Considerando as matrizes:
\mathbf{A}=\left[\begin{array}{cc}3 & 0 \\\\ -1 & 2 \\\\ 1 & 1\end{array}\right], \quad \mathbf{B}=\left[\begin{array}{cc}4 & -1 \\\\ 0 & 2\end{array}\right], \quad \mathbf{C}=\left[\begin{array}{lll}1 & 4 & 2 \\\\ 3 & 1 & 5\end{array}\right], \quad \mathbf{D}=\left[\begin{array}{ccc}1 & 5 & 2 \\\\ -1 & 0 & 1 \\\\ 3 & 2 & 4\end{array}\right], \quad \mathbf{E}=\left[\begin{array}{ccc}6 & 1 & 3 \\\\ -1 & 1 & 2 \\\\ 4 & 1 & 3\end{array}\right], calcule os seguintes (quando possível):
d)
\mathbf{B}^{T}\left(\mathbf{C} \mathbf{C}^{T}-\mathbf{A}^{T} \mathbf{A}\right)

Passo 1

O enunciado pede para calcular
\mathbf{B}^{T}\bigl(\mathbf{C}\,\mathbf{C}^{T}-\mathbf{A}^{T}\mathbf{A}\bigr)
onde \mathbf{A} é 3\times2, \mathbf{C} é 2\times3 e \mathbf{B} é 2\times2. Vamos primeiro achar \mathbf{C}\,\mathbf{C}^{T}, depois \mathbf{A}^{T}\mathbf{A}, subtrair, e então multiplicar à esquerda por \mathbf{B}^{T}.

Passo 2

Calcular \mathbf{C}\,\mathbf{C}^{T}.
Temos
\mathbf{C} = \begin{pmatrix} 1 & 4 & 2\\\\ 3 & 1 & 5 \end{pmatrix},
então
\mathbf{C}\,\mathbf{C}^{T} = \begin{pmatrix} 1\cdot1 + 4\cdot4 + 2\cdot2 & 1\cdot3 + 4\cdot1 + 2\cdot5 \\[6pt] 3\cdot1 + 1\cdot4 + 5\cdot2 & 3\cdot3 + 1\cdot1 + 5\cdot5 \end{pmatrix} = \begin{pmatrix} 21 & 17\\\\ 17 & 35 \end{pmatrix}.]

Passo 3

Calcular \mathbf{A}^{T}\mathbf{A}.
Como
\mathbf{A} = \begin{pmatrix} 3 & 0\\\\ -1 & 2\\\\ 1 & 1 \end{pmatrix},
temos
\mathbf{A}^{T}\mathbf{A} = \begin{pmatrix} 3^2 + (-1)^2 + 1^2 & 3\cdot0 + (-1)\cdot2 + 1\cdot1 \\[6pt] 3\cdot0 + (-1)\cdot2 + 1\cdot1 & 0^2 + 2^2 + 1^2 \end{pmatrix} = \begin{pmatrix} 11 & -1\\\\ -1 & 5 \end{pmatrix}.]

Passo 4

Formar a diferença
\mathbf{C}\,\mathbf{C}^{T} \;-\;\mathbf{A}^{T}\mathbf{A} = \begin{pmatrix} 21 & 17\\\\ 17 & 35 \end{pmatrix} - \begin{pmatrix} 11 & -1\\\\ -1 & 5 \end{pmatrix} = \begin{pmatrix} 10 & 18\\\\ 18 & 30 \end{pmatrix}.

Passo 5

Multiplicar \mathbf{B}^{T} por essa diferença. Primeiro
\mathbf{B} = \begin{pmatrix}4 & -1\\\ 0 & 2\end{pmatrix} \quad\Longrightarrow\quad \mathbf{B}^{T} = \begin{pmatrix}4 & 0\\\ -1 & 2\end{pmatrix}.
Então
\mathbf{B}^{T} \begin{pmatrix} 10 & 18\\\\ 18 & 30 \end{pmatrix} = \begin{pmatrix} 4\cdot10 + 0\cdot18 & 4\cdot18 + 0\cdot30\\[4pt] (-1)\cdot10 + 2\cdot18 & (-1)\cdot18 + 2\cdot30 \end{pmatrix} = \begin{pmatrix} 40 & 72\\\\ 26 & 42 \end{pmatrix}.]

Resposta

\mathbf{B}^{T}\bigl(\mathbf{C}\,\mathbf{C}^{T}-\mathbf{A}^{T}\mathbf{A}\bigr) = \begin{pmatrix} 40 & 72\\\\ 26 & 42 \end{pmatrix}.

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