Question

2) Dada a matriz A =[ -302 -1 4 3
211] determine:
a) cof(a12)
b) cof(a31)
c) cof(a22)
d) cof(a13)
e) cof(a23)
f) cof(233)​

Answer 1

Temos a matriz:

$\begin{array}{l}\sf A=\begin{bmatrix}\sf \!\! -3&\sf0&\sf2 \\ \sf \!\! -1&\sf4&\sf3 \\ \sf2&\sf1&\sf1 \end{bmatrix}\end{array}$

Para encontrar o cofator, aplicaremos a fórmula:

$\boxed{\boxed{\begin{array}{l}\\ \quad\sf cof(a_{ij})=(-1)^{i+j}\cdot D_{ij}\quad \\\\ \end{array}}}$

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Este Dij é o determinante de uma matriz 2x2 formada a partir da eliminação  da linha e da coluna de um elemento aij

• Lembrando que i = linha, j = coluna

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A) cof(a12)

$\begin{array}{l}\sf cof(a_{12})=(-1)^{1+2}\cdot D_{12}\end{array}$

a12 => linha 1, coluna 2. Logo eliminando a linha e a coluna do elemento 0 formaremos uma nova matriz, assim calcule seu determinante

$\begin{array}{l}\sf cof(a_{12})=(-1)^{3}\cdot \begin{vmatrix}\sf -1&\sf3 \\ \sf2&\sf1\end{vmatrix} \\\\ \sf cof(a_{12})=-1\cdot [(-1\cdot1)-(3\cdot2)] \\\\ \sf cof(a_{12})=-1\cdot[-1-6] \\\\ \sf cof(a_{12})=-1\cdot[-7] \\\\ \!\boxed{\sf cof(a_{12})=7}\end{array}$

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B) cof(a31)

$\begin{array}{l}\sf cof(a_{31})=(-1)^{3+1}\cdot D_{31}\end{array}$

a31 => linha 3, coluna 1

$\begin{array}{l}\sf cof(a_{31})=(-1)^{4}\cdot \begin{vmatrix}\sf 0&\sf2 \\ \sf4&\sf3\end{vmatrix} \\\\ \sf cof(a_{31})=1\cdot [(0\cdot3)-(2\cdot4)] \\\\ \sf cof(a_{31})=0-8 \\\\ \!\boxed{\sf cof(a_{31})=-8}\end{array}$

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C) cof(a22)

$\begin{array}{l}\sf cof(a_{22})=(-1)^{2+2}\cdot D_{22} \end{array}$

a22 => linha 2, coluna 2

$\begin{array}{l}\sf cof(a_{22})=(-1)^{4}\cdot \begin{vmatrix}\sf -3&\sf2 \\ \sf2&\sf1\end{vmatrix} \\\\ \sf cof(a_{22})=1\cdot [(-3\cdot1)-(2\cdot2)] \\\\ \sf cof(a_{22})=-3-4 \\\\ \!\boxed{\sf cof(a_{22})=-7}\end{array}$

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D) cof(a13)

$\begin{array}{l}\sf cof(a_{13})=(-1)^{1+3}\cdot D_{13} \end{array}$

a13 => linha 1, coluna 3

$\begin{array}{l}\sf cof(a_{13})=(-1)^{4}\cdot \begin{vmatrix}\sf -1&\sf4 \\ \sf2&\sf1\end{vmatrix} \\\\ \sf cof(a_{13})=1\cdot [(-1\cdot1)-(4\cdot2)] \\\\ \sf cof(a_{13})=-1-8 \\\\ \!\boxed{\sf cof(a_{13})=-9}\end{array}$

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E) cof(a23)

$\begin{array}{l}\sf cof(a_{23})=(-1)^{2+3}\cdot D_{23} \end{array}$

a23 => linha 2, coluna 3

$\begin{array}{l}\sf cof(a_{23})=(-1)^{5}\cdot \begin{vmatrix}\sf -3&\sf0 \\ \sf2&\sf1\end{vmatrix} \\\\ \sf cof(a_{23})=-1\cdot [(-3\cdot1)-(0\cdot2)] \\\\ \sf cof(a_{23})=-1\cdot[-3-0] \\\\ \sf cof(a_{23})=-1\cdot[-3] \\\\ \!\boxed{\sf cof(a_{23})=3}\end{array}$

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F) cof(a33)

$\begin{array}{l}\sf cof(a_{33})=(-1)^{3+3}\cdot D_{33} \end{array}$

a33 => linha 3, coluna 3

$\begin{array}{l}\sf cof(a_{33})=(-1)^{6}\cdot \begin{vmatrix}\sf -3&\sf0 \\ \sf -1&\sf4\end{vmatrix} \\\\ \sf cof(a_{33})=1\cdot [(-3\cdot4)-(0\cdot(-1))] \\\\ \sf cof(a_{33})=-12-0 \\\\ \!\boxed{\sf cof(a_{33})=-12}\end{array}$

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Att. Nasgovaskov

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