Differences

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

--- lecture_11 [2016/03/01 10:57] – [Example] rupert
+++ lecture_11 [2017/02/28 12:07] (current) – rupert
@@ Line 23: / Line 23: @@
 So there is not a unique solution to $AX=B$, for $B$ the zero matrix. If $A$ was invertible, this would contradict the uniqueness statement of the last Proposition. So $A$ cannot be invertible. ■
-==== Example ====
+==== Examples ====
+  * We can now see why the matrix $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}A=\mat{1&2\\-3&-6}$ is not invertible. If $X=\mat{-2\\1}$ and $K=\mat{2\\-1}$, then $K$ is non-zero, but $AK=0_{2\times 1}$. So $A$ is not invertible, by the Corollary.
+  * $A=\mat{1&4&5\\2&5&7\\3&6&9}$ is not invertible, since $K=\mat{1\\1\\-1}$ is non-zero and $AK=0_{3\times 1}$.
+===== $2\times 2$ matrices: determinants and invertibility =====
+==== Question ====
+Which $2\times 2$ matrices are invertible? For the invertible matrices, can we find their inverse?
+==== Lemma ====
+If $A=\mat{a&b\\c&d}$ and $J=\mat{d&-b\\-c&a}$, then we have
+\[ AJ=\delta I_2=JA\]
+where $\delta=ad-bc$.
+=== Proof ===
+This is a calculation (done in the lectures; you should also check it yourself). ■
+==== Definition: the determinant of a $2\times 2$ matrix ====
+{{page>determinant of a 2x2 matrix}}
+==== Theorem: the determinant determines the invertibility (and inverse) of a $2\times 2$ matrix ====
+Let $A=\mat{a&b\\c&d}$ be a $2\times 2$ matrix.
+  - $A$ is invertible if and only if $\det(A)\ne0$.
+  - If $A$ is invertible, then $A^{-1}=\frac{1}{\det(A)}\mat{d&-b\\-c&a}$.
+=== Proof ===
+If $A=0_{2\times 2}$, then $\det(A)=0$ and $A$ is not invertible. So the statement is true is this special case.
+Now assume that $A\ne0_{2\times 2}$ and let $J=\mat{d&-b\\-c&a}$.
+By the previous lemma, we have \[AJ=(\det(A))I_2=JA.\]
+If $\det(A)\ne0$, then multiplying this equation through by the scalar $\frac1{\det(A)}$, we get
+\[ A\left(\frac1{\det(A)}J\right)=I_2=\left(\frac1{\det(A)}J\right) A,\]
+so if we write $B=\frac1{\det(A)}J$ to make this look simpler, then  we obtain
+\[ AB=I_2=BA,\]
+so in this case $A$ is invertible with inverse $B=\frac1{\det(A)}J=\frac1{\det(A)}\mat{d&-b\\-c&a}$.
+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.  ■
-We can now see why the matrix $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}A=\mat{1&2\\-3&-6}$ is not invertible. If $X=\mat{-2\\1}$ and $K=\mat{2\\-1}$, then $K$ is non-zero, but $AK=0_{2\times 1}$. So $A$ is not invertible, by the Corollary.