- #26

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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.