How to Find an Inverse Matrix with Integer Entries

An inverse matrix will have only integer entries if the original matrix has determinant 1. So, the task reduces to finding matrices with determinant equal to 1.

2x2 Matrices with Determinant 1

Consider a general 2x2 matrix with unit determinant:

$$ \det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = 1 $$

Since the determinant is

$$ ad - bc = 1 $$

we need integer values of \(a, b, c, d\) that satisfy this equation.

For instance, if \( ad = 9 \) and \( bc = 8 \), the condition holds. Choosing \( a = 3 \) and \( d = 3 \) gives:

$$ M = \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix} $$

Here \( b = 2 \) and \( c = 4 \) yield \( bc = 8 \). The resulting matrix has determinant 1:

$$ \det M = \det \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix} = 1 $$

Because its determinant is 1, the matrix is invertible, and its inverse contains only integers:

$$ M^{-1 } = \begin{pmatrix} 3 & -2 \\ -4 & 3 \end{pmatrix} $$

Check. Multiply M by its inverse: $$ M \cdot M^{-1} = \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix} \cdot \begin{pmatrix} 3 & -2 \\ -4 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$ As expected, the product is the identity matrix.

3x3 Matrices with Determinant 1

Now take arbitrary integers \( a, b, c, d, e, f \) and form the product:

$$ \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ d & 1 & 0 \\ e & f & 1 \end{pmatrix} $$

The result is a 3x3 matrix with determinant 1:

$$ \det \left( \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ d & 1 & 0 \\ e & f & 1 \end{pmatrix} \right) = 1 $$

This guarantees an invertible matrix with an integer inverse.

For example, with \( a = 1, b = 2, c = 3, d = 4, e = 5, f = 6 \):

$$ M = \begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 5 & 6 & 1 \end{pmatrix} $$

$$ M = \begin{pmatrix} 15 & 13 & 2 \\ 19 & 19 & 3 \\ 5 & 6 & 1 \end{pmatrix} $$

This matrix also has determinant 1:

$$ \det M = \det \begin{pmatrix} 15 & 13 & 2 \\ 19 & 19 & 3 \\ 5 & 6 & 1 \end{pmatrix} = 1 $$

Therefore, its inverse consists entirely of integers:

$$ M^{-1} = \begin{pmatrix} 1 & -1 & 1 \\ -4 & 5 & -7 \\ 19 & -25 & 38 \end{pmatrix} $$

And so on - the same reasoning applies to higher dimensions.