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--- lecture_11 [2016/03/01 11:01] – rupert
+++ lecture_11 [2017/02/28 12:07] (current) – rupert
@@ Line 70: / Line 70: @@
 If $\det(A)=0$, then $AJ=0_{2\times 2}$  and $J\ne 0_{2\times2}$ (since $A\ne0_{2\times2}$, and $J$ is obtained from $A$ by swapping two entries and multiplying the others by $-1$). Hence by the previous corollary, $A$ is not invertible in this case.  ■
-==== Example ====
-Let's solve the matrix equation $\mat{1&5\\3&-2}X=\mat{4&1&0\\0&2&1}$ for $X$.
-Write $A=\mat{1&5\\3&-2}$. Then $\det(A)=1(-2)-5(3)=-2-15=-17$ which isn't zero, so $A$ is invertible. And $A^{-1}=\frac1{-17}\mat{-2&-5\\-3&1}=\frac1{17}\mat{2&5\\3&-1}$.
-Hence the solution is $X=A^{-1}\mat{4&1&0\\0&2&1}=\frac1{17}\mat{2&5\\3&-1}\mat{4&1&0\\0&2&1}=\frac1{17}\mat{8&12&5\\12&1&-1}$.