Invertible Matrices Over Any Ring

Purpose
The purpose of this page is to give examples of determining when a matrix is invertible and how to find it’s inverse.

Main Result: Let R be a commutative ring. Then,

    \[M_2(R)^\times = \left\{A=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \mid \det(A)\in R^\times\right\}\]

Let R=\mathbb{Z}, recall \mathbb{Z}^\times=\{\pm 1\}.

Example 1: Let

    \[A = \begin{pmatrix} 2 & 3 \\ 1 & 2\end{pmatrix}, B = \begin{pmatrix} 5 & 4 \\ 5 & 2\end{pmatrix}, C= \begin{pmatrix} 3 & 2 \\ 2 & 1\end{pmatrix} \]


Then,

    \[\det(A)=2\cdot 2 - 3\cdot 1 = 4-3=1\]


Therefore, A is invertible in M_2(\mathbb{Z}). The inverse is given by:

    \[A^{-1} = \det(A)^{-1}\begin{pmatrix}2 & -3 \\ -1 & 2 \end{pmatrix}=1 \cdot \begin{pmatrix}2 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix}2 & -3 \\ -1 & 2 \end{pmatrix}\]



    \[\det(B)=5\cdot 2 - 4 \cdot 5 = 10-20 = -10\]


Therefore, B is not invertible in M_2(\mathbb{Z}).

    \[\det(C)=3\cdot 1 - 2\cdot 2 = -1\]


Therefore, C is invertible in M_2(\mathbb{Z}. The inverse is given by:

    \[C^{-1}=\det(C)^{-1}\begin{pmatrix}1 &-2 \\ -2 & 3 \end{pmatrix}=-1\cdot \begin{pmatrix}1 &-2 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix}-1 &2 \\ 2 & 03 \end{pmatrix}\]

Exercises: Determine if the following matrices are invertible in M_2(\mathbb{Z}):

    \[A = \begin{pmatrix} 7 & 2 \\ 10 & 3 \end{pmatrix}, B = \begin{pmatrix} 5 & 2 \\ 10 & 4 \end{pmatrix}\]

Solution for A We have \det(A)=7\cdot 3 - 10\cdot 2 = 21 -20=1. Therefore, A is invertible. The inverse is given by:

    \[A^{-1}=\det(A)^{-1}\begin{pmatrix} 3 & -2 \\ -10 & 7\end{pmatrix}=1\cdot \begin{pmatrix} 3 & -2 \\ -10 & 7\end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -10 & 7\end{pmatrix}\]

Solution for B We have \det(B)=5\cdot 4 - 10\cdot 2 = 20 -20=0. Therefore, B is not invertible.

Let R=\mathbb{Z}/12\mathbb{Z}. The group of units is \mathbb{Z}/12\mathbb{Z}^\times=\{1,5,7,11\}.

Example 2: Let

    \[A = \begin{pmatrix} 5 + 12\mathbb{Z} & 1+ 12\mathbb{Z} \\ 7+ 12\mathbb{Z} & 6+ 12\mathbb{Z}\end{pmatrix} \]


Then, \det(A) = (5\cdot 6 - 1\cdot 7)+ 12\mathbb{Z}=23+ 12\mathbb{Z}\equiv 11+ 12\mathbb{Z}. Note that (11+ 12\mathbb{Z})^{-1}=11+ 12\mathbb{Z}. Therefore,

    \begin{align*} A^{-1} &= \det(A)^{-1}\begin{pmatrix} 6+ 12\mathbb{Z} & -1+ 12\mathbb{Z} \\ -7+ 12\mathbb{Z} & 5+12\mathbb{Z} \end{pmatrix} \\ &= 11\cdot \begin{pmatrix} 6+ 12\mathbb{Z} & -1+ 12\mathbb{Z} \\ -7+12\mathbb{Z} & 5+ 12\mathbb{Z} \end{pmatrix} \\ &= \begin{pmatrix} 66+12\mathbb{Z} & -11+12\mathbb{Z} \\ -77+ 12\mathbb{Z} & 55+ 12\mathbb{Z} \end{pmatrix} \\ &= \begin{pmatrix} 6+ 12\mathbb{Z} & 1+ 12\mathbb{Z} \\ 7+ 12\mathbb{Z} & 7+12\mathbb{Z} \end{pmatrix} \end{align*}

Exercises: Determine if the following matrices are invertible in M_2(\mathbb{Z}/15\mathbb{Z}):

    \[A = \begin{pmatrix} 8+15\mathbb{Z} & 4+15\mathbb{Z} \\ 2+15\mathbb{Z} & 2 + 15\mathbb{Z} \end{pmatrix} \text{ and } B = \begin{pmatrix} 7+15\mathbb{Z} & 2+15\mathbb{Z} \\ 2+15\mathbb{Z} & 2 + 15\mathbb{Z} \end{pmatrix}\]

Solution for A We have \det(A)=(8\cdot 2 - 4\cdot 2) + 15\mathbb{Z} = (16-8)+ + 15\mathbb{Z} = 8 + 15\mathbb{Z}. Therefore, A is invertible. The inverse is given by:

    \begin{align*} A^{-1} &= \det(A)^{-1}\begin{pmatrix} 2 + 15\mathbb{Z} & -4 + 15\mathbb{Z} \\ -2 + 15\mathbb{Z} & 8 + 15\mathbb{Z} \end{pmatrix} \\ &= 2 \cdot \begin{pmatrix} 2 + 15\mathbb{Z} & -4 + 15\mathbb{Z} \\ -2 + 15\mathbb{Z} & 8 + 15\mathbb{Z} \end{pmatrix} \\ &= \begin{pmatrix} 4 + 15\mathbb{Z} & -8 + 15\mathbb{Z} \\ -4 + 15\mathbb{Z} & 16 + 15\mathbb{Z} \end{pmatrix} \\ &= \begin{pmatrix} 4 + 15\mathbb{Z} & 7 + 15\mathbb{Z} \\ 11 + 15\mathbb{Z} & 1 + 15\mathbb{Z} \end{pmatrix} \end{align*}

Solution for B We have \det(B)=(7\cdot 2 - 2\cdot 2)+15\mathbb{Z} = (14-4)+15\mathbb{Z}=10+15\mathbb{Z}. Therefore, B is not invertible.