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--- lecture_15_slides [2017/03/27 17:24] – [Examples] rupert
+++ lecture_15_slides [2017/03/27 17:31] (current) – [Example: $n=3$] rupert
@@ Line 110: / Line 110: @@
 ==== Example: $n=3$ ====
-Let $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$.
+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 $A^{-1}$.
   * Matrix of signs: $\mat{+&-&+\\-&+&-\\+&-&+}$
-  * Matrix of cofactors: $C=\mat{\vm{-4&3\\4&-2}&-\vm{-2&3\\5&-2}&\vm{-2&-4\\5&4}\\-\vm{1&0\\4&-2}&\vm{3&0\\5&-2}&-\vm{3&1\\5&4}\\\vm{1&0\\-4&3}&-\vm{3&0\\-2&3}&\vm{3&1\\-2&-4}}= \mat{-4&11&12\\2&-6&-7\\3&-9&-10}$
+  * Matrix of cofactors: $C=\def\vm#1{\left|\begin{smallmatrix}#1\end{smallmatrix}\right|}\mat{\vm{-4&3\\4&-2}&-\vm{-2&3\\5&-2}&\vm{-2&-4\\5&4}\\-\vm{1&0\\4&-2}&\vm{3&0\\5&-2}&-\vm{3&1\\5&4}\\\vm{1&0\\-4&3}&-\vm{3&0\\-2&3}&\vm{3&1\\-2&-4}}= \mat{-4&11&12\\2&-6&-7\\3&-9&-10}$
-  * So the adjoint of $A$ is $J=C^T=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$.
 ==== ====
-  * Adjoint of $A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$ is $J=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$
+  * Adjoint of $A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$ is $J=C^T=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$
   * $AJ=\mat{3&1&0\\-2&-4&3\\5&4&-2}\mat{-4&2&3\\11&-6&-9\\12&-7&-10}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$
   * $JA=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}\mat{3&1&0\\-2&-4&3\\5&4&-2}=\mat{-1&0&0\\0&-1&0\\0&0&-1}=-1\cdot I_3$
@@ Line 133: / Line 132: @@
 ==== 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 $A$.
+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 $\det A$.  ■
+  * Divide the equation $AJ=(\det A)I_n=JA$ by $\det A$.
+  * $A(\frac1{\det A}J)=I_n=(\frac1{\det A})JA$
+  * So $A^{-1}=\frac1{\det A} J$.  ■
+==== 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 $AA^{-1}=I_4=A^{-1}A$.)
-==== 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 $A^{-1}=\frac1{-1}J=-J=\mat{4&-2&-3\\-11&6&9\\-12&7&10}$.