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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 $A^{-1}=\frac1{-1}J=-J=\mat{4&-2&-3\\-11&6&9\\-12&7&10}$.

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 $A^{-1}=\left[\mat{4&-2&3\\-11&6&9\\-12&7&10}\right]$.