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--- lecture_14 [2016/03/09 14:48] – [Corollary] 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 77: / Line 90: @@
   - If $B$ is another $n\times n$ matrix, then $\det(AB)=\det(A)\det(B)$
-==== Theorem: row/column operations and determinants ====
+==== Corollary on invertibility ====
-Let $A$ be an $n\times n$ matrix, let $c$ be a scalar and let $i\ne j$.
+  - $A^T$ is invertible if and only if $A$ is invertible
+  - $AB$ is invertible if and only if **both** $A$ and $B$ are invertible
-$A_{Ri\to x}$ means $A$ but with row $i$ replaced by $x$.
-  - If $i\ne j$, then $\det(A_{Ri\leftrightarrow Rj})=-\det(A)$ (swapping two rows changes the sign of det).
-  - $\det(A_{Ri\to c Ri}) = c\det(A)$ (scaling one row scales $\det(A)$ in the same way)
-  - $\det(A_{Ri\to Ri + c Rj}) = \det(A)$ (adding a multiple of one row to another row doesn't change $\det(A)$)
-  * Also, these properties all hold if you change "row" into "column" throughout.
-==== Corollary ====
-If an $n\times n$ matrix $A$ has two equal rows (or columns), then $\det(A)=0$, and $A$ is not invertible.
 === Proof ===
+  - 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.
-If $A$ has two equal rows, row $i$ and row $j$, then $A=A_{Ri\leftrightarrow Rj}$
+  - 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. ■
-So $\det(A)=\det(A_{Ri\leftrightarrow Rj}) = -\det(A)$, so $2\det(A)=0$, so $\det(A)=0$.
-If $A$ has two equal columns, then $A^T$ has two equal rows, so $\det(A)=\det(A^T)=0$.
-In either case, $\det(A)=0$. So $A$ is not invertible.■
-=== 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*}
-  * $\det(A_{R1\to R1-R4})=\det(A)$, so \begin{align*}\vm{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4} &=\vm{ 0&0&0&1\\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 \begin{align*}\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.\end{align*}
-==== Corollary ===
-If $\def\row{\text{row}}\row_j(A)=c\cdot \row_i(A)$ for some $i\ne j$ and some $c\in \mathbb{R}$, then $\det(A)=0$.
-=== Proof ===
-Let $B$ have the same rows as $A$, except with $\row_i(A)$ in both row $i$ and row $j$. Observe that
-  * row $j$ of $B$ is $c$ times row $j$ of $A$, and all the other rows are equal; and
-  * $A$ has two equal rows.
-Hence using property 3 in the theorem and the previous corollary, we have $\det(B)=c\cdot \det(A)=c\cdot 0=0.$ ■
-==== Corollary ====
-If $E$ has the same rows as $A$ except in row $j$, and $\row_j(E)=\row_j(A)+c\cdot \row_i(A)$ for some $i\ne j$ and some $c\in \mathbb{R}$, then $\det(E)=\det(A)$.
-=== Proof ===
-Let $B$ have the same rows as $A$, except with $c\cdot \row_i(A)$ in row $j$. Observe that
-  * $\det(B)=0$, by the previous corollary; and
-  * except in row $j$, the rows of $E$, $A$ and $B$ are all equal, and $\row_j(E)=\row_j(A)+\row_j(B)$.
-Hence by property 4 of the theorem, we have $\det(E)=\det(A)+\det(B)=\det(A)+0=\det(A)$. ■
-We have now seen the effect of each of the three types of [[ERO]] on the determinant of a matrix:
-  - changing the order of the rows of the matrix multiplies the determinant by $-1$;
-  - multiplying one of the rows of the matrix by $c\in \mathbb{R}$ multiplies the determinant by $c$; and
-  - replacing row $j$ by "row $j$ ${}+{}$ $c\times {}$ (row $i$)", where $c$ is a non-zero real number and $i\ne j$ does not change the determinant.
-Moreover, since $\det(A)=\det(A^T)$, this all applies equally to columns instead of rows.
-We can use EROs to put a matrix into upper triangular form, and then finding the determinant is easy: just multiply the diagonal entries together. We just have to keep track of how the determinant is changed by the EROs of types 1 and 2.
-==== Example: using EROs to find the determinant ====
-\begin{align*}\def\vm#1{\begin{vmatrix}#1\end{vmatrix}} \vm{1&3&1&3\\\color{red}4&\color{red}8&\color{red}0&\color{red}{12}\\0&1&3&6\\2&2&1&6}&= \color{red}{4}\vm{1&3&1&\color{blue}3\\1&2&0&\color{blue}3\\0&1&3&\color{blue}6\\2&2&1&\color{blue}6}\\&=4\cdot \color{blue}3\vm{\color{green}1&3&1&1\\\color{red}1&2&0&1\\\color{red}0&1&3&2\\\color{red}2&2&1&2}
-\\&=12\vm{1&3&1&1\\\color{blue}0&\color{blue}{-1}&\color{blue}{-1}&\color{blue}{0}\\\color{blue}0&\color{blue}1&\color{blue}3&\color{blue}2\\0&-4&-1&-0}
-\\&=\color{blue}{-}12\vm{1&3&1&1\\0&\color{green}1&3&2\\0&\color{red}{-1}&{-1}&{0}\\0&\color{red}{-4}&-1&0}
-\\&=-12\vm{1&3&1&1\\0&1&3&2\\0&0&\color{green}2&2\\0&0&\color{red}{11}&8}
-\\&=-12\vm{1&3&1&1\\0&1&3&2\\0&0&2&2\\0&0&0&-3}
-\\&=-12(1)(1)(2)(-3)=72.
-\end{align*}