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 $A$ be an $n\times n$ matrix.

The $(i, j)$ cofactor $C_{ij}$ of $A$ is defined to be
\[C_{ij}=(-1)^{ij}\det(M_{ij}),\] where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column.

Then consider the $n\times n$ matrix $C=(C_{ij})$, and define the $n\times n$ matrix $\Adj(A)=C^{\trans}$.
The matrix $\Adj(A)$ is called the adjoint matrix of $A$.

When $A$ is invertible, then its inverse can be obtained by the formula

For each of the following matrices, determine whether it is invertible, and if so, then find the invertible matrix using the above formula.

(a) $A=\begin{bmatrix}
1 & 5 & 2 \\
0 &-1 &2 \\
0 & 0 & 1
\end{bmatrix}$.

 
(b) $B=\begin{bmatrix}
1 & 0 & 2 \\
0 &1 &4 \\
3 & 0 & 1
\end{bmatrix}$.

Solution.

(a) The Inverse Matrix of $A$.

Since $A$ is an upper triangular matrix, the determinant of $A$ is the product of diagonal entries.
Thus we have $\det(A)=-1\neq 0$, and hence $A$ is invertible.

To find the inverse using the formula, we first determine the cofactors $C_{ij}$ of $A$.
We have
\begin{align*}
C_{11}&=\begin{vmatrix}
-1 & 2\\
0& 1
\end{vmatrix}=-1,\quad C_{12}=-\begin{vmatrix}
0 & 2\\
0& 1
\end{vmatrix}=0, \quad C_{13}=\begin{vmatrix}
0 & -1\\
0& 0
\end{vmatrix}=0\\[6pt] C_{21}&=-\begin{vmatrix}
5 & 2\\
0& 1
\end{vmatrix}=-5, \quad C_{22}=\begin{vmatrix}
1 & 2\\
0& 1
\end{vmatrix}=1, \quad C_{23}=-\begin{vmatrix}
1 & 5\\
0& 0
\end{vmatrix}=0 \\[6pt] C_{31}&=\begin{vmatrix}
5 & 2\\
-1& 2
\end{vmatrix}=12, \quad C_{32}=-\begin{vmatrix}
1 & 2\\
0& 2
\end{vmatrix}=-2, \quad C_{33}=\begin{vmatrix}
1 & 5\\
0& -1
\end{vmatrix}=-1.
\end{align*}
The the adjoint matrix of $A$ is
\begin{align*}
\Adj(A)=C^{\trans}=\begin{bmatrix}
-1 & -5 & 12 \\
0 &1 &-2 \\
0 & 0 & -1
\end{bmatrix}.
\end{align*}

Using the formula, we obtain the inverse matrix
\[A^{-1}=\frac{1}{\det(A)}\Adj(A)=\begin{bmatrix}
1 & 5 & -12 \\
0 &-1 &2 \\
0 & 0 & 1
\end{bmatrix}.\]

(b) The Inverse Matrix of $B$.

To check the invertibility of the matrix $B$, we compute the determinant of $B$.
The second column cofactor expansion yields that
\begin{align*}
\det(B)=\begin{vmatrix}
1 & 2\\
3& 1
\end{vmatrix}=-5 \neq 0.
\end{align*}
So the matrix $B$ is invertible.

Now the cofactors $C_{ij}$ of $B$ are
\begin{align*}
C_{11}&=\begin{vmatrix}
1 & 4\\
0& 1
\end{vmatrix}=1, \quad C_{12}=-\begin{vmatrix}
0 & 4\\
3& 1
\end{vmatrix}=12. \quad C_{13}=\begin{vmatrix}
0 & 1\\
3& 0
\end{vmatrix}=-3 \\[6pt] C_{21}&=-\begin{vmatrix}
0 & 2\\
0& 1
\end{vmatrix}=0, \quad C_{22}=\begin{vmatrix}
1 & 2\\
3& 1
\end{vmatrix}=-5, \quad C_{23}=-\begin{vmatrix}
1 & 0\\
3& 0
\end{vmatrix}=0 \\[6pt] C_{31}&=\begin{vmatrix}
0 & 2\\
1& 4
\end{vmatrix}=-2, \quad C_{32}=-\begin{vmatrix}
1 & 2\\
0& 4
\end{vmatrix}=-4, \quad C_{33}=\begin{vmatrix}
1 & 0\\
0& 1
\end{vmatrix}=1.
\end{align*}
Hence the adjoint matrix of $B$ is
\[\Adj(B)=C^{\trans}=\begin{bmatrix}
1 & 0 & -2 \\
12 &-5 &-4 \\
-3 & 0 & 1
\end{bmatrix}.\] It follows from the formula that the inverse matrix of $B$ is
\[B^{-1}=\frac{1}{\det(B)}\Adj(B)=\frac{1}{5}\begin{bmatrix}
-1 & 0 & 2 \\
-12 &5 &4 \\
3 & 0 & -1
\end{bmatrix}.\] Click here if solved 127

