Differences

This shows you the differences between two versions of the page.

--- lecture_11_slides [2017/02/27 12:08] – [Lemma] rupert
+++ lecture_11_slides [2017/02/27 12:35] (current) – [Proof] rupert
@@ Line 103: / Line 103: @@
 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 an easy calculation!
-  * Note that $\delta I_n=\delta\mat{1&0\\0&1}=\mat{\delta &0\\0&\delta}=\mat{ad-bc&0\\0&ad-bc}$
+  * 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).
@@ Line 123: / Line 122: @@
 ==== 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 $\det(A)=0$:
   * $AJ=0_{2\times 2}$
-  * 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 ====