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Llev sokucions d'exercicis desapareguts (#18)

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Xavier B. 1 week ago
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      10-solucions.tex
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10-solucions.tex View File

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\begin{itemize}
\item[\ref{exercici:det-1}] \begin{enumerate*}[label=\emph{\alph*})] \item $13$, \item $73$ \item $-12$ \item $18$ \item $-256$ \end{enumerate*}

\item[\ref{exercici:det-7}] \begin{enumerate*}[label=\emph{\alph*})] \item $256$ \item $-32$ \item $296$ \item $0$ \item $-200$ \end{enumerate*}

\item[\ref{exercici:det-2}] \begin{enumerate*}[label=\emph{\alph*})] \item $-\left\vert \begin{array}{cc}
l & m \\
n & p%
\end{array}\right\vert = - (-13) = 13$, \item $36 \cdot \left\vert\begin{array}{cc}
n & p \\
l & m%
\end{array}\right\vert = 36 \cdot 13 = 468$ (aplicant \autoref{nota:extraccio-factor-comu} de \autoref{seccio:propietats-dels-determinants} dues vegades i l'apartat anterior), \item $4 \cdot (-13) = -52$ (aplicant \autoref{nota:extraccio-factor-comu} de \autoref{seccio:propietats-dels-determinants}) \end{enumerate*}

\item[\ref{exercici:det-3}] \begin{enumerate*}[label=\emph{\alph*})] \item sí (\autoref{item:propietat-3} de \autoref{seccio:propietats-dels-determinants}), \item sí (en els dos casos, $3 \cdot 3 \cdot \left\vert
\begin{array}{cc}
1 & 2 \\
3 & 3%
\end{array}
\right\vert$) \item no ($3 \cdot 3 \cdot \left\vert
\begin{array}{cc}
1 & 2 \\
3 & 3%
\end{array}
\right\vert \neq 3 \cdot \left\vert\begin{array}{cc}
1 & 2 \\
3 & 3%
\end{array}
\right\vert$)
\end{enumerate*}

\item[\ref{exercici:det-4}]\begin{enumerate*}[label=\emph{\alph*})] \item $-\left\vert
\begin{array}{cc}
m & p \\
n & q%
\end{array}
\right\vert = - (mq - np) = -\left\vert
\begin{array}{cc}
m & n \\
p & q%
\end{array}
\right\vert = 5$\footnote{Es podria haver vist aquest resultat usant que els determinants d'una matriu quadrada i de la seva transposta són iguals (vegi's \autoref{prop:determinant-matriu-transposta} de \autoref{subseccio:propietats-matrius-determinants}).}, \item $\left\vert
\begin{array}{cc}
m & p \\
n & q%
\end{array}
\right\vert + \left\vert
\begin{array}{cc}
3n & 3q \\
n & q%
\end{array}
\right\vert$ $= -5 + 3 \cdot 0 = -5$, \item $-3 \cdot \left\vert
\begin{array}{cc}
n & m \\
q & p%
\end{array}
\right\vert = 3 \left\vert
\begin{array}{cc}
m & n \\
p & q%
\end{array}
\right\vert = -15$ \item $2 \cdot \left\vert
\begin{array}{cc}
p & m \\
q & n%
\end{array}
\right\vert = 2 \cdot 5 = 10$, \item $1/m \cdot \left\vert
\begin{array}{cc}
m & n \\
mp & mq%
\end{array}
\right\vert = \left\vert
\begin{array}{cc}
m & n \\
p & q%
\end{array}
\right\vert = -5$, \item $5 \cdot \left\vert
\begin{array}{cc}
m & m \\
p & p%
\end{array}
\right\vert = 0$
\end{enumerate*}

\item[\ref{exercici:det-9}]\begin{enumerate}[label=\emph{\alph*})]
\item \begin{align*}
\left\vert
\begin{array}{ccc}
1 & 1 & 1 \\
a+7 & b+7 & c+7\\
\frac{x}{2} & \frac{y}{2} & \frac{z}{2}
\end{array}
\right\vert & = \frac{1}{2} \left\vert
\begin{array}{ccc}
1 & 1 & 1 \\
a+7 & b+7 & c+7\\
x & y & z
\end{array}
\right\vert\\
& = \frac{1}{2} \cdot \left(\left\vert
\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c\\
x & y & z
\end{array}
\right\vert + \left\vert
\begin{array}{ccc}
1 & 1 & 1 \\
7 & 7 & 7\\
x & y & z
\end{array}
\right\vert\right)\\
& = \frac{1}{2} \cdot (5 + 0) = \frac{5}{2}
\end{align*}

\item \begin{equation*}
\left\vert
\begin{array}{ccc}
a & b & c\\
x & y & z\\
1 & 1 & 1
\end{array}
\right\vert = - \left\vert
\begin{array}{ccc}
1 & 1 & 1\\
x & y & z\\
a & b & c
\end{array}
\right\vert = \left\vert
\begin{array}{ccc}
1 & 1 & 1\\
a & b & c\\
x & y & z
\end{array}\right\vert = 5
\end{equation*}

\item \begin{align*}
\left\vert
\begin{array}{ccc}
1-x & 1-y & 1-z\\
a+2x & b+2y & c+2z\\
2x & 2y & 2z
\end{array}\right\vert = & \left\vert
\begin{array}{ccc}
1-x & 1-y & 1-z\\
a & b & c\\
2x & 2y & 2z
\end{array}\right\vert\\
= 2 \left\vert
\begin{array}{ccc}
1-x & 1-y & 1-z\\
a & b & c\\
x & y & z
\end{array}\right\vert & = 2\left\vert
\begin{array}{ccc}
1 & 1 & 1\\
a & b & c\\
x & y & z
\end{array}\right\vert = 10
\end{align*}

\end{enumerate}


\item[\ref{exercici:det-5}]\begin{enumerate*}[label=\emph{\alph*})] \item Aplicant la regla de Sarrus, tenim que el determinant és igual a $x^3 - x = x(x^2 -1)=0$ si, i només si, $x=0$ o $x=\pm 1$, \item Fent Sarrus, tenim que $-2a + 2 = 0$ si, i només si, $a=1$ \item $\left\vert
\begin{array}{ccc}
a & 1 & -1 \\
0 & a+6 & 3\\
a & 2 & 0
\end{array}
\right\vert$ $+$ $\left\vert
\begin{array}{ccc}
-1 & 1 & -1 \\
0 & a+6 & 3\\
-1 & 2 & 0
\end{array}
\right\vert = (a-1) \Delta$, on $\Delta = \left\vert
\begin{array}{ccc}
1 & 1 & -1 \\
0 & a+6 & 3\\
1 & 2 & 0
\end{array}
\right\vert = (a+3)$. Per tant, $a = 1$ o $a=-3$.
\end{enumerate*}

\item[\ref{exercici:det-6}] Aplicant la regla de Sarrus, obtenim que el determinant val $x-x^3 = x (1-x^2)$. Per tant, el determinant val zero si, i només si, $x=0$ o bé $x= \pm 1$.



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