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How to find the inverse of the matrix over $\mathbb Z_5$

$$ \left( \begin{matrix} 1 & 2& 0\\ 0 &2& 4 \\ 0& 0& 3\\ \end {matrix} \right) $$

Martin Sleziak
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jolkol piu
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  • I know how to get inverse by adj / det operation.I am not familiar about solving this kind of questions.How should I approach by taking 9 variables matrix and solving the equations? I am confused. – jolkol piu Jan 23 '15 at 06:51
  • In that case, compute the inverse by the adj/det operation. Where do you get stuck trying to do so? – anomaly Jan 23 '15 at 06:52
  • When you compute the inverse normally, just figure out what each element is mod 5, yes? – A. Thomas Yerger Jan 23 '15 at 07:00
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    Note that the determinant is $1 \cdot 2 \cdot 3 = 6 = 1!!!! \mod!! 5$ so the inverse in this case is the adjoint matrix, since $1/\text{det} = 1/1 = 1 $ ( all $!!!!\mod 5$ ). – orangeskid Jan 23 '15 at 07:28
  • You can find several similar questions on the site (for example, have a look at the list of related question on the right). Just randomly picking two similar questions: http://math.stackexchange.com/questions/280522/how-to-find-inverse-of-a-matrix-in-bbb-z-5 and http://math.stackexchange.com/questions/85753/i-dont-understand-why-the-inverse-is-this – Martin Sleziak Jan 23 '15 at 08:11

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Hint: One can proceed as usual by taking the given matrix $A$ and row-reducing the adjoined matrix $[\,A\,|\,\mathbb{I}\,]$. Note that the matrix is upper triangular, which makes available a very efficient method, because it implies the following about the inverse: (1) the inverse is upper triangular, and (2) the diagonal entries of the inverse are reciprocals of the diagonal entries of $A$. In other words, so the inverse has the form $$\begin{pmatrix}1 & a & b \\ 0 & 3 & c \\ 0 & 0 & 2\end{pmatrix}.$$ Then, one can simply expand the appropriate components of the equation $A A^{-1} = 0$ to solve for the three remaining unknown components $a, b, c$.

Travis Willse
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