Differences

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

--- lecture_16 [2016/03/30 12:02] – 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: $n=3$ ===
@@ Line 30: / Line 36: @@
 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 $A^{-1}$ =====
@@ Line 70: / Line 87: @@
 \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 $A^{-1}=\left[\mat{4&-2&3\\-11&6&9\\-12&7&10}\right]$.