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--- lecture_15 [2015/03/24 12:08] – [Corollary] rupert
+++ lecture_15 [2017/03/28 11:16] (current) – [Definition: the adjoint of a square matrix] rupert
@@ Line 1: / Line 1: @@
-==== Step 4: the determinant of an $n\times n$ matrix ====
+==== Theorem: row/column operations and determinants ====
+Let $A$ be an $n\times n$ matrix, let $c$ be a scalar and let $i\ne j$.
-===Definition===
+$A_{Ri\to x}$ means $A$ but with row $i$ replaced by $x$.
-{{page>determinant of an nxn matrix}}
+  - 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 $\det(A)$)
-===Example===
+  * Also, these properties all hold if you change "row" into "column" throughout.
-\begin{align*}
+==== Corollary ====
-\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}
-\vm{\color{red}1&\color{red}0&\color{red}2&\color{red}3\\0&2&1&-1\\2&0&0&1\\3&0&4&2} &= \color{red}1\vm{\color{blue}2&\color{blue}1&\color{blue}-1\\0&0&1\\0&4&2}-\color{red}0\vm{0&1&-1\\2&0&1\\3&4&2}+\color{red}2\vm{\color{orange}0&\color{orange}2&\color{orange}{-1}\\2&0&1\\3&0&2}-\color{red}3\vm{\color{purple}0&\color{purple}2&\color{purple}1\\2&0&0\\3&0&4}\\
-&= 1\left(\color{blue}2\vm{0&1\\4&2}-\color{blue}1\vm{0&1\\0&2}\color{blue}{-1}\vm{0&0\\0&4}\right)-0+2\left(\color{orange}0-\color{orange}{2}\vm{2&1\\3&2}\color{orange}{-1}\vm{2&0\\3&0}\right)-3\left(\color{purple}0-\color{purple}2\vm{2&0\\3&4}+\color{purple}1\vm{2&0\\3&0}\right)\\
-&=1(2(-4)-0-0)+2(-2(1)-0)-3(-2(8)+0)\\
-&=-8-4+48\\
-&=36.
-\end{align*}
-==== Theorem: Laplace expansion along any row or column gives the determinant ====
+If an $n\times n$ matrix $A$ has two equal rows (or columns), then $\det(A)=0$, and $A$ is not invertible.
-  - For any fixed $i$: $\det(A)=a_{i1}C_{i1}+a_{i2}C_{i2}+\dots+a_{in}C_{in}$ (Laplace expansion along row $i$)
+=== Proof ===
-  - For any fixed $j$: $\det(A)=a_{1j}C_{1j}+a_{2j}C_{2j}+\dots+a_{nj}C_{nj}$ (Laplace expansion along column $j$)
-=== Example ===
-We can make life easier by choosing expansion rows or columns with lots of zeros, if possible. Let's redo the previous example with this in mind:
+If $A$ has two equal rows, row $i$ and row $j$, then $A=A_{Ri\leftrightarrow Rj}$
+So $\det(A)=\det(A_{Ri\leftrightarrow Rj}) = -\det(A)$, so $2\det(A)=0$, so $\det(A)=0$.
-\begin{align*}
+If $A$ has two equal columns, then $A^T$ has two equal rows, so $\det(A)=\det(A^T)=0$.
-\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}
-\vm{1&\color{red}0&2&3\\0&\color{red}2&1&-1\\2&\color{red}0&0&1\\3&\color{red}0&4&2} &=
--\color{red}0+\color{red}2\vm{1&2&3\\\color{purple}2&\color{purple}0&\color{purple}1\\3&4&2}-\color{red}0+\color{red}0\\
-&=2\left(-\color{purple}2\vm{2&3\\4&2}+\color{purple}0-\color{purple}1\vm{1&2\\3&4}\right)\\
-&=2(-2(-8)-(-2))\\
-&=36.
-\end{align*}
-==== Definition: upper triangular matrices ====
+In either case, $\det(A)=0$. So $A$ is not invertible.■
-An $n\times n$ matrix $A$ is **upper triangular** if all the entries below the main diagonal are zero.
+=== 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*}
-==== Definition: diagonal matrices ====
+==== Corollary ===
-An $n\times n$ matrix $A$ is **diagonal** if the only non-zero entries are on its main diagonal.
