=== Example: $n=2$, general case ===

If $A=\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\mat{a&b\\c&d}$, then $C=\mat{d&-c\\-b&a}$, so the adjoint of $A$ is $J=C^T=\mat{d&-b\\-c&a}$. 

Recall that $AJ=(\det A)I_2=JA$; we calculated this earlier when we looked at the inverse of a $2\times 2$ matrix. Hence for a $2\times 2$ matrix $A$, if $\det A\ne0$, then $A^{-1}=\frac1{\det A}J$.

=== Example: $n=3$ ===

If $\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$, then the matrix of signs is $\mat{+&-&+\\-&+&-\\+&-&+}$, so
\[\def\vm#1{\begin{vmatrix}#1\end{vmatrix}} C=\mat{
\vm{-4&3\\4&-2}&-\vm{-2&3\\5&-2}&\vm{-2&-4\\5&4}\\
-\vm{1&0\\4&-2}&\vm{3&0\\5&-2}&-\vm{3&1\\5&4}\\
\vm{1&0\\-4&3}&-\vm{3&0\\-2&3}&\vm{3&1\\-2&-4}}
= \mat{-4&11&12\\2&-6&-7\\3&-9&-10}\]
so the adjoint of $A$ is
\[ J=C^T=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}.\]

Observe that $AJ=\mat{3&1&0\\-2&-4&3\\5&4&-2}\mat{-4&2&3\\11&-6&-9\\12&-7&-10}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$, and $JA=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}\mat{3&1&0\\-2&-4&3\\5&4&-2}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$; and $\det(A)=-1$.

This is an illustration of the following theorem, whose proof is omitted:

==== Theorem: key property of the adjoint of a square matrix ====

If $A$ is any $n\times n$ matrix and $J$ is its [[adjoint]], then $AJ=(\det A)I_n=JA$.

==== Corollary: a formula for the inverse of a square matrix ====

If $A$ is any $n\times n$ matrix with $\det(A)\ne 0$, then $A$ is invertible and \[A^{-1}=\frac1{\det A}J\] where $J$ is the adjoint of $A$.

=== Proof ===

Divide the equation $AJ=(\det A)I_n=JA$ by $\det A$.  ■ 

==== Example ====

If again we take $A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$, then $J=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$ and $\det(A)=-1$, so $A$ is invertible and $A^{-1}=\frac1{-1}J=-J=\mat{4&-2&-3\\-11&6&9\\-12&7&10}$.

==== Example ($n=4$) ====

Let $A=\mat{1&0&0&0\\1&2&0&0\\1&2&3&0\\1&2&3&4}$.

Recall that a matrix with a repeated row or a zero row has determinant zero.
We have \[C=\mat{+\vm{2&0&0\\2&3&0\\2&3&4}&-\vm{1&0&0\\1&3&0\\1&3&4}&+0&-0\\-0&+\vm{1&0&0\\1&3&0\\1&3&4}&-\vm{1&0&0\\1&2&0\\1&2&4}&+0\\+0&-0&+\vm{1&0&0\\1&2&0\\1&2&4}&-\vm{1&0&0\\1&2&0\\1&2&3}\\-0&+0&-0&+\vm{1&0&0\\1&2&0\\1&2&3}}=\mat{24&-12&0&0\\0&12&-8&0\\0&0&8&-6\\0&0&0&6}\]
so \[J=C^T=\mat{24&0&0&0\\-12&12&0&0\\0&-8&8&0\\0&0&-6&6}.\]
Since $A$ is lower triangular, its determinant is given by multiplying together its diagonal entries: $\det(A)=1\times 2\times 3\times 4=24$. (Note that even if $A$ was not triangular, $\det A$ can be easily found from the matrix of cofactors $C$ by summing the entries of $A$ multiplied by the entries of $C$ (i.e., the minors) along any row or column.)

