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Prove that $gcd(det(A),26)=1$ if a $2x2$ matrix $A$ with entries in $Z_{26}$ is invertible.

We know that $A$ is invertible. Let the inverse of $A$ be $A^{-1}=(det(A))^{-1}A^{*}$, where $A^{*}$ is the adjoint matrix of $A$.

However, this is where I am stuck, I cannot see how the formula for $A^{-1}$ is useful.

Any help is greatly appreciated

Bernard
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Silvia Rossi
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1 Answers1

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You can't write $(\det A)^{-1}$ as long as you don't know $\det A$ is a unit in $\mathbf Z/26\mathbf Z$.

Suppose $A$ is invertible: $\;A A^{-1}=A^{-1} A= I$, where $I$ is the unit matrix of dimension $2$. Taking the determinant, we deduce $$\det(A A^{-1})=\det A\,\det A^{-1}=\det I=1$$ Thus $\det A$ is a unit in $\mathbf Z/26\mathbf Z$. Now the units of $\mathbf Z/n\mathbf Z$ are the congruence classes of numbers which a coprime with $n$. So we conclude that $\;\gcd(\det A,26)=1$.

Note:

It is a general fact that, in any commutative ring $R$, a matrix $A\in\mathcal M_n(R)$ is invertible if and only if $\det A\in R^{\times}$.

Bernard
  • 175,478