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--- lecture_14 [2016/03/09 11:29] – rupert
+++ lecture_14 [2017/03/09 10:49] (current) – [Corollary on invertibility] rupert
@@ Line 1: / Line 1: @@
+=== Example ===
+\begin{align*}\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\det\mat{1&2&3\\7&8&9\\11&12&13} &= 1\cdot C_{11} + 2 C_{12} + 3 C_{13}\\
+&= 1 \cdot (+M_{11}) + 2 \cdot (-M_{12}) + 3 \cdot(+M_{13})\\
+&= M_{11}-2M_{12}+3M_{13}\\
+&= \det\mat{8&9\\12&13} -2\det\mat{7&9\\11&13} + 3\det\mat{7&8\\11&12}\\
+&= (8\cdot 13-9\cdot 12) -2(7\cdot 13-9\cdot 11)+3(7\cdot 12-8\cdot 11)\\
+&=-4 -2(-8)+3(-4)\\
+&=-4+16-12\\
+&=0.\end{align*}
+From this, we can conclude that $\mat{1&2&3\\7&8&9\\11&12&13}$ is //not// invertible.
 === Notation ===
@@ Line 74: / Line 87: @@
   - $A$ is invertible if and only if $\det(A)\ne0$.
-  - If $A'$ is the same as $A$, except with two rows swapped, then $\det(A')=-\det(A)$.
+  - $\det(A^T)=\det(A)$
-  - If $c$ is a scalar and $A'$ is the same as $A$ except with one row multiplied by $c$, then $\det(A')=c\det(A)$.
+  - If $B$ is another $n\times n$ matrix, then $\det(AB)=\det(A)\det(B)$
-  - If $A'$ and $A''$ are the same as $A$ except in row $i$, and $\text{row}_i(A'')=\text{row}_i(A)+\text{row}_i(A')$, then $\det(A'')=\det(A)+\det(A')$.
-  - $\det(A^T)=\det(A)$. So we can swap "row" with "column" in these properties.
-  - If $B$ is another $n\times n$ matrix, then $\det(AB)=\det(A)\det(B)$.
-==== Corollary ====
+==== Corollary on invertibility ====
-If an $n\times n$ matrix $A$ has two equal rows (or columns), then $\det(A)=0$, and $A$ is not invertible.
+  - $A^T$ is invertible if and only if $A$ is invertible
+  - $AB$ is invertible if and only if **both** $A$ and $B$ are invertible
 === Proof ===
-Suppose $A$ has two equal rows. Let $A'$ be $A$ with the two equal rows swapped. By property 2, we have $\det(A')=-\det(A)$. But since these two rows are equal, we have $A'=A$! So $\det(A)=-\det(A)$. Solving this for $\det(A)$ gives $\det(A)=0$, and hence $A$ is not invertible, by property 1.
+  - We have $\det(A^T)=\det(A)$. So $A^T$ is invertible $\iff$ $\det(A^T)\ne0$ $\iff$ $\det(A)\ne 0$ $\iff$ $A$ is invertible.
+  - We have $\det(AB)=\det(A)\det(B)$. So $AB$ is invertible $\iff$ $\det(AB)\ne 0$ $\iff$ $\det(A)\det(B)\ne0$ $\iff$ $\det(A)\ne0$ and $\det(B)\ne 0$ $ \iff$ $A$ is invertible and $B$ is invertible. ■
-If $A$ has two equal columns, then $A^T$ has two equal rows, so $\det(A^T)=0$, so $\det(A)=0$ by property 5. ■
-=== Examples ===
-  * Swapping two rows changes the sign, so $\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\vm{0&0&2\\0&3&0\\4&0&0} = -\vm{4&0&0\\0&3&0\\0&0&2}=-4\cdot 3\cdot 2 = -24$.
-  * Multiplying a row or a column by a constant multiplies the determinant by that constant, so \begin{align*}\vm{ 2&4&6&10\\5&0&0&-10\\9&0&81&99\\1&2&3&4} &= 2\vm{ 1&2&3&5\\5&0&0&-10\\9&0&81&99\\1&2&3&4} \\&= 2\cdot 5\vm{ 1&2&3&5\\1&0&0&-2\\9&0&81&99\\1&2&3&4}\\&=  2\cdot 5\cdot 9 \vm{ 1&2&3&5\\1&0&0&-2\\1&0&9&11\\1&2&3&4}\\&=2\cdot 5\cdot 9\cdot 2 \vm{ 1&1&3&5\\1&0&0&-2\\1&0&9&11\\1&1&3&4}\\&=2\cdot 5\cdot 9\cdot 2\cdot 3 \vm{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4}.\end{align*}
-  * The matrices $\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}A''=\mat{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4}$, $A=\mat{ 0&0&0&1\\1&0&0&-2\\1&0&3&11\\1&1&1&4}$ and $A'=\mat{ 1&1&1&4\\1&0&0&-2\\1&0&3&11\\1&1&1&4}$ are the same except in row $1$, and $\text{row}_1(A'')=\text{row}_1(A)+\text{row}_1(A')$. Moreover, $A'$ has two equal rows, so has determinant zero. Hence by property 4 in the theorem, \begin{align*}\vm{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4} &=\det(A'')=\det(A)+\det(A')\\&=\vm{ 0&0&0&1\\1&0&0&-2\\1&0&3&11\\1&1&1&4}+\vm{ 1&1&1&4\\1&0&0&-2\\1&0&3&11\\1&1&1&4}\\&=-1\vm{1&0&0\\1&0&3\\1&1&1}+0\\&=-\vm{0&3\\1&1} = -(-3)=3.\end{align*}
-  * Hence \[\vm{ 2&4&6&10\\5&0&0&-10\\9&0&81&99\\1&2&3&4} = 2\cdot 5\cdot 9\cdot 2\cdot 3 \vm{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4} = 2\cdot 5\cdot 9\cdot 2\cdot 3 \cdot 3 = 1620.\]