The post Find Inverse Matrices Using Adjoint Matrices first appeared on Problems in Mathematics.

Calculate the determinants of the following $n\times n$ matrices.
\[A=\begin{bmatrix}
1 & 0 & 0 & \dots & 0 & 0 &1 \\
1 & 1 & 0 & \dots & 0 & 0 & 0 \\
0 & 1 & 1 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \dots & \dots & \ddots & \vdots \\
0 & 0 & 0 &\dots & 1 & 1 & 0\\
0 & 0 & 0 &\dots & 0 & 1 & 1
\end{bmatrix}\]

The entries of $A$ is $1$ at the diagonal entries, entries below the diagonal, and $(1, n)$-entry.
The other entries are zero.
\[B=\begin{bmatrix}
1 & 0 & 0 & \dots & 0 & 0 & -1 \\
-1 & 1 & 0 & \dots & 0 & 0 & 0 \\
0 & -1 & 1 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \dots & \dots & \ddots & \vdots \\
0 & 0 & 0 &\dots & -1 & 1 & 0\\
0 & 0 & 0 &\dots & 0 & -1 & 1
\end{bmatrix}.\]

The entries of $B$ is $1$ at the diagonal entries.
The entries below the diagonal and $(1,n)$-entry are $-1$.
The other entries are zero.

Hint.

1. Calculate the first row cofactor expansion.
2. The determinant of a triangular matrix is the product of its diagonal entries.

Solution.

Apply the cofactor expansion corresponding to the first row. We obtain
\begin{align*}
\det(A)&=
\begin{vmatrix}
1 & 0 & \dots & 0 & 0 & 0 \\
1 & 1 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \dots & \ddots & \vdots & \vdots \\
0 & 0 &\dots & 1 & 1 & 0\\
0 & 0 &\dots & 0 & 1 & 1
\end{vmatrix}
+(-1)^{n+1}
\begin{vmatrix}
1 & 1 & 0 & \dots & 0 & 0 \\
0 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 &\dots & 1 & 1 \\
0 & 0 & 0 &\dots & 0 & 1
\end{vmatrix}
\end{align*}
The two smaller (minor) $n-1 \times n-1$ matrices are both triangular.
The determinant of a triangular matrix is the product of its diagonal entries.
Thus we see that
\begin{align*}
\det(A)&=1+(-1)^{n+1} \\
&= \begin{cases}
2 & \text{ if } n \text{ is odd}\\
0 & \text{ if } n \text{ is even}.
\end{cases}
\end{align*}
Next we calculate $\det(B)$. By the first row cofactor expansion , we obtain
\begin{align*}
\det(B)&=\\
&\begin{vmatrix}
1 & 0 & \dots & 0 & 0 & 0 \\
-1 & 1 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \dots & \ddots & \vdots & \vdots \\
0 & 0 &\dots & -1 & 1 & 0\\
0 & 0 &\dots & 0 & -1 & 1
\end{vmatrix}
+(-1)^{n+1}(-1)
\begin{vmatrix}
-1 & 1 & 0 & \dots & 0 & 0 \\
0 & -1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 &\dots & -1 & 1 \\
0 & 0 & 0 &\dots & 0 & -1
\end{vmatrix}.
\end{align*}
The two minor matrices are both triangular.
All the diagonal entries of the first minor matrix are $1$ and those of the second minor matrix are $-1$.
Thus we have
\begin{align*}
\det(B)&=1+(-1)^{n}(-1)^{n-1}=0.
\end{align*}

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