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--- lecture_13_slides [2017/03/06 17:39] – [Using the inverse to solve a matrix equation] rupert
+++ lecture_13_slides [2017/03/06 17:48] (current) – [Example] rupert
@@ Line 93: / Line 93: @@
   * Short version: to find $M_{ij}$: delete the row and col containing $(i,j)$ entry, then take determinant.
-==== Examples of minors ====
+==== Examples of minors (1) ====
 Short version: to find $M_{ij}$: delete the row and col containing $(i,j)$ entry, then take determinant.
-  - If $A=\mat{3&5\\-4&7}$, then
+Example: for $A=\mat{3&5\\-4&7}$:
-    * $M_{11}=\det[7]=7$
+  * $M_{11}=\det[7]=7$
-    * $M_{12}=\det[-4]=-4$
+  * $M_{12}=\det[-4]=-4$
-    * $M_{21}=5$
+  * $M_{21}=5$
-    * $M_{22}=3$.
+  * $M_{22}=3$.
-  - If $A=\mat{1&2&3\\7&8&9\\11&12&13}$, then
-    * $M_{23}=\det\mat{1&2\\11&12}=1\cdot 12-2\cdot 11=-10$
+==== Examples of minors (2) ====
-    * $M_{32}=\det\mat{1&3\\7&9}=-12$.
+Short version: to find $M_{ij}$: delete the row and col containing $(i,j)$ entry, then take determinant.
+Example: for $A=\mat{1&2&3\\7&8&9\\11&12&13}$:
+  * $M_{23}=\det\mat{1&2\\11&12}=1\cdot 12-2\cdot 11=-10$
+  * $M_{32}=\det\mat{1&3\\7&9}=-12$
+  * etc
 ==== Step 2: cofactors ====
@@ Line 117: / Line 123: @@
   * Short version: minors with some sign changes (according to matrix of signs).
-====Examples of cofactors====
+====Examples of cofactors (1)====
 Short version: minors with sign changes $\mat{+&-&+&-&\dots\\-&+&-&+&\dots\\+&-&+&-&\dots\\\vdots&\vdots&\vdots&\vdots&\ddots}$
-  - If $A=\mat{3&5\\-4&7}$, then
+If $A=\mat{3&5\\-4&7}$, then
-    * $C_{11}=+M_{11}=\det[7]=7$
+  * $C_{11}=+M_{11}=\det[7]=7$
-    * $C_{12}=-M_{12}=-\det[-4]=4$
+  * $C_{12}=-M_{12}=-\det[-4]=4$
-    * $C_{21}=-5$, and $C_{22}=3$.
+  * $C_{21}=-5$, and $C_{22}=3$.
-  - If $A=\mat{1&2&3\\7&8&9\\11&12&13}$, then
-    * $C_{23}=-M_{23}=-(-10)=10$
+====Examples of cofactors (2)====
-    * $C_{33}=+M_{33}=\det\mat{1&2\\7&8}=-6$.
+Short version: minors with sign changes $\mat{+&-&+&-&\dots\\-&+&-&+&\dots\\+&-&+&-&\dots\\\vdots&\vdots&\vdots&\vdots&\ddots}$
+If $A=\mat{1&2&3\\7&8&9\\11&12&13}$, then
+  * $C_{23}=-M_{23}=-(-10)=10$
+  * $C_{33}=+M_{33}=\det\mat{1&2\\7&8}=-6$
+  * etc
 ==== Step 3: the determinant of a $3\times 3$ matrix ====
@@ Line 148: / Line 160: @@
   * = $-4 -2(-8)+3(-4)$
   * = $0$.
+  * (So this matrix isn't invertible!)
-  * $\mat{1&2&3\\7&8&9\\11&12&13}$ has zero determinant
-  * So this matrix is //not// invertible.
 ==== Notation ====
@@ Line 173: / Line 183: @@
 \begin{align*}
-\vm{\color{red}1&\color{red}0&\color{red}2&\color{red}3\\0&2&1&-1\\2&0&0&1\\3&0&4&2} &= \color{red}1\vm{\color{blue}2&\color{blue}1&\color{blue}-1\\0&0&1\\0&4&2}-\color{red}0\vm{0&1&-1\\2&0&1\\3&4&2}+\color{red}2\vm{\color{orange}0&\color{orange}2&\color{orange}{-1}\\2&0&1\\3&0&2}-\color{red}3\vm{\color{purple}0&\color{purple}2&\color{purple}1\\2&0&0\\3&0&4}\\
+\def\vm#1{\left|\begin{smallmatrix}#1\end{smallmatrix}\right|}
-&= 1\left(\color{blue}2\vm{0&1\\4&2}-\color{blue}1\vm{0&1\\0&2}\color{blue}{-1}\vm{0&0\\0&4}\right)\\&\quad +2\left(\color{orange}0-\color{orange}{2}\vm{2&1\\3&2}\color{orange}{-1}\vm{2&0\\3&0}\right)\\&\quad - 3\left(\color{purple}0-\color{purple}2\vm{2&0\\3&4}+\color{purple}1\vm{2&0\\3&0}\right)\\
+\vm{\color{red}1&\color{red}0&\color{red}2&\color{red}3\\0&2&1&-1\\2&0&0&1\\3&0&4&2}
-&=1(2(-4)-0-0)+2(-2(1)-0)-3(-2(8)+0)\\
+&= \color{red}1\vm{\color{blue}2&\color{blue}1&\color{blue}-1\\0&0&1\\0&4&2}-\color{red}0\vm{0&1&-1\\2&0&1\\3&4&2}+\color{red}2\vm{\color{orange}0&\color{orange}2&\color{orange}{-1}\\2&0&1\\3&0&2}-\color{red}3\vm{\color{purple}0&\color{purple}2&\color{purple}1\\2&0&0\\3&0&4}
-&=-8-4+48\\
+\\&= 1\left(\color{blue}2\vm{0&1\\4&2}-\color{blue}1\vm{0&1\\0&2}\color{blue}{-1}\vm{0&0\\0&4}\right)
+\\&\quad -0+2\left(\color{orange}0-\color{orange}{2}\vm{2&1\\3&2}\color{orange}{-1}\vm{2&0\\3&0}\right)
+\\&\quad -3\left(\color{purple}0-\color{purple}2\vm{2&0\\3&4}+\color{purple}1\vm{2&0\\3&0}\right)
+\\&=1(2(-4)-0-0)+2(-2(1)-0)-3(-2(8)+0)
+\\&=-8-4+48\\
 &=36.
 \end{align*}