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lecture_11_slides

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lecture_11_slides [2017/02/27 12:03] rupertlecture_11_slides [2017/02/27 12:35] (current) – [Proof] rupert
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   * e.g. $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\mat{2&4\\0&1}$ is invertible, with inverse $C=\mat{0.5&-2\\0&1}$   * e.g. $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\mat{2&4\\0&1}$ is invertible, with inverse $C=\mat{0.5&-2\\0&1}$
-  * if $a$ is a non-zero scalar, then $C=[\tfrac 1a]$ is an inverse+  * if $[a]$ is $1\times 1$ and $a\ne 0$, then $C=[\tfrac 1a]$ is an inverse
   * $I_n$ is its own inverse for any $n$   * $I_n$ is its own inverse for any $n$
   * $0_{n\times n}$, $\mat{1&0\\0&0}$ and $\mat{1&2\\-3&-6}$ aren't invertible    * $0_{n\times n}$, $\mat{1&0\\0&0}$ and $\mat{1&2\\-3&-6}$ aren't invertible 
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 If $A=\mat{a&b\\c&d}$ and $J=\mat{d&-b\\-c&a}$, then we have If $A=\mat{a&b\\c&d}$ and $J=\mat{d&-b\\-c&a}$, then we have
-\[ AJ=\delta I_2=JA\] +\[ AJ=(ad-bc) I_2=JA.\]
-where $\delta=ad-bc$.+
  
-  * Proof is calculation!+  * Proof is an easy calculation! 
 +  * Note that $(ad-bc) I_2=(ad-bc)\mat{1&0\\0&1}=\mat{ad-bc&0\\0&ad-bc}$ 
 +  * Now just show that $AJ$ and $JA$ both give the same matrix (exercise).
  
 ==== Definition: the determinant of a $2\times 2$ matrix ==== ==== Definition: the determinant of a $2\times 2$ matrix ====
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 ==== Proof ==== ==== Proof ====
  
-  * Let $J=\mat{d&-b\\-c&a}$ and write $\delta=\det(A)$. +  * Let $J=\mat{d&-b\\-c&a}$, so $AJ=\det(A) I_2=JA$ (Lemma).
-  * By the previous lemma, $AJ=\delta I_2=JA$.+
  
-If $\delta\ne 0$: +If $\det(A)\ne 0$: 
-  * Multiply by $\frac1{\delta}$: $\quad A(\tfrac1{\delta}J)=I_2=(\tfrac1{\delta}J) A+  * Multiply by $\frac1{\det(A)}$: $\quad A(\tfrac1{\det(A)}J)=I_2=(\tfrac1{\det(A)}J) A$ 
-  * So $ AB=I_2=BA$, where $B=\tfrac1{\delta}J+  * So $A$ invertible with $A^{-1}=\frac1{\det(A)}J=\frac1{\det(A)}\mat{d&-b\\-c&a}$.
-  * So $A$ invertible with $A^{-1}=B=\frac1{\delta}J=\frac1{\det(A)}\mat{d&-b\\-c&a}$.+
  
-If $\delta=0$:+If $\det(A)=0$:
   * $AJ=0_{2\times 2}$   * $AJ=0_{2\times 2}$
-  * If $J=0_{2\times 2}$ then $A=0_{2\times 2}$ so $A$ isn't invertible +  * If $J\ne 0_{2\times 2}$ then (by corollary) $A$ isn't invertible 
-  * If $J\ne 0_{2\times 2}$ then by the previous corollary, $A$ isn't invertible.  ■  +  * If $J=0_{2\times 2}$ then $A=0_{2\times 2}$, which isn't invertible.■ 
 ==== Using the inverse to solve a matrix equation ==== ==== Using the inverse to solve a matrix equation ====
  
lecture_11_slides.1488197039.txt.gz · Last modified: by rupert

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