So \[A^{-1}=\frac1{\det A}J = \frac1{24}\mat{24&0&0&0\\-12&12&0&0\\0&-8&8&0\\0&0&-6&6}=\mat{1&0&0&0\\-1/2&1/2&0&0\\0&-1/3&1/3&0\\0&0&-1/4&1/4}.\]
You should check that this really is the inverse, by checking that $AA^{-1}=I_4=A^{-1}A$.
===== A more efficient way to find $A^{-1}$ =====

Given an $n\times n$ matrix $A$, form the $n\times 2n$ matrix
\[ \def\m#1{\left[
\begin{array}{@{} c|c {}@} % it does autodetection
#1
\end{array}
\right]}\m{A&I_n}\]
and use [[EROs]] to put this matrix into [[RREF]]. One of two things can happen:

  * Either you get a row of the form $[0~0~\dots~0~|~*~*~\dots~*]$ which starts with $n$ zeros. You can then conclude that $A$ is not invertible.
  * Or you end up with a matrix of the form $\m{I_n&B}$ for some $n\times n$ matrix $B$. You can then conclude that $A$ is invertible, and $A^{-1}=B$.

==== Examples ====

  * Consider $A=\def\mat#1{\begin{matrix}#1\end{matrix}}\left[\mat{1&3\\2&6}\right]$. \begin{align*}\m{A&I_2}&=\m{\mat{1&3\\2&6}&\mat{1&0\\0&1}}
\def\go#1#2{\m{\mat{#1}&\mat{#2}}}
\def\ar#1{\\[6pt]\xrightarrow{#1}&}
\ar{R2\to R2-2R1}\go{1&3\\0&0}{1&0\\-2&1}
\end{align*} Conclusion: $A$ is not invertible.
  * Consider $A=\left[\mat{1&3\\2&7}\right]$.\begin{align*}\m{A&I_2}&=\m{\mat{1&3\\2&7}&\mat{1&0\\0&1}}
\ar{R2\to R2-2R1}\go{1&3\\0&1}{1&0\\-2&1}
\ar{R1\to R1-3R1}\go{1&0\\0&1}{7&-3\\-2&1}
\end{align*} Conclusion: $A$ is invertible and $A^{-1}=\left[\mat{7&-3\\-2&1}\right]$.
  * Consider $A=\left[\mat{3&1&0\\-2&-4&3\\5&4&-2}\right]$.\begin{align*}\m{A&I_3}&=\go{3&1&0\\-2&-4&3\\5&4&-2}{1&0&0\\0&1&0\\0&0&1}
\ar{R1\to R1+R2}
\go{1&-3&3\\-2&-4&3\\5&4&-2}{1&1&0\\0&1&0\\0&0&1}
\ar{R2\to R2+2R1,\ R3\to R3-5R1}
\go{1&-3&3\\0&-10&9\\0&19&-17}{1&1&0\\2&3&0\\-5&-5&1}
\ar{R3\leftrightarrow R2}
\go{1&-3&3\\0&19&-17\\0&-10&9}{1&1&0\\-5&-5&1\\2&3&0}
\ar{R2\to R2+2R3}
\go{1&-3&3\\0&-1&1\\0&-10&9}{1&1&0\\-1&1&1\\2&3&0}
\ar{R1\to R1+3R2,\ R3\to R3-10R2}
\go{1&0&0\\0&-1&1\\0&0&-1}{4&-2&3\\-1&1&1\\12&-7&-10}
\ar{R2\to R2+R3}
\go{1&0&0\\0&-1&0\\0&0&-1}{4&-2&3\\11&-6&-9\\12&-7&-10}
\ar{R2\to -R2,\ R3\to -R3}
\go{1&0&0\\0&1&0\\0&0&1}{4&-2&3\\-11&6&9\\-12&7&10}
\end{align*} Conclusion: $A$ is invertible, and $A^{-1}=\left[\mat{4&-2&3\\-11&6&9\\-12&7&10}\right]$.