+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 $\det(A)=0$.
-==== Corollary: the determinant of upper triangular matrices and diagonal matrices ====
-  - The determinant of an upper triangular $n\times n$ matrix is the product of its diagonal entries: $\det(A)=a_{11}a_{22}\dots a_{nn}$.
-  - The determinant of an $n\times n$ diagonal matrix is the product of its diagonal entries: $\det(A)=a_{11}a_{22}\dots a_{nn}$.
 === Proof ===
-  - This is true for $n=1$, trivially. For $n>1$, assume inductively that it is true for $(n-1)\times (n-1)$ matrices and use the Laplace expansion of an upper triangular $n\times n$ matrix $A$ along the first column of $A$ to see that $\det(A)=a_{11}C_{11}+0+\dots+0=a_{11}C_{11}$. Now $C_{11}$ is the determinant of the $(n-1)\times (n-1)$ matrix formed by removing the first row and and column of $A$, and this matrix is upper triangular with diagonal entries $a_{22},a_{33},\dots,a_{nn}$. By our inductive assumption, we have $C_{11}=a_{22}a_{33}\dots a_{nn}$. So $\det(A)=a_{11}C_{11}=a_{11}a_{22}a_{33}\dots a_{nn}$ as desired.
+Note that $\row_i(A)-c \cdot\row_j(A)=0$. So $A_{Ri\to Ri-c\,Rj}$ has a zero row, and by Laplace expansion along this row we obtain $\det(A_{Ri\to Ri-c\,Rj})=0$.  So $\det(A)=\det(A_{Ri\to Ri-c\,Rj})=0$.■
-  - Any diagonal matrix is upper triangular, so this is a special case of statement 1. ■
-==== Examples ====
+==== The effect of EROs on the determinant ====
-  - For any $n$, we have $\det(I_n)=1\cdot 1\cdots 1 = 1$.
+We have now seen the effect of each of the three types of [[ERO]] on the determinant of a matrix:
-  - For any $n$, we have $\det(5I_n)=5^n$.
-  - $\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\vm{1&9&43&23434&4&132\\0&3&43&2&-1423&-12\\0&0&7&19&23&132\\0&0&0&2&0&0\\0&0&0&0&-1&-903\\0&0&0&0&0&6}=1\cdot3\cdot7\cdot2\cdot(-1)\cdot6 = 252$.
+  - swapping two rows of the matrix multiplies the determinant by $-1$. By swapping rows repeatedly, we are able to shuffle the rows in an arbitrary fashion, and the determinant will either remain unchanged (if we used an even number of swaps) or be multiplied by $-1$ (if we used an odd number of swaps).
+  - 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.
-==== Theorem: important properties of the determinant ====
+Moreover, since $\det(A)=\det(A^T)$, this all applies equally to columns instead of rows.
-Let $A$ be an $n\times n$ matrix.
+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.
-  - $A$ is invertible if and only if $\det(A)\ne0$.
+==== Example: using EROs to find the determinant ====
-  - If $A'$ is the same as $A$, except with two rows swapped, then $\det(A')=-\det(A)$.
-  - 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 $A'$ 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 ====
+\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*}
-If an $n\times n$ matrix $A$ has two equal rows (or columns), then $\det(A)=0$, and $A$ is not invertible.
+===== Finding the inverse of an invertible $n\times n$ matrix =====
-=== Proof ===
+==== Definition: the adjoint of a square matrix ====
-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.
+{{page>adjoint}}
-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. ■
+=== Example: $n=2$ ===
-=== Examples ===
+If $A=\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\mat{1&2\\3&4}$, then $C_{11}=+4$, $C_{12}=-3$, $C_{21}=-2$, $C_{22}=+1$. So the matrix of cofactors is $C=\mat{4&-3\\-2&1}$, so the adjoint of $A$ is $J=C^T=\mat{4&-2\\-3&1}$.
-  * 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 \[\vm{ 2&4&6&10\\5&0&0&-10\\9&0&81&99\\1&2&3&4} = 2\cdot 5\cdot 9\cdot 2\cdot 3 \vm{ 1&1&1&5\\1&0&0&-2\\1&0&3&11\\1&1&1&4} = 2\cdot 5\cdot 9\cdot 2\cdot 3 \cdot 3 = 1620.\]