Let $A$ be an $n\times n$ nonsingular matrix with integer entries.

Prove that the inverse matrix $A^{-1}$ contains only integer entries if and only if $\det(A)=\pm 1$.

Hint.

• If $B$ is a square matrix whose entries are integers, then the determinant of $B$ is an integer.
• The inverse matrix of $A$ can be computed by the formula
  \[A^{-1}=\frac{1}{\det(A)}\Adj(A).\]

Proof.

Let $I$ be the $n\times n$ identity matrix.

$(\implies)$: If $A^{-1}$ is an integer matrix, then $\det(A)=\pm 1$

Suppose that every entry of the inverse matrix $A^{-1}$ is an integer.
It follows that $\det(A)$ and $\det(A^{-1})$ are both integers.
Since we have
\begin{align*}
\det(A)\det(A^{-1})=\det(AA^{-1})=\det(I)=1,
\end{align*}
we must have $\det(A)=\pm 1$.

$(\impliedby)$: If $\det(A)=\pm 1$, then $A^{-1}$ is an integer matrix

Suppose that $\det(A)=\pm 1$. The inverse matrix of $A$ is given by the formula
\[A^{-1}=\frac{1}{\det(A)}\Adj(A),\] where $\Adj(A)$ is the adjoint matrix of $A$.
Thus, we have
\[A^{-1}=\pm \Adj(A).\] Note that each entry of $\Adj(A)$ is a cofactor of $A$, which is an integer.

(Recall that a cofactor is of the form $\pm \det(M_{ij})$, where $M_{ij}$ is the $(i, j)$-minor matrix of $A$, hence entries of $M_{ij}$ are integers.)

Therefore, the inverse matrix $A^{-1}$ contains only integer entries.

The post Inverse Matrix Contains Only Integers if and only if the Determinant is $\pm 1$ first appeared on Problems in Mathematics.

Let $R$ be the ring of all $2\times 2$ matrices with integer coefficients:
\[R=\left\{\, \begin{bmatrix}
a & b\\
c& d
\end{bmatrix} \quad \middle| \quad a, b, c, d\in \Z \,\right\}.\]

Let $S$ be the subset of $R$ given by
\[S=\left\{\, \begin{bmatrix}
s & 0\\
0& s
\end{bmatrix} \quad \middle | \quad s\in \Z \,\right\}.\]

(a) True or False: $S$ is a subring of $R$.

(b) True or False: $S$ is an ideal of $R$.

Solution.

(a) True or False: $S$ is a subring of $R$.

True.

In fact, let
\[A= \begin{bmatrix}
t & 0\\
0& t
\end{bmatrix}\text{ and } B=\begin{bmatrix}
s & 0\\
0& s
\end{bmatrix}\] be arbitrary elements in $S$ with $t, s\in \Z$.

Then we have
\begin{align*}
A+B=\begin{bmatrix}
t+s & 0\\
0& t+s
\end{bmatrix} \in S
\end{align*}
and
\begin{align*}
AB=\begin{bmatrix}
ts & 0\\
0& ts
\end{bmatrix} \in S.
\end{align*}
Hence $S$ is closed under addition and multiplication.

Note that the $2\times 2$ identity matrix is the unity element of $R$ as well as the unity element of $S$.

Thus, the subset $S$ is a subring of $R$.

(b) True or False: $S$ is an ideal of $R$.

False.

To see that $S$ is not an ideal of $R$, consider the element
\[\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix} \in R\] and the element
\[\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix} \in S.\] Then we have
\begin{align*}
\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}=\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix},
\end{align*}
which is not in $S$.

This implies that $S$ is not an ideal of $R$.
(If $S$ were an ideal of $R$, then an element of $S$ multiplied by an element of $R$ would stay in $S$.)

The post Is the Given Subset of The Ring of Integer Matrices an Ideal? first appeared on Problems in Mathematics.