Differences

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

--- lecture_16 [2016/03/30 12:00] – rupert
+++ lecture_16 [2017/03/30 09:20] (current) – [Chapter 3: Vectors and geometry] rupert
@@ Line 1: / Line 1: @@
+=== Example: $n=2$, general case ===
+If $A=\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\def\vm#1{\begin{vmatrix}#1\end{vmatrix}}\mat{a&b\\c&d}$, then $C=\mat{d&-c\\-b&a}$, so the adjoint of $A$ is $J=C^T=\mat{d&-b\\-c&a}$.
+Recall that $AJ=(\det A)I_2=JA$; we calculated this earlier when we looked at the inverse of a $2\times 2$ matrix. Hence for a $2\times 2$ matrix $A$, if $\det A\ne0$, then $A^{-1}=\frac1{\det A}J$.
 === Example: $n=3$ ===
 If $\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$, then the matrix of signs is $\mat{+&-&+\\-&+&-\\+&-&+}$, so
-\[ C=\mat{
+\[\def\vm#1{\begin{vmatrix}#1\end{vmatrix}} C=\mat{
 \vm{-4&3\\4&-2}&-\vm{-2&3\\5&-2}&\vm{-2&-4\\5&4}\\
 -\vm{1&0\\4&-2}&\vm{3&0\\5&-2}&-\vm{3&1\\5&4}\\
@@ Line 29: / Line 35: @@
 If again we take $A=\mat{3&1&0\\-2&-4&3\\5&4&-2}$, then $J=\mat{-4&2&3\\11&-6&-9\\12&-7&-10}$ and $\det(A)=-1$, so $A$ is invertible and $A^{-1}=\frac1{-1}J=-J=\mat{4&-2&-3\\-11&6&9\\-12&7&10}$.
+==== Example ($n=4$) ====
+Let $A=\mat{1&0&0&0\\1&2&0&0\\1&2&3&0\\1&2&3&4}$.
+Recall that a matrix with a repeated row or a zero row has determinant zero.
+We have \[C=\mat{+\vm{2&0&0\\2&3&0\\2&3&4}&-\vm{1&0&0\\1&3&0\\1&3&4}&+0&-0\\-0&+\vm{1&0&0\\1&3&0\\1&3&4}&-\vm{1&0&0\\1&2&0\\1&2&4}&+0\\+0&-0&+\vm{1&0&0\\1&2&0\\1&2&4}&-\vm{1&0&0\\1&2&0\\1&2&3}\\-0&+0&-0&+\vm{1&0&0\\1&2&0\\1&2&3}}=\mat{24&-12&0&0\\0&12&-8&0\\0&0&8&-6\\0&0&0&6}\]
+so \[J=C^T=\mat{24&0&0&0\\-12&12&0&0\\0&-8&8&0\\0&0&-6&6}.\]
+Since $A$ is lower triangular, its determinant is given by multiplying together its diagonal entries: $\det(A)=1\times 2\times 3\times 4=24$. (Note that even if $A$ was not triangular, $\det A$ can be easily found from the matrix of cofactors $C$ by summing the entries of $A$ multiplied by the entries of $C$ (i.e., the minors) along any row or column.)
+So \[A^{-1}=\frac1{\det A}J = \frac1{24}\mat{24&0&0&0\\-12&12&0&0\\0&-8&8&0\\0&0&-6&6}=\mat{1&0&0&0\\-1/2&1/2&0&0\\0&-1/3&1/3&0\\0&0&-1/4&1/4}.\]
+You should check that this really is the inverse, by checking that $AA^{-1}=I_4=A^{-1}A$.
+===== A more efficient way to find $A^{-1}$ =====
+Given an $n\times n$ matrix $A$, form the $n\times 2n$ matrix
+\[ \def\m#1{\left[
+\begin{array}{@{} c|c {}@} % it does autodetection
+#1
+\end{array}
+\right]}\m{A&I_n}\]
+and use [[EROs]] to put this matrix into [[RREF]]. One of two things can happen:
+  * Either you get a row of the form $[0~0~\dots~0~|~*~*~\dots~*]$ which starts with $n$ zeros. You can then conclude that $A$ is not invertible.
+  * Or you end up with a matrix of the form $\m{I_n&B}$ for some $n\times n$ matrix $B$. You can then conclude that $A$ is invertible, and $A^{-1}=B$.
+==== Examples ====
+  * Consider $A=\def\mat#1{\begin{matrix}#1\end{matrix}}\left[\mat{1&3\\2&6}\right]$. \begin{align*}\m{A&I_2}&=\m{\mat{1&3\\2&6}&\mat{1&0\\0&1}}
+\def\go#1#2{\m{\mat{#1}&\mat{#2}}}
+\def\ar#1{\\[6pt]\xrightarrow{#1}&}
+\ar{R2\to R2-2R1}\go{1&3\\0&0}{1&0\\-2&1}
+\end{align*} Conclusion: $A$ is not invertible.
+  * Consider $A=\left[\mat{1&3\\2&7}\right]$.\begin{align*}\m{A&I_2}&=\m{\mat{1&3\\2&7}&\mat{1&0\\0&1}}
+\ar{R2\to R2-2R1}\go{1&3\\0&1}{1&0\\-2&1}
+\ar{R1\to R1-3R1}\go{1&0\\0&1}{7&-3\\-2&1}
+\end{align*} Conclusion: $A$ is invertible and $A^{-1}=\left[\mat{7&-3\\-2&1}\right]$.
+  * Consider $A=\left[\mat{3&1&0\\-2&-4&3\\5&4&-2}\right]$.\begin{align*}\m{A&I_3}&=\go{3&1&0\\-2&-4&3\\5&4&-2}{1&0&0\\0&1&0\\0&0&1}
+\ar{R1\to R1+R2}
+\go{1&-3&3\\-2&-4&3\\5&4&-2}{1&1&0\\0&1&0\\0&0&1}
+\ar{R2\to R2+2R1,\ R3\to R3-5R1}
+\go{1&-3&3\\0&-10&9\\0&19&-17}{1&1&0\\2&3&0\\-5&-5&1}
+\ar{R3\leftrightarrow R2}
+\go{1&-3&3\\0&19&-17\\0&-10&9}{1&1&0\\-5&-5&1\\2&3&0}
+\ar{R2\to R2+2R3}
+\go{1&-3&3\\0&-1&1\\0&-10&9}{1&1&0\\-1&1&1\\2&3&0}
+\ar{R1\to R1+3R2,\ R3\to R3-10R2}
+\go{1&0&0\\0&-1&1\\0&0&-1}{4&-2&3\\-1&1&1\\12&-7&-10}
+\ar{R2\to R2+R3}
+\go{1&0&0\\0&-1&0\\0&0&-1}{4&-2&3\\11&-6&-9\\12&-7&-10}
+\ar{R2\to -R2,\ R3\to -R3}
+\go{1&0&0\\0&1&0\\0&0&1}{4&-2&3\\-11&6&9\\-12&7&10}
+\end{align*} Conclusion: $A$ is invertible, and $A^{-1}=\left[\mat{4&-2&3\\-11&6&9\\-12&7&10}\right]$.