Question

Determinante!!!!

Se A = (aij)3x3 tal que aij=i-j, calcule det A e det A

*obs: após o último A tem um t elevado.*

Answer 1

Montar matriz:
$\displaystyle A_{(3x3)}\longrightarrow\{a_{ij}=i-j\}\rightarrow \left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right] \longrightarrow\\\\ \left[\begin{array}{ccc}(1-1)&(1-2)&(1-3)\\(2-1)&(2-2)&(2-3)\\(3-1)&(3-2)&(3-3)\end{array}\right] = \left[\begin{array}{ccc}0&-1&-2\\1&0&-1\\-2&1&0\end{array}\right] =A\\\\ A^t= \left[\begin{array}{ccc}0&1&-2\\-1&0&1\\-2&-1&0\end{array}\right]$
Calcular determinante de A:
$\displaystyle \det(A)= \left|\begin{array}{ccccc}0&-1&-2&0&-1\\1&0&-1&1&0\\-2&1&0&-2&1\end{array}\right|=\\ \ [ (0.0.0)+(-1.-1.-2)+(-2.1.1)]-[(-2.\ 0.-2)+(0.-1.1)+\\(-1.1.0)]=[0-2-2]-[0+0+0]=\boxed{-4}$


Calcular determinante de A transposta:
$\displaystyle \det(A^t)= \left|\begin{array}{ccccc}0&1&-2&0&1\\-1&0&1&-1&0\\-2&-1&0&-2&-1\end{array}\right|=\\ \ [ (0.0.0)+(1.1.-2)+(-2.-1.-1) ]-[(-2.0.-2)+(0.1.-1)\\(1.-1.0)]=[0-2-2]-[0+0+0]=\boxed{-4}$


$\boxed{\det (A)=-4}\\\\ \boxed{\det(A^t)=-4}$