IFRR Superior 2015/1

Precisamos calcular o determinante de \(C\), onde

\[C = A^{\mathrm{T}} + B^{-1}\]

As matrizes dadas s (ordem \(2\times2\)):

• \(A=(a_{ij})\) com \(a_{ij}=3i+2j\)
• \(B=(b_{ij})\) com \(b_{ij}=i+j\)

1. Construindo \(A\) e \(A^{\mathrm{T}}\)

Para \(i,j\in\{1,2\}\):

\(a_{11}=3\cdot1+2\cdot1=5\)
\(a_{12}=3\cdot1+2\cdot2=7\)
\(a_{21}=3\cdot2+2\cdot1=8\)
\(a_{22}=3\cdot2+2\cdot2=10\)

Logo

\[A=\begin{bmatrix}5&7\\8&10\end{bmatrix}\quad\Rightarrow\quad A^{\mathrm{T}}=\begin{bmatrix}5&8\\7&10\end{bmatrix}.\]

2. Construindo \(B\) e \(B^{-1}\)

Para \(B\):

\(b_{11}=2\ ,\ b_{12}=3\ ,\ b_{21}=3\ ,\ b_{22}=4\)

\[B=\begin{bmatrix}2&3\\3&4\end{bmatrix}.\]

Determinante de \(B\): \(\det(B)=2\cdot4-3\cdot3=8-9=-1\).

Para matriz \(2\times2\),

\[B^{-1}=\frac1{\det(B)}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}=\frac1{-1}\begin{bmatrix}4&-3\\-3&2\end{bmatrix}=\begin{bmatrix}-4&3\\3&-2\end{bmatrix}.\]

3. Matriz \(C\)

Somando termo a termo:

\[C=A^{\mathrm{T}}+B^{-1}=\begin{bmatrix}5&8\\7&10\end{bmatrix}+\begin{bmatrix}-4&3\\3&-2\end{bmatrix}=\begin{bmatrix}1&11\\10&8\end{bmatrix}.\]

4. Determinante de \(C\)

Para \(2\times2\): \(\det(C)=1\cdot8-11\cdot10=8-110=-102\).

Resposta

\(\det(C)=-102\). Alternativa A.