Question

ME AJUDEM POR FAVOR.......
dada a matriz M= $\left[\begin{array}{ccc}2&0&1\\0&1&3\\2&1&5\end{array}\right]$ , calcular:

a) o determinante de M;

b) a matriz dos cofatores de M;

c) a matriz adjunta de M;

d) a matriz inversa de M;

Answer 1

A)

O determinante dessa matriz iremos calcular por "Sarrus"

$\\ D = det \left[\begin{array}{ccc}2&0&1\\0&1&3\\2&1&5\end{array}\right] \left\begin{array}{ccc}2&0\\0&1\\2&1\end{array}\right] \\ \\ D = (2*1*5)+(0*3*2)+(1*0*1)- [ (2*1*1)+(1*3*2)+0] \\ \\ D = 10 - [ 2+6] \\ \\ D = 2$
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B)

$\\ a_{11} = det\left[\begin{array}{ccc}1&3\\1&5\end{array}\right] = 1*5-3*1 =2 \\ \\ \\ a_{12} = -det\left[\begin{array}{ccc}0&3\\2&5\end{array}\right] = -(0*5-2*3) = 6 \\ \\ \\ a_{13} =det\left[\begin{array}{ccc}0&1\\2&1\end{array}\right] = 0*1-2*1 = -2 \\ \\ \\ a_{21} = -det\left[\begin{array}{ccc}0&1\\1&5\end{array}\right] = -(0*5-1*1) = 1 \\ \\ \\ a_{22} = det\left[\begin{array}{ccc}2&1\\2&5\end{array}\right] =2*5-2*1 = 8$

$\\ a_{23} = -det\left[\begin{array}{ccc}2&0\\2&1\end{array}\right] = -(2*1-2*0) =- 2 \\ \\ \\ a_{31} = det\left[\begin{array}{ccc}0&1\\1&3\end{array}\right] = 0*3-1*1 = -1 \\ \\ \\ a_{32} = -det\left[\begin{array}{ccc}2&1\\0&3\end{array}\right] =-(2*3-1*0) = -6 \\ \\ \\ a_{33} = det\left[\begin{array}{ccc}2&0\\0&1\end{array}\right] = 2*1-0*0 =2$

A matriz das cofatoras será:

$Cof_{M} = \left[\begin{array}{ccc}2&6&-2\\1&8&-2\\-1&-6&2\end{array}\right]$
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C)

A matriz adjunta é a transposta da matriz cofatora...

Então:

$\\ Adj_{M} = (Cof_{M} )^T \\ \\ Adj_{M} = \left[\begin{array}{ccc}2&6&-2\\1&8&-2\\-1&-6&2\end{array}\right] ^T \\ \\ \\ Adj_{M} = \left[\begin{array}{ccc}2&1&-1\\6&8&-6\\-2&-2&2\end{array}\right]$
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A matriz inversa será a divisão da matriz adjunta de "M" pelo determinante de M

$\\ M^-^1 = \frac{ Adj_{M} }{D} \\ \\ \\ M^-^1 = \frac{ \left[\begin{array}{ccc}2&1&-1\\6&8&-6\\-2&-2&2\end{array}\right] }{2} \\ \\ \\ M^-^1 = \left[\begin{array}{ccc}1& \frac{1}{2} &- \frac{1}{2} \\3&4&-3\\-1&-1&1\end{array}\right]$