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--- lecture_11_slides [2017/02/27 12:09] – [Lemma] rupert
+++ lecture_11_slides [2017/02/27 12:35] (current) – [Proof] rupert
@@ Line 122: / 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 ====