Question

determine quando possível, as matrizes inversas de :

Answer 1

a)

$A= \left[\begin{array}{ccc}-1&5\\3&-15\end{array}\right]\\\\ detA=(-1).(-15)-5.3\\ detA=15-15=0 \to\ \nexists A^{-1}$

b)

$B=\left[\begin{array}{ccc}-1&2&1\\2&3&5\\1&-1&2\end{array}\right]\\\\ detB=-6+10-2-3-5-8\\ detB=-14\ \to\ \exists B^{-1}\ \to\ B.B^{-1}=I_3$

$\left[\begin{array}{ccc}-1&2&1\\2&3&5\\1&-1&2\end{array}\right].\left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right]=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\\\\ \left[\begin{array}{ccc}-a+2d+g&-b+2e+h&-c+2f+i\\2a+3d+5g&2b+3e+5h&2c+3f+5i\\a-d+2g&b-e+2h&c-f+2i\end{array}\right]=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

$-a+2d+g=1\\ 2a+3d+5g=0\\ a-d+2g=0\\\\ X= \left[\begin{array}{ccc}-1&2&1\\2&3&5\\1&-1&2\end{array}\right]=\left[\begin{array}{ccc}-14\end{array}\right]\\\\ X_a=\left[\begin{array}{ccc}1&2&1\\0&3&5\\0&-1&2\end{array}\right]=\left[\begin{array}{ccc}11\end{array}\right]\\\\ X_d=\left[\begin{array}{ccc}-1&1&1\\2&0&5\\1&0&2\end{array}\right]=\left[\begin{array}{ccc}1\end{array}\right]\\\\ X_g=\left[\begin{array}{ccc}-1&2&1\\2&3&0\\1&-1&0\end{array}\right]=\left[\begin{array}{ccc}-5\end{array}\right]$

$a=\frac{detX_a}{detX}= \frac{11}{-14}= \frac{-11}{14}\\\\ d= \frac{detX_d}{detX}=\frac{1}{-14}=\frac{-1}{14}\\\\ g=\frac{detX_g}{detX}=\frac{-5}{-14}=\frac{5}{14}$

$-b+2e+h=0\\ 2b+3e+5h=1\\ b-e+2h=0\\\\ Y= \left[\begin{array}{ccc}-1&2&1\\2&3&5\\1&-1&2\end{array}\right]=\left[\begin{array}{ccc}-14\end{array}\right]\\\\ Y_b=\left[\begin{array}{ccc}0&2&1\\1&3&5\\0&-1&2\end{array}\right]=\left[\begin{array}{ccc}-5\end{array}\right]\\\\ Y_e=\left[\begin{array}{ccc}-1&0&1\\2&1&5\\1&0&2\end{array}\right]=\left[\begin{array}{ccc}-3\end{array}\right]\\\\ Y_h=\left[\begin{array}{ccc}-1&2&0\\2&3&1\\1&-1&0\end{array}\right]=\left[\begin{array}{ccc}1\end{array}\right]$

$b= \frac{detY_b}{detY}=\frac{-5}{-14}=\frac{5}{14} \\\\ e=\frac{detY_e}{detY}=\frac{-3}{-14}=\frac{3}{14}\\\\ h=\frac{detY_h}{detY}=\frac{1}{-14}=\frac{-1}{14}$

$-c+2f+i=0\\ 2c+3f+5i=0\\ c-f+2i=1\\\\ Z= \left[\begin{array}{ccc}-1&2&1\\2&3&5\\1&-1&2\end{array}\right]=\left[\begin{array}{ccc}-14\end{array}\right]\\\\ Z_c=\left[\begin{array}{ccc}0&2&1\\0&3&5\\1&-1&2\end{array}\right]=\left[\begin{array}{ccc}7\end{array}\right]\\\\ Z_f=\left[\begin{array}{ccc}-1&0&1\\2&0&5\\1&1&2\end{array}\right]=\left[\begin{array}{ccc}7\end{array}\right]\\\\ Z_i=\left[\begin{array}{ccc}-1&2&0\\2&3&0\\1&-1&1\end{array}\right]=\left[\begin{array}{ccc}-7\end{array}\right]$

$c= \frac{detZ_c}{detZ}=\frac{7}{-14}=\frac{-1}{2}\\\\ f=\frac{detZ_f}{detZ}=\frac{7}{-14}=\frac{-1}{2}\\\\ i= \frac{detZ_i}{detZ}=\frac{-7}{-14}=\frac{1}{2}$

$Portanto:\\\\ B^{-1}= \left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right]\\\\ B^{-1}=\left[\begin{array}{ccc}-11/14&5/14&-1/2\\-1/14&3/14&-1/2\\5/14&-1/14&1/2\end{array}\right]$