Question

Inversa matricei A(-1) in M3(IR),uregent!

Answer 1

$A(a)=\begin{pmatrix} a& 1 &1 \\ 1& a & 1\\ 1& 1 & a \end{pmatrix}$

$A( - 1)=\begin{pmatrix} - 1& 1 &1 \\ 1& - 1 & 1\\ 1& 1 & - 1 \end{pmatrix}$

$detA(-1)=\begin{vmatrix} -1& 1 & 1\\ 1& -1 & 1\\ 1& 1& -1 \end{vmatrix}$

$detA(-1)=-1 \times (-1) \times (-1)+1 \times 1 \times 1+1 \times 1 \times 1-1 \times (-1) \times 1-1 \times 1 \times (-1)-(-1) \times 1 \times 1$

$detA( - 1)=-1+1+1+1+1+1$

$detA( - 1) = 4 \: \neq \: 0 = > A( - 1) \:este \: inversabila=>\exists\:{A(-1)}^{-1}$

${A(-1)}^{t}=\begin{pmatrix} - 1& 1& 1\\ 1 & - 1& 1\\ 1& 1 & - 1 \end{pmatrix} = {A(-1)}^{t}=A( - 1)$

$A(-1)_{11}={(-1)}^{1+1}\times\begin{vmatrix} - 1& 1\\ 1 & - 1 \end{vmatrix} = 1 \times (1 - 1) = 1 \times 0 = 0$

$A(-1)_{12}={(-1)}^{1+2}\times\begin{vmatrix} 1&1 \\ 1& - 1 \end{vmatrix} = - 1 \times ( - 2) = 2$

$A(-1)_{13}={(-1)}^{1+3}\times\begin{vmatrix} 1& - 1\\ 1 & 1 \end{vmatrix} = 1 \times( 1 + 1) = 1 \times 2 = 2$

$A(-1)_{21}={(-1)}^{2+1}\times\begin{vmatrix} 1& 1\\ 1& - 1 \end{vmatrix} = - 1 \times ( - 1 - 1) = - 1 \times ( - 2) = 2$

$A(-1)_{22}={(-1)}^{2+2}\times\begin{vmatrix} - 1&1 \\ 1& - 1 \end{vmatrix} = 1 \times (1 - 1) = 1 \times 0 = 0$

$A(-1)_{23}={(-1)}^{2+3}\times\begin{vmatrix} - 1& 1\\ 1 & 1 \end{vmatrix} = - 1 \times ( - 1 - 1) = - 1 \times ( - 2) = 2$

$A(-1)_{31}={(-1)}^{3+1}\times\begin{vmatrix} 1& 1\\ - 1& 1 \end{vmatrix} = 1 \times (1 + 1) = 1 \times 2 = 2$

$A(-1)_{32}={(-1)}^{3+2}\times\begin{vmatrix} - 1&1 \\ 1& 1 \end{vmatrix} = - 1 \times ( - 1 - 1) = - 1 \times ( - 2) = 2$

$A(-1)_{33}={(-1)}^{3+3}\times\begin{vmatrix} - 1& 1\\ 1& - 1 \end{vmatrix} = 1 \times (1 - 1) = 1 \times 0 = 0$

${A(-1)}^{*}=\begin{pmatrix} A(-1)_{11}& A(-1)_{12} &A(-1)_{13} \\ A(-1)_{21} & A(-1)_{22} &A(-1)_{23} \\ A(-1)_{31} &A(-1)_{32} & A(-1)_{33} \end{pmatrix}$

${A(-1)}^{*} = \begin{pmatrix} 0& 2 & 2\\ 2& 0 &2 \\ 2& 2 & 0 \end{pmatrix}$

${A(-1)}^{-1}=\frac{1}{detA(-1)}\times{A(-1)}^{*}$

${A(-1)}^{-1} = \frac{1}{4} \begin{pmatrix} 0& 2 & 2\\ 2& 0 &2 \\ 2& 2 & 0 \end{pmatrix}$

${A(-1)}^{-1} = \begin{pmatrix} 0& \frac{2}{4} & \frac{2}{4} \\ \frac{2}{4} & 0 & \frac{2}{4} \\ \frac{2}{4} & \frac{2}{4} & 0 \end{pmatrix}$

${A(-1)}^{-1} = \begin{pmatrix} 0& \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 \end{pmatrix}$