# Matrix inverse in mod6

Does the matrix $$\left[\begin{array}{cc} 5 & 2\\ 0 & 3\end{array}\right]$$ have a multiplicative inverse belonging to integer mod 6?

• For this to be so, the determinant would have to be coprime to 6. – hardmath Mar 22 '15 at 18:05
• Why does it have to be coprime? – user204450 Mar 22 '15 at 18:06
• well I know the inverse is $\left[\begin{array}{cc}\frac{1}{5} & \frac{-2}{5}\\ 0 & \frac{1}{3}\end{array}\right]$ and because it doesnt belong to integer mod 6. What is thestuff about the coprime? – user204450 Mar 22 '15 at 18:10
• I don't think you understand the Question yourself. If you were asking about the matrix inverse in the rational number field, then that would not work either. Try multiplying the two matrices. More generally the determinant of a product is the product of the two determinants, and the determinant of the identity matrix is one. – hardmath Mar 22 '15 at 18:15
• I am deleting this question to alleviate a potential problem of academic integrity. Because there is an upvoted answer, the deletion will be only TEMPORARY. Undeleting on April 12th. – Jyrki Lahtonen Apr 5 '15 at 6:35

## 2 Answers

Integer mod 6, or $\mathbb{Z}_6$, is not a field...and so some elements in it has no multiplicative inverse. The integer $3$ is one example: we need $x \in \{1,2,3,4,5\}$ such that \begin{equation} 3\cdot x = 1\mod 6,\end{equation} but as you can see such an $x$ does not exist.

Now check the last row of the given matrix - for this matrix to have an inverse we must find a $2 \times 2$ matrix with entries from $\mathbb{Z}_6$ such that it's second column has the property: \begin{equation} \begin{bmatrix}0 & 3\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}=1 (\mod 6). \end{equation} This means we must have $3 \cdot y =1 (\mod 6)$, which is not possible, as we have seen above.

• By the way, how do you know that integer mod 6 is not a field? – user204450 Mar 22 '15 at 19:54
• See for example the wikipedia entry on field. By definition, or by the axioms defining a field, every nonzero element in a field has a multiplicative inverse. For 'integer mod x' rings this is only true if x is a prime number. – Christiaan Hattingh Mar 22 '15 at 20:07
• I am deleting this question to alleviate a potential problem of academic integrity. Because there is an upvoted answer, the deletion will be only TEMPORARY. Undeleting on April 12th. @-ping me or flag a moderator, if I forget. +1 to the answer. – Jyrki Lahtonen Apr 5 '15 at 6:36

Instead of thinking about the matrix $\begin{bmatrix}1/5 & -2/5\\ 0& 1/3\end{bmatrix}$, you should think about the matrix $$\begin{bmatrix}1\cdot5^{-1} & -2\cdot 5^{-1}\\ 0& 1\cdot3^{-1}\end{bmatrix},$$ where the inverses are computed in the ring $\mathbb{Z}/\mathbb{Z}_6$ (or $\mathbb{Z}_6$, if you prefer that notation).

• Per my comment, this isn't the right rational matrix inverse. The nonzero off-diagonal element should have been $-2/15$. – hardmath Mar 23 '15 at 0:51