Determinante Matriz

Usando o teoream fundamental de Laplace:

det M = a1j*A1j + a2j*A2J + a3j*A3j +...+ anj*Anj, ou seja, escolhemos a primeira coluna da matriz:

det M = 0*A11 + 0*A21 + a*A31 + 1*A41, agora vamos calcular A31 2 A41:

A31= (-1)^[³+¹]*D31 = (-1)^4* det ⌈ a b 1 = 1*a = a;

                                                       1 0 0

                                                       b a 0⌋

ou seja A31= a;

agora calculando A41= (-1)^[4+1]*D41 = (-1)^5*D41 = -1*  ⌈ a b 1 = -1*-b^2 = b^2;

                                                                                           1 0 0

                                                                                           a 0 b⌉

por fim;

det M = a*A31 + 1*A41 = a*a + b^2 = a^2 + b^2

O determinante de uma matriz qualquer é:

\(\Longrightarrow \det M = \det \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \\ \end{bmatrix}\)

\(\Longrightarrow \det M = a_{11} \cdot \det \begin{bmatrix} a_{22} & a_{23} & a_{24} \\ a_{32} & a_{33} & a_{34} \\ a_{42} & a_{43} & a_{44} \\ \end{bmatrix} - a_{12} \cdot \det \begin{bmatrix} a_{21} & a_{23} & a_{24} \\ a_{31} & a_{33} & a_{34} \\ a_{41} & a_{43} & a_{44} \\ \end{bmatrix} + a_{13} \cdot \det \begin{bmatrix} a_{21} & a_{22} & a_{24} \\ a_{31} & a_{32} & a_{34} \\ a_{41} & a_{42} & a_{44} \\ \end{bmatrix} - a_{14} \cdot \det \begin{bmatrix} a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ a_{41} & a_{42} & a_{43} \\ \end{bmatrix}\)

Portanto, o determinante da matriz \(M\) é:

\(\Longrightarrow \det M = \det \begin{bmatrix} 0 & a & b & 1 \\ 0 & 1 & 0 & 0 \\ a & a & 0 & b \\ 1 & b & a & 0 \end{bmatrix}\)

\(\Longrightarrow \det M =0 \cdot \det \begin{bmatrix} 1 & 0 & 0 \\ a & 0 & b \\ b & a & 0 \end{bmatrix} - a \cdot \det \begin{bmatrix} 0 & 0 & 0 \\ a & 0 & b \\ 1 & a & 0 \end{bmatrix} + b \cdot \det \begin{bmatrix} 0 & 1 & 0 \\ a & a & b \\ 1 & b & 0 \end{bmatrix} - 1 \cdot \det \begin{bmatrix} 0 & 1 & 0 \\ a & a & 0 \\ 1 & b & a \end{bmatrix}\)

\(\Longrightarrow \det M =0 - a \cdot 0 + b \cdot \Bigg ( -1 \cdot \det \begin{bmatrix} a & b \\ 1 & 0 \end{bmatrix} \Bigg )- 1 \cdot \Bigg ( -1 \cdot \det \begin{bmatrix} a & 0 \\ 1 & a \end{bmatrix} \bigg )\)

\(\Longrightarrow \det M =b \cdot (-1) \cdot \Bigg ( a \cdot 0 - b \cdot 1 \Bigg )- 1 \cdot (-1) \cdot \Bigg ( a \cdot a - 0 \cdot 1 \bigg )\)

\(\Longrightarrow \det M =-b \cdot ( 0 - b )+( a^2 - 0 )\)

\(\Longrightarrow \det M =b^2 + a^2 \)

Resposta correta: \(\fbox {$ a^2 + b^2 $}\)