Question

(Unirio) Dada a matriz representada na figura adiante
$\left[\begin{array}{cc}-5&-3\\3&-2\\\end{array}\right]$
Determine o valor de A⁻¹ + A transposta - I₂

Answer 1

Oi . Fiz a questão em duas figuras. 
Na figura 1 tem o cálculo da matriz inversa.
Na figura 2 tem o resultado das somas das matrizes conforme manda o enunciado.

Espero que goste. 
:)

Answer 2

$\large\boxed{\begin{array}{l}\sf A=\begin{bmatrix}\sf-5&\sf-3\\\sf3&\sf-2\end{bmatrix}\\\sf det~A=(-5)\cdot(-2)-3\cdot(-3)\\\sf det~A=10+9=19\\\\\sf cof~A=\begin{bmatrix}\sf-2&\sf-3\\\sf3&\sf-5\end{bmatrix}\\\\\sf adj(A)=cof A^T\\\sf adj(A)=\begin{bmatrix}\sf-2&\sf3\\\sf-3&\sf-5\end{bmatrix}\\\\\sf A^{-1}=\dfrac{1}{det~A}\cdot adj(A)\\\\\sf A^{-1}=\dfrac{1}{19}\cdot\begin{bmatrix}\sf-2&\sf3\\\sf3&\sf-5\end{bmatrix}\\\\\end{array}}$

$\large\boxed{\begin{array}{l}\sf A^{-1}=\begin{bmatrix}\sf-\dfrac{2}{19}&\sf\dfrac{3}{19}\\\\\sf-\dfrac{3}{19}&\sf-\dfrac{5}{19}\end{bmatrix}\\\\\sf A^T=\begin{bmatrix}\sf-5&\sf3\\\sf-3&\sf-2\end{bmatrix}\\\\\sf I_2=\begin{bmatrix}\sf1&\sf0\\\sf0&\sf1\end{bmatrix}\end{array}}$

$\large\boxed{\begin{array}{l}\sf A^{-1}+A^T-I_2=\begin{bmatrix}\sf-\dfrac{2}{19}-5-1&\sf\dfrac{3}{19}+3-0\\\\\sf-\dfrac{3}{19}-3-0&\sf-\dfrac{5}{19}-2-1\end{bmatrix}\\\\\sf A^{-1}+A^T-I_2=\begin{bmatrix}\sf-\dfrac{116}{19}&\sf\dfrac{60}{19}\\\\\sf-\dfrac{60}{19}&\sf-\dfrac{62}{19}\end{bmatrix}\end{array}}$