# Wolfram|Alpha refuses to find the inverse of a large 6x6 matrix.

Just to be clear, this isn't a question on how to find the inverse of a matrix, I just don't want to find the inverse by hand (I hope you see why).

$$\begin{pmatrix} 1 & 2006 & 2006^2 & 2006^3 & 2006^4 & 2006^5 \\ 1 & 2013 & 2013^2 & 2013^3 & 2013^4 & 2013^5 \\ 1 & 2028 & 2028^2 & 2028^3 & 2028^4 & 2028^5 \\ 1 & 2062 & 2062^2 & 2062^3 & 2062^4 & 2062^5 \\ 1 & 2076 & 2076^2 & 2076^3 & 2076^4 & 2076^5 \\ 1 & 2086 & 2086^2 & 2086^3 & 2086^4 & 2086^5 \\ \end{pmatrix}^{-1}$$

I started with my usual method of solving the inverse of matrices: TI-84

In the past, I have had no problem with finding the inverse of 2x2 and 3x3 matrices. I thought it would be easy, and would only require more time to fill in all the elements. The output?

ERR:SINGULAR MAT

That couldn't be right. This matrix most definitely has an inverse. I tried changing all these x-values by a couple numbers. The output?

ERR:SINGULAR MAT

Confused, I went to my secondary source for solving matrices: Wolfram|Alpha

Input:

inv {{1, 2006, 2006^2, 2006^3, 2006^4, 2006^5}, {1, 2013, 2013^2, 2013^3, 2013^4, 2013^5}, {1, 2028, 2028^2, 2028^3, 2028^4, 2028^5}, {1, 2062, 2062^2, 2062^3, 2062^4, 2062^5}, {1, 2076, 2076^2, 2076^3, 2076^4, 2076^5}, {1, 2086, 2086^2, 2086^3, 2086^4, 2086^5}}

WA normally makes the best of any input, but not this time. The output?

Showing instead result for query: inv 2006^3

Just to make sure I wasn't doing anything wrong, I cut out my 6th row and column to see if WA could solve for the inverse of a 5x5 matrix.

Input:

inv {{1, 2006, 2006^2, 2006^3, 2006^4}, {1, 2013, 2013^2, 2013^3, 2013^4}, {1, 2028, 2028^2, 2028^3, 2028^4}, {1, 2062, 2062^2, 2062^3, 2062^4}, {1, 2076, 2076^2, 2076^3, 2076^4}}

Output:

{{49646092509/1715, -1934961730624/36015, 641928437/20, -31251517011/2401, 117265112107/20580}, {-1068474357/18865, 1263054496/12005, -2826145691/44880, 1046649267/40817, -925614889/82320}, {12542367/301840, -25041572/324135, 6237353/134640, -12371091/653072, 6164381/740880}, {-8179/603680, 908/36015, -2719/179520, 8123/1306144, -901/329280}, {1/603680, -1/324135, 1/538560, -1/1306144, 1/2963520}}

I have no idea why the technology I so desperately rely on is failing me now. I have two questions for you:

1. How can I find the inverse of my matrix?
2. What is the inverse of my matrix?
• " I have no idea why the technology I so desperately rely on is failing me now." relying on wolframalpha is for noobs. Feb 12 '16 at 5:37
• @TheKindCat Do you have a solution for my dilemma? Feb 12 '16 at 5:38
• Issac Beeman, you can compute it with Matlab/Mathematica/Maple. All will give the correct answer. Relying on W|A could be an overestimation of its capabilities. Would you like me to post an answer with the inverse? Feb 12 '16 at 5:39
• @Galc127 Yes, that would be helpful Feb 12 '16 at 5:42
• That would not fit in a comment. Feb 12 '16 at 5:42

$\left[ \begin {array}{cccccc} {\frac {7397267783841}{9800}}&-{\frac { 576618595725952}{375585}}&{\frac {669531359791}{580}}&-{\frac { 1552158678213}{1372}}&{\frac {17472501703943}{14700}}&-{\frac { 182581779550599}{423400}}\\ -{\frac {2774944526301}{ 1509200}}&{\frac {9839156766592}{2629095}}&-{\frac {3667913662027}{ 1301520}}&{\frac {904862053383}{326536}}&-{\frac {1199946553441}{ 411600}}&{\frac {896488859247}{846800}}\\ {\frac { 21629483637}{12073600}}&-{\frac {86339190584}{23661855}}&{\frac { 21489555431}{7809120}}&-{\frac {7092080683}{2612288}}&{\frac { 21189432767}{7408800}}&-{\frac {7042596319}{6774400}} \\ -{\frac {2634133}{3018400}}&{\frac {42088364}{ 23661855}}&-{\frac {723533}{538560}}&{\frac {5210845}{3918432}}&-{ \frac {20786449}{14817600}}&{\frac {44711}{87600}} \\ {\frac {2053}{9658880}}&-{\frac {10258}{23661855} }&{\frac {10243}{31236480}}&-{\frac {3403}{10449152}}&{\frac {2039}{ 5927040}}&-{\frac {679}{5419520}}\\ -{\frac {1}{ 48294400}}&{\frac {1}{23661855}}&-{\frac {1}{31236480}}&{\frac {1}{ 31347456}}&-{\frac {1}{29635200}}&{\frac {1}{81292800}}\end {array}\right]$ (via Maple)
This is probably an ill-conditioned matrix; the ratio of the largest eigenvalue to the smallest eigenvalue (in magnitude) is about $4\cdot 10^{16}$. Such matrices are extremely susceptible to round-off error, and the inverse might be found to "not exist". (Hilbert matrices are another example of ill-conditioned matrices.)
• The reason I needed the inverse matrix was so I could find $a=A^{-1}b$, if $b= \begin{pmatrix} 531.37\\ 484.13\\ 543.71\\ 53.06\\ 91.88\\ 11.94\\ \end{pmatrix}$ what is $a$? Feb 12 '16 at 5:57
• Could someone use the $A^{-1}$ and $b$ matrices to find $a$? Feb 12 '16 at 6:04
• @IsaacBeeman, using Matlab/Maple I got $$\begin{pmatrix}3.29499956258977 \cdot 10^{11}\\-8.05664400043738\cdot 10^8\\7.87927924129122\cdot 10^5\\-385.267456025648\\0.0941849856314081\\-0.00000920946793108401\end{pmatrix}$$ Feb 12 '16 at 14:27