Matemáticas CCSS · Matrices y Determinantes · Comunidad Valenciana · 2023

Ejercicio resuelto de Matrices y Determinantes · Matemáticas CCSS

Solución

#x27; in math mode at position 69: …^2)]^t,\quad \ $̲ltx|A^2| = \beg…" style="color:#cc0000"> (A^2)^{-1}=\frac{1}{\det(A^2)},[\operatorname{Adj}(A^2)]^t,\quad \ $ltx|A^2| = \begin{vmatrix} 1 & 3 & -2 \\ 2 & 4 & -6 \\ -2 & -1 & 1 \end{vmatrix} = 4 + 36 + 4 - 16 - 6 - 6 = 16 \neq 0 \rightarrow \text{Existe la inversa}

$\mathrm{Adj}(A^2)=\left(\begin{array}{ccc}4& -6& 2\\ -4& 1& -2\\ -3& -2& 1\end{array}\right)\Rightarrow [\mathrm{Adj}(A^2){]}^t=\left(\begin{array}{ccc}4& -4& -3\\ -6& 1& -2\\ 2& -2& 1\end{array}\right)$

$(A^2{)}^{-1}=\frac{1}{16}\left(\begin{array}{ccc}-2& -1& -10\\ 10& -3& 2\\ 6& -5& -2\end{array}\right)$



$2A^{2}X=4B\Rightarrow A^{2}X=2B\Rightarrow (A^2{)}^{-1}{A}^{2}X=(A^2{)}^{-1}(2B)\Rightarrow X=(A^2{)}^{-1}(2B)$


$X=(A^2{)}^{-1}(2B)=\frac{1}{16}\left(\begin{array}{ccc}-2& -1& -10\\ 10& -3& 2\\ 6& -5& -2\end{array}\right)\left(\begin{array}{ccc}6& -4& 2\\ -2& 8& 4\\ 0& 0& 2\end{array}\right)=\frac{1}{16}\left(\begin{array}{ccc}-10& 0& -28\\ 66& -64& 12\\ 46& -64& -12\end{array}\right)=\left(\begin{array}{ccc}-\frac{5}{8}& 0& -\frac{7}{4}\\ \frac{33}{8}& -4& \frac{3}{4}\\ \frac{23}{8}& -4& -\frac{3}{4}\end{array}\right)$