~~REVEAL~~


===== Finding the inverse of an invertible $n\times n$ matrix =====

==== The adjoint of a square matrix ====

{{page>adjoint}}

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

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

=== Proof ===

  * Omitted

==== 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 $\det A$.
  * $A(\frac1{\det A}J)=I_n=(\frac1{\det A}J)A$
  * So $A^{-1}=\frac1{\det A} J$.  ■ 

==== Example: $n=2$ ====

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 $J=C^T=\mat{d&-b\\-c&a}$. 

  * So if $\det A\ne0$, then $A^{-1}=\frac1{\det A}J=\frac1{ad-bc}\mat{d&-b\\-c&a}$.
  * Same as previous formula!

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

Find $J$, the adjoint of $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$, and compute $AJ$, $JA$, $\det(A)$ and $A^{-1}$.
  * Matrix of signs: $\mat{+&-&+\\-&+&-\\+&-&+}$
  * Matrix of cofactors: $C=\def\vm#1{\left|\begin{smallmatrix}#1\end{smallmatrix}\right|}\mat{\vm{-4&3\\4&-2}&-\vm{-2&3\\5&-2}&\vm{-2&-4\\5&4}\\-\vm{1&0\\4&-2}&\vm{3&0\\5&-2}&-\vm{3&1\\5&4}\\\vm{1&0\\-4&3}&-\vm{3&0\\-2&3}&\vm{3&1\\-2&-4}}= \mat{-4&11&12\\2&-6&-7\\3&-9&-10}$

==== ====
  * Adjoint of $A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$ is $J=C^T=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$
  * $AJ=\mat{3&1&0\\-2&-4&3\\5&4&-2}\mat{-4&2&3\\11&-6&-9\\12&-7&-10}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$
  * $JA=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}\mat{3&1&0\\-2&-4&3\\5&4&-2}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$
  * So $\det(A)=-1$.
  * So $A^{-1}=\frac1{\det(A)}J=-J=\mat{4&-11&-12\\-2&6&7\\-3&9&10}$.

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

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

  * Reminder: repeated row or zero row gives determinant zero
  * $C=\mat{+\vm{2&0&0\\2&3&0\\2&3&4}&-\vm{1&0&0\\1&3&0\\1&3&4}&+0&-0\\-0&+\vm{1&0&0\\1&3&0\\1&3&4}&-\vm{1&0&0\\1&2&0\\1&2&4}&+0\\+0&-0&+\vm{1&0&0\\1&2&0\\1&2&4}&-\vm{1&0&0\\1&2&0\\1&2&3}\\-0&+0&-0&+\vm{1&0&0\\1&2&0\\1&2&3}}=\mat{24&-12&0&0\\0&12&-8&0\\0&0&8&-6\\0&0&0&6}$
==== Example ($n=4$) ====

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

  * $C=\mat{24&-12&0&0\\0&12&-8&0\\0&0&8&-6\\0&0&0&6}$ so $J=C^T=\mat{24&0&0&0\\-12&12&0&0\\0&-8&8&0\\0&0&-6&6}$.
  * So $A^{-1}=\frac1{\det A}J = \frac1{24}\mat{24&0&0&0\\-12&12&0&0\\0&-8&8&0\\0&0&-6&6}=\mat{1&0&0&0\\-1/2&1/2&0&0\\0&-1/3&1/3&0\\0&0&-1/4&1/4}$.
  * (Easy to check 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  $\def\m#1{\small\left[\begin{array}{@{} c|c {}@}#1\end{array}\right]}\m{A&I_n}$.
  * Use [[EROs]] to put this into [[RREF]]. 
  * One of two things can happen:
    * 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 get 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}\end{align*}
==== ====
\begin{align*}
\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}\end{align*}
==== ====
\begin{align*}
\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}\end{align*}
==== ====
\begin{align*}
\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*}
==== ====
  * So $\m{A&I_3}\xrightarrow{\text{EROs}}\go{1&0&0\\0&1&0\\0&0&1}{4&-2&3\\-11&6&9\\-12&7&10}$
  * Conclusion: $A$ is invertible, and $A^{-1}=\left[\mat{4&-2&3\\-11&6&9\\-12&7&10}\right]$.