Question

resposta certa por favor obrigada​

Answer 1

Resposta correta:

$\quad {\because \quad \boxed{\boxed{\begin{array}{lr}\red{ {\space} \det A = 2 \quad {;} \quad \det A^{2} = 4 {\space} }\end{array}}}}$

$\bf\large\gray{\underline{\qquad \qquad\qquad \qquad \qquad \qquad \qquad \quad }}$

$\quad\quad\huge\mid{\boxed{\bf{\blue{ Matem\acute{a}tica }}}\mid}$

Solução passo a passo

$~$

Matriz de Ordem 2 ou matriz 2 x 2.

ex.:

$\boxed{~~A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} \rightarrow\det A=a_{11}\:\cdot\:a_{22}-a_{21}\:\cdot\:a_{12}~~}$

Como vimos acima, para resolver, primeiro multiplica os valores constantes nas diagonais, um principal e um secundário, em seguida, subtrair os resultados obtidos a partir desta multiplicação.

$\\ \\ \large\underline{ \overline{\boxed{\begin{array}{clr}\\ \displaystyle{ {~}A=\begin{bmatrix}2&4\\1&3\end{bmatrix} {~} }\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\end{array}}}}\\ \\ \tiny \vdash \text{forma \ simplificada} \dashv$

Resolvendo $\underline{\gray{\det\:A}}$

$~$

Usando a fórmula:

$\boxed{\qquad\qquad{\det\begin{array}{|lr|}a&b\\c&d\end{array}=ad- bc}\qquad\qquad}$

${\quad}{\space}\large\gray{ \downarrow }{\quad}{\quad}\displaystyle\begin{aligned} =\det\begin{array}{|lr|} 2&4\\ 1&3 \end{array} \end{aligned}$

${\quad}{\space}\large\gray{ \downarrow }{\quad}{\quad}\displaystyle\begin{aligned}=2 \: \cdot \: 3 - 4\: \cdot \: 1\end{aligned}$

${\quad}{\space}\large\gray{ \downarrow }{\quad} {\quad}\displaystyle\begin{aligned} = 6 -4\: \cdot \: 1\end{aligned}$

${\quad}{\space}\large\gray{ \downarrow }{\quad} {\quad}\displaystyle\begin{aligned} = 6 - 4 \end{aligned}$

$\begin{gathered}{{\quad}{\space}\large\gray{ \uparrow }{\quad} {\quad} \displaystyle\text{ \begin{aligned}\boxed{~~\orange{= 2}~~} \end{aligned} }}\\ \\ \end{gathered}$

Resolvendo $\underline{\gray{\det\:A^{2}}}$

${\quad}{\space}\large\gray{ \downarrow }{\quad} {\quad} \displaystyle\begin{aligned} =\begin{bmatrix}2&4\\ 1&3\end{bmatrix}^{2} \end{aligned}$

${\quad}{\space}\large\gray{ \downarrow }{\quad} {\quad} \displaystyle\begin{aligned} = \begin{bmatrix}2&4\\1&3\end{bmatrix}\begin{bmatrix}2&4\\1&3\end{bmatrix}\end{aligned}$

${\quad}{\space}\large\gray{ \downarrow }{\quad} {\quad} \displaystyle\begin{aligned} =\begin{bmatrix}4 + 4&2 \: \cdot \: 4 + 4 \:\cdot \: 3\\ 1 \: \cdot \: 2 + 3 \: \cdot \: 1&1 \: \cdot \: 4 + 3 \: \cdot \: 3\end{bmatrix}\end{aligned}$

${\quad}{\space} \large\gray{ \downarrow }{\quad} {\quad} \displaystyle\begin{aligned} =\begin{bmatrix}8&20 \\ 1 \: \cdot \: 2 + 3 \: \cdot \: 1&1 \: \cdot \: 4 + 3 \: \cdot \: 3\end{bmatrix}\end{aligned}$

${\quad}{\space}\large\gray{ \downarrow }{\quad} {\quad} \displaystyle\begin{aligned} =\begin{bmatrix}8&20 \\ 5&4 + 3 \: \cdot \: 3\end{bmatrix}\end{aligned}$

${\quad}{\space} \large\gray{ \downarrow }{\quad} {\quad} \displaystyle\begin{aligned} =\begin{bmatrix}8&20 \\ 5&13\end{bmatrix}\end{aligned}$

Usando a fórmula:

$\boxed{\qquad\qquad{\det\begin{array}{|lr|}a&b\\c&d\end{array}=ad- bc}\qquad\qquad}$

${\quad}{\space}\large\gray{ \downarrow }{\quad}{\quad} \displaystyle\begin{aligned} = \det \begin{array}{|lr|}8&20\\5&13\end{array}\end{aligned}$

${\quad}{\space}\large\gray{ \downarrow }{\quad} {\quad} \displaystyle\begin{aligned} = 104 - 20 \: \cdot \: 5 \end{aligned}$

$\begin{gathered}{{\quad}{\space}\large\gray{ \uparrow }{\quad} {\quad} \displaystyle\text{ \begin{aligned}\boxed{~~\orange{= 4}~~} \end{aligned} }}\\ \\\end{gathered}$

Observações:

1. $\sf O\:determinante\:de\:uma\: matriz\:quadrada \\ \sf \acute{e}\:um\: \acute{u}nico\:valor\:num\acute{e}rico \:ou\:um \\ \sf valor\:de\:resumo \: que\:representa\:todo\:o \\ \sf conjunto\:de\:elementos\:da\: matriz{.}\\ ~ \\ \bf \large\gray{\underline{\qquad \qquad\qquad \qquad \qquad \qquad \qquad \quad \quad}} \\ ~ \\ ~$
2. $\sf Como\:vimos{,}\:o\: determinante\:para\:a\:matriz\: \\ \sf quadrada\:de\:ordem\:2x2\:pode\: ser\:facilmente \\ \sf calculado \: usando\:a\:f\acute{o}rmula{.}$

Solução:

$\quad {\therefore \quad\boxed{\boxed{\begin{array}{lr}\red{ {\space} \det A = 2\quad {;} \quad \det A^{2} = 4 {\space} }\end{array}}}}$

$\quad \text{\underline{Att.}}$

${\huge\boxed { {\bf{M}}}\boxed { \red {\bf{y}}} \boxed { \blue {\bf{t}}} \boxed { \gray{\bf{h}}} \boxed { \red {\bf{}}} \boxed { \orange {\bf{P}}} \boxed {\bf{i}}}$

$\quad {\vdots\quad\large\boxed {\boxed{02:10h}~29.10.21}}$

$\bf\large\gray{\underline{\qquad \qquad\qquad \qquad \qquad \qquad \qquad \quad }}$

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