Commit 73d2b6e1 authored by Jim Hefferon's avatar Jim Hefferon

finished gr3

parent 954061eb
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......@@ -3618,75 +3618,6 @@
\end{ans}
\begin{ans}{One.III.2.19}
\begin{exparts}
\partsitem
In the equation
\begin{equation*}
\rho_i=c_1\rho_1+c_2\rho_2+\cdots+c_{i-1}\rho_{i-1}+
c_{i+1}\rho_{i+1}+\cdots+c_m\rho_m
\end{equation*}
we already know that $c_1=0$.
Let $\ell_2$ be the column number of the leading entry of the
second row.
Consider the prior equation on entries in that column.
\begin{equation*}
\rho_{i,\ell_1}=c_2\rho_{2,\ell_2}+\cdots+c_{i-1}\rho_{i-1,\ell_2}
+c_{i+1}\rho_{i+1,\ell_2}+\cdots+c_m\rho_{m,\ell_2}
\end{equation*}
Because $\ell_2$ is the column of the leading entry in the second
row, $\rho_{i,\ell_2}=0$ for $i>2$.
Thus the equation reduces to
\begin{equation*}
0=c_2\rho_{2,\ell_2}+0+\cdots+0
\end{equation*}
and since $\rho_{2,\ell_2}$ is not $0$ we have that $c_2=0$.
\partsitem
In the equation
\begin{equation*}
\rho_i=c_1\rho_1+c_2\rho_2+\cdots+c_{i-1}\rho_{i-1}+
c_{i+1}\rho_{i+1}+\cdots+c_m\rho_m
\end{equation*}
we already know that $0=c_1=c_2=\cdots =c_n$.
Let $\ell_{n+1}$ be the column number of the leading entry of
row~$n+1$.
Consider the above equation on entries in that column.
\begin{equation*}
\rho_{i,\ell_{n+1}}=c_{n+1}\rho_{n+1,\ell_{n+1}}+\cdots
+c_{i-1}\rho_{i-1,\ell_{n+1}}
+c_{i+1}\rho_{i+1,\ell_{n+1}}
+\cdots+c_m\rho_{m,\ell_{n+1}}
\end{equation*}
Because $\ell_{n+1}$ is the column of the leading entry in the
row $n+1$, we have that $\rho_{j,\ell_{n+1}}=0$ for $j>{n+1}$.
Thus the equation reduces to
\begin{equation*}
0=c_{n+1}\rho_{n+1,\ell_{n+1}}+0+\cdots+0
\end{equation*}
and since $\rho_{n+1,\ell_{n+1}}$ is not $0$ we have that $c_{n+1}=0$.
\partsitem
From the prior item in this exercise we know that in the equation
\begin{equation*}
\rho_i=c_1\rho_1+c_2\rho_2+\cdots+c_{i-1}\rho_{i-1}+
c_{i+1}\rho_{i+1}+\cdots+c_m\rho_m
\end{equation*}
we already know that $0=c_1=c_2=\cdots =c_{i-1}$.
Let $\ell_{i}$ be the column number of the leading entry of
row~$i$.
Rewrite the above equation on entries in that column.
\begin{equation*}
\rho_{i,\ell_{i}}=c_{i+1}\rho_{i+1,\ell_{i}}
+\cdots+c_m\rho_{m,\ell_{i}}
\end{equation*}
Because $\ell_{i}$ is the column of the leading entry in the
row $i$, we have that $\rho_{j,\ell_{i}}=0$ for $j>i$.
That makes the right side of the equation sum to $0$, but the
left side is not $0$ since it is the leading entry of the row.
That's the contradiction.
\end{exparts}
\end{ans}
\begin{ans}{One.III.2.20}
We know that $4s+c+10d=8.45$ and that $3s+c+7d=6.30$, and we'd like to
know what $s+c+d$ is.
Fortunately, $s+c+d$ is a linear combination of $4s+c+10d$ and $3s+c+7d$.
......@@ -3713,7 +3644,7 @@
The price paid is \$$2.00$.
\end{ans}
\begin{ans}{One.III.2.21}
\begin{ans}{One.III.2.20}
\begin{enumerate}
\item An easy answer is this:
\begin{equation*}
......@@ -3752,13 +3683,13 @@
\end{enumerate}
\end{ans}
\begin{ans}{One.III.2.22}
\begin{ans}{One.III.2.21}
Define linear systems to be equivalent if their augmented
matrices are row equivalent.
The proof that equivalent systems have the same solution set is easy.
\end{ans}
\begin{ans}{One.III.2.23}
\begin{ans}{One.III.2.22}
\begin{exparts}
\partsitem The three possible row swaps are easy,
as are the three possible rescalings.
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