### 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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