Questão 24 sejam A=\left(\begin{array}{rrr}1 & 0 & 1 \\\\ 2 & 3 & 2 \\\\ 0 & 1 & -2\end{array}\right) e B=\left(\begin{array}{rcr}9 & -8 & -1 \\\\ -7 & -3 & 0 \\\\ -2 & 0 & 0\end{array}\right)
a) Mostre que as matrizes A e B são inversíveis.
b) Calcule:
i) \operatorname{det}\left(A^{t} \cdot \operatorname{Tr}\left(\frac{1}{4} A\right)^{t} \cdot A^{-2}\right)
ii) \operatorname{det}\left(\left(A^{t}(\operatorname{Tr} A) A\right)^{-1}\right)
iii) \operatorname{det}\left(-2 A^{t}\left(\left(-\frac{2}{3} B^{3}\right)^{-1}\right)\left(\left(B^{t}\right)^{-1} A\right)^{-1}\right)

Passo 1

Resumindo, temos as matrizes
A=\begin{pmatrix}1 & 0 & 1\\\ 2 & 3 & 2\\\ 0 & 1 & -2\end{pmatrix}, \quad B=\begin{pmatrix}9 & -8 & -1\\\ -7 & -3 & 0\\\ -2 & 0 & 0\end{pmatrix}
e queremos, em (a), mostrar que ambas são inversíveis (ou seja, determinante \neq 0) e, em (b), calcular três determinantes de expressões envolvendo A, B, transposta, traço e inversas. Vamos usar propriedades de determinantes:
  1. \det(PQ)=\det(P)\det(Q),
  2. \det(M^{-1})=\det(M)^{-1},
  3. \det(M^t)=\det(M),
  4. \det(cM)=c^n\det(M) para matriz n\times n,
  5. \Tr(\alpha M)=\alpha\Tr(M).

Passo 2

Para (a), calculamos
\det(A) =1\cdot\begin{vmatrix}3&2\\\ 1&-2\end{vmatrix} -0\cdot\begin{vmatrix}2&2\\\ 0&-2\end{vmatrix} +1\cdot\begin{vmatrix}2&3\\\ 0&1\end{vmatrix} =(3\cdot(-2)-2\cdot1)+(2\cdot1-3\cdot0) =-6-2+2=-6\neq0.
Logo A é inversível.

Passo 3

Ainda em (a),
\det(B) =9\cdot\begin{vmatrix}-3&0\\\ 0&0\end{vmatrix} -(-8)\cdot\begin{vmatrix}-7&0\\\ -2&0\end{vmatrix} +(-1)\cdot\begin{vmatrix}-7&-3\\\ -2&0\end{vmatrix} =0+0+6=6\neq0,
logo B também é inversível.

Passo 4

Agora (b). Já temos \det(A)=-6 e \det(B)=6.

Passo 5

(i) Queremos \det\bigl(A^{t}\cdot(\Tr(\tfrac14A))^{t}\cdot A^{-2}\bigr).
Note que \Tr(\tfrac14A)=\tfrac14\Tr(A)=\tfrac14\cdot2=\tfrac12
e, sendo escalar, (\tfrac12)^{t}=\tfrac12. Assim
\det\bigl(A^{t}(\tfrac12I)A^{-2}\bigr) =\det(A^{t})\;\det(\tfrac12I)\;\det(A^{-2}) =\det(A)\;( \tfrac12)^3\;\det(A)^{-2} =\tfrac18\det(A)^{-1} =\tfrac18\cdot(-6)^{-1} =-\tfrac1{48}.

Passo 6

(ii) Queremos \det\bigl((A^{t}(\Tr A)A)^{-1}\bigr).
Como \Tr A=2, temos A^{t}(\Tr A)A=2A^{t}A e
\det\bigl((2A^{t}A)^{-1}\bigr) =\det(2A^{t}A)^{-1} =\bigl(2^3\det(A^{t})\det(A)\bigr)^{-1} =\bigl(8\det(A)^2\bigr)^{-1} =\tfrac1{8\cdot(-6)^2} =\tfrac1{288}.

Passo 7

(iii) Queremos \det\bigl(-2\,A^{t}\,( -\tfrac23B^{3} )^{-1}\,\bigl((B^{t})^{-1}A\bigr)^{-1}\bigr).
Tratamos -2 como escalar: \det(-2M)=(-2)^3\det(M)=-8\det(M).
Além disso
\det\bigl((-\tfrac23B^{3})^{-1}\bigr) =\bigl((-\tfrac23)^3\det(B^3)\bigr)^{-1} =-\frac{27}{8\,\det(B)^3}, \quad \det\bigl((B^{t})^{-1}A\bigr)^{-1} =\frac{\det(B)}{\det(A)}.
Como \det(A^{t})=\det(A)=-6 e \det(B)=6, junta-se tudo:
(-2)^3\cdot(-6)\cdot\Bigl(-\tfrac{27}{8\cdot6^3}\Bigr)\cdot\Bigl(\tfrac{6}{-6}\Bigr) =\tfrac34.

Resposta

(i) -\tfrac1{48}
(ii) \tfrac1{288}
(iii) \tfrac34

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