc} \frac{ x \cdot x}{x \cdot x -N \overline{x}^2} & \frac{-\overline{x }N}{x \cdot x - N \overline{x} ^2} \\ \frac{-\overline{x} N}{x \cdot x - N \overline{x}^2} & \frac{N}{x \cdot x - N \overline{x}^2}\end{array} \right][/tex]

Your matrix multiplication is not fine however. You do not have the terms matched up properly. Review your matrix multiplication rules and give this another try. The expressions for a and b will not be very "simple" in any case, but you could help yourself by leaving the 1/det(A) term out in front of the matrix product, since it is a common factor in every term. It will also be a common factor in a and b.
 
  • #27
OlderDan
Science Advisor
Homework Helper
3,021
2
Mathman23 said:
Hi

I try again.


Is its correct now?

[tex]\left[ \begin{array}{c} a \\ b \end{array} \right] = \left[ \begin{array}{*{40} cc} \frac{ x \cdot x}{x \cdot x -N \overline{x}^2} & \frac{-\overline{x }N}{x \cdot x - N \overline{x} ^2} \\ \frac{-\overline{x} N}{x \cdot x - N \overline{x}^2} & \frac{N}{x \cdot x - N \overline{x}^2} \end{array} \right] \left[ \begin{array}{c} \overline{y} \\ \frac{x \cdot y}{N} \end{array} \right] = \left[ \begin{array}{c} \frac{( x \cdot x) \overline{y}}{x \cdot x -N \overline{x}^2} + \frac{-\overline{x }N (x \cdot y)}{(x \cdot x)N - N^2 \overline{x} ^2} \\ \frac{-\overline{x }N(\overline{y})}{x \cdot x - N \overline{x} ^2} + \frac{N (x \cdot y)}{N(x \cdot x) - N^2 \overline{x}^2} \end{array} \right] [/tex]

Sincerely and Best Regards

Fred

It looks correct, but it should be simplified. You again have factors of N that you can cancel in the second term of a and the second term of b. You should be combining the two terms in each case to write them over common denominators. The common denominator will be there if you get rid of the unnecessary N factors. Not only that, it will be the same denominator in b that you have in a. That would be obvious if you left the 1/Det(A) term out in front, and it would save you a lot of writing.
 

Related Threads on Linear Algebra Problem

  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
7
Views
1K
Replies
6
Views
3K
  • Last Post
Replies
3
Views
1K
Replies
2
Views
925
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
4
Views
946
  • Last Post
Replies
4
Views
8K
Top