Question

Determine a matriz inversidade de:

a) $A \ = \ \left[\begin{array}{ccc}1&-1\\2&0\\\end{array}\right]$

b) $B \ = \ \left[\begin{array}{ccc}5&8\\2&3\\\end{array}\right]$

Answer 1

a)

$A \ \bullet A^{-1} \ = \ I_{2}$

$\left[\begin{array}{ccc}1&-1\\2& \ \ 0 \\\end{array}\right] \ \bullet \ \left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] \ = \ \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$

$1^{a} \ linha \ \ x \ \ 1^{a} \ coluna \ = \boxed{a \ - \ c}$

$1^{a} \ linha \ \ x \ \ 2^{a} \ coluna \ = \boxed{b \ - \ d}$

$2^{a} \ linha \ \ x \ \ 1^{a} \ coluna \ = \boxed{2 \ a }$

$2^{a} \ linha \ \ x \ \ 2^{a} \ coluna \ = \boxed{2 \ b}$

$\left[\begin{array}{ccc}a - c&b - 1\\2a&2b\\\end{array}\right] \ = \ \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$

$\left \{ {{a \ - \ c \ = \ 1} \atop {2 \ a \ = \ 0 \ \Rightarrow \boxed{a = 0}}} \right.$

$\bullet \ a - c = 1$

$0 - c = 1$

$\boxed{c = -1}$

$\left \{ {{b - b \ = \ 0} \atop {2b \ = \ 1 \ \Rightarrow \ \boxed{b = \frac{1}{2} }}} \right.$

$\bullet \ b - d = 0$

$\frac{1}{2} \ - d \ = 0$

$\boxed{d \ = \ \frac{1}{2} }$

$\boxed{\boxed{ A^{-1} \ = \ \left[\begin{array}{ccc} \ \ \ 0& \frac{1}{2} \\-1& \frac{1}{2} \\\end{array}\right] }}$

b)

$B \ = \ \left[\begin{array}{ccc}5&8\\2&3\\\end{array}\right]$

$B \ = \ \left[\begin{array}{ccc}5&8\\2&3\\\end{array}\right] \ \bullet \ \left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] \ = \ \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$

$1^{a} \ linha \ \ x \ \ 1^{a} \ coluna \ = \boxed{5a \ + \ 8c}$

$1^{a} \ linha \ \ x \ \ 2^{a} \ coluna \ = \boxed{5b \ + \ 8d}$

$2^{a} \ linha \ \ x \ \ 1^{a} \ coluna \ = \boxed{2a \ + \ 3c}$

$2^{a} \ linha \ \ x \ \ 2^{a} \ coluna \ = \boxed{2b \ + \ 3d}$

$\left[\begin{array}{ccc}\ 5a \ + \ 8c \ &5b \ + \ 8d\\ 2a \ + \ 3c&2b \ + \ 3d\\\end{array}\right] \ = \ \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$

$\left \{ {{5a \ + \ 8c \ = \ 1 \ (2)} \atop {2a \ + \ 3c \ = \ 0 \ (-5)}} \right.$

$\left \{ {{10a \ + \ 16c \ = \ 2} \atop {10a \ - \ 15c \ = \ 0}} \right.$

$\boxed{c = 2}$

$\bullet \ 2a \ + \ 3c \ = \ 0$

$2a \ + \ 6 \ = 0$

$2a \ = \ - \ 6$

$\boxed{a \ = \ - \ 3}$

$\left \{ {{5b \ + \ 8d \ = \ 0 \ (12)} \atop {2b \ + \ 3d \ = \ 1 \ (-5)}} \right.$

$\left \{ {{10b \ + 16d \ = \ 0} \atop {- \ 10b \ - \ 15 \ = \ - \ 5}} \right.$

$\boxed{d \ = \ - \ 5}$

$\bullet \ 2b \ - \ 15 \ = \ 1$

$2b \ = \ 16$

$\boxed{b \ = \ 8}$

$\boxed{\boxed{ B^{-1} \ = \ \left[\begin{array}{ccc}-3& \ \ \ \ 8\\ \ \ \ 2& \ - \ 5\\\end{array}\right] }}$