Get just a few values of invertible matrix

**bumcheekcity** · December 10th 2008, 01:33 AM

$\left(\begin{array}{ccccc}4&1&0&0&0\\\end{array}\r ight) \left(\begin{array}{ccccc}s&-1&0&0&0\\0&s&-1&0&0\\0&0&s&-1&0\\0&0&0&s&-1\\0&0&4&4&s-1\\\end{array}\right)^{-1} \left(\begin{array}{c}0\\0\\0\\0\\1\\\end{array}\r ight)$

If I want to multiply the above 3 matrices, then I only need to find the top-right, and the entry below it in the inverse matrix (as all others will be zero).

How do I go about finding only a few entries in a large inverted matrix like this?

**Opalg** · December 10th 2008, 12:31 PM

Call the matrix S, and let $v = \begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}$ . If $x = S^{-1}v$ then $Sx = v$ . if $x = \begin{bmatrix}a\\b\\c\\d\\e\end{bmatrix}$ then the equation $Sx = v$ becomes $\begin{bmatrix}sa-b\\ sb-c\\ sc-d\\ sd-e\\ 4c+4d+(s-1)e\end{bmatrix} = \begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}$ . That gives you five easily solved equations telling you that $a = \frac1{s^5-s^4+4s^3+4s^2}$ .

The expression $\begin{bmatrix}4&1&0&0&0\end{bmatrix}x$ then becomes $a(4+s) = \frac{s+4}{s^5-s^4+4s^3+4s^2}$ .

**CaptainBlack** · December 10th 2008, 01:59 PM

Originally Posted by bumcheekcity

$\left(\begin{array}{ccccc}4&1&0&0&0\\\end{array}\r ight) \left(\begin{array}{ccccc}s&-1&0&0&0\\0&s&-1&0&0\\0&0&s&-1&0\\0&0&0&s&-1\\0&0&4&4&s-1\\\end{array}\right)^{-1} \left(\begin{array}{c}0\\0\\0\\0\\1\\\end{array}\r ight)$

If I want to multiply the above 3 matrices, then I only need to find the top-right, and the entry below it in the inverse matrix (as all others will be zero).

How do I go about finding only a few entries in a large inverted matrix like this?

google for block matrix inverse

CB

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