(2, 0 pontos) Sejam os vetores \boldsymbol{u}=(1,0,-2), \boldsymbol{v}=(1,3,1) e \boldsymbol{w}=(-1,1,1) e considere as seguintes m at rizes:
U=\left[\begin{array}{lll} x & 1 & 2 \\\\ 0 & 3 & 5 \\\\ 2 & 2 & 2 \end{array}\right], \quad V=\left[\begin{array}{lll} 1 & 1 & 2 \\\\ 0 & y & 1 \\\\ 1 & 3 & 2 \end{array}\right], \text { e } W=\left[\begin{array}{lll} 2 & 1 & 2 \\\\ 1 & 4 & 1 \\\\ 2 & 3 & z \end{array}\right] .
Utilizando a regra de Cramer, determine x, y e z de forma que os vetores \tilde{\boldsymbol{u}}=U \boldsymbol{u}, \tilde{\boldsymbol{v}}=V \boldsymbol{v} e \tilde{\boldsymbol{w}}=W \boldsymbol{w} sejam ortogonais.
2022, 2023, 2022 - TEMOS EM MONITORIA GRAVADA

Passo 1

O enunciado dá os vetores
\boldsymbol{u}=(1,0,-2), \boldsymbol{v}=(1,3,1), \boldsymbol{w}=(-1,1,1)
e as matrizes
U=\begin{pmatrix}x & 1 & 2\\\ 0 & 3 & 5\\\ 2 & 2 & 2\end{pmatrix},\quad V=\begin{pmatrix}1 & 1 & 2\\\ 0 & y & 1\\\ 1 & 3 & 2\end{pmatrix},\quad W=\begin{pmatrix}2 & 1 & 2\\\ 1 & 4 & 1\\\ 2 & 3 & z\end{pmatrix}
e pede x,y,z tais que
\tilde{\boldsymbol{u}}=U\boldsymbol{u}, \tilde{\boldsymbol{v}}=V\boldsymbol{v} e \tilde{\boldsymbol{w}}=W\boldsymbol{w}
sejam mutuamente ortogonais. Vamos usar produto escalar para montar um sistema linear e depois a regra de Cramer para resolvê-lo.

Passo 2

Calcular
\tilde{\boldsymbol{u}}=U\boldsymbol{u}.
\tilde{\boldsymbol{u}} =\begin{pmatrix}x&1&2\\\ 0&3&5\\\ 2&2&2\end{pmatrix} \begin{pmatrix}1\\\ 0\\\ -2\end{pmatrix} =\begin{pmatrix} x\cdot1+1\cdot0+2\cdot(-2)\\\\ 0\cdot1+3\cdot0+5\cdot(-2)\\\\ 2\cdot1+2\cdot0+2\cdot(-2) \end{pmatrix} =\begin{pmatrix}x-4\\\ -10\\\ -2\end{pmatrix}.

Passo 3

Calcular
\tilde{\boldsymbol{v}}=V\boldsymbol{v}.
\tilde{\boldsymbol{v}} =\begin{pmatrix}1&1&2\\\ 0&y&1\\\ 1&3&2\end{pmatrix} \begin{pmatrix}1\\\ 3\\\ 1\end{pmatrix} =\begin{pmatrix} 1+3+2\\\\ 0+3y+1\\\\ 1+9+2 \end{pmatrix} =\begin{pmatrix}6\\\ 3y+1\\\ 12\end{pmatrix}.

Passo 4

Calcular
\tilde{\boldsymbol{w}}=W\boldsymbol{w}.
\tilde{\boldsymbol{w}} =\begin{pmatrix}2&1&2\\\ 1&4&1\\\ 2&3&z\end{pmatrix} \begin{pmatrix}-1\\\ 1\\\ 1\end{pmatrix} =\begin{pmatrix} -2+1+2\\\\ -1+4+1\\\\ -2+3+z \end{pmatrix} =\begin{pmatrix}1\\\ 4\z+1\end{pmatrix}.

Passo 5

Impor ortogonalidade ( \cdot é produto escalar):
\tilde{\boldsymbol{u}}\cdot\tilde{\boldsymbol{v}}=0,\quad \tilde{\boldsymbol{u}}\cdot\tilde{\boldsymbol{w}}=0,\quad \tilde{\boldsymbol{v}}\cdot\tilde{\boldsymbol{w}}=0.
Isso gera as equações
6(x-4)-10(3y+1)-2\cdot12=0 \quad\Longrightarrow\quad x-5y=\tfrac{29}{3},
(x-4)\cdot1-10\cdot4-2(z+1)=0 \quad\Longrightarrow\quad x-2z=46,
6\cdot1+4(3y+1)+12(z+1)=0 \quad\Longrightarrow\quad y+z=-\tfrac{11}{6}.
Em forma matricial \,(x,y,z) obedece
\begin{pmatrix} 1 & -5 & 0\\\\ 1 & 0 & -2\\\\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix}x\y\z\end{pmatrix} = \begin{pmatrix}29/3\\\ 46\\\ -11/6\end{pmatrix}.

Passo 6

Aplicar regra de Cramer. Seja
D=\det\begin{pmatrix}1&-5&0\\\ 1&0&-2\\\ 0&1&1\end{pmatrix},\quad D_x=\det\begin{pmatrix}29/3&-5&0\\\ 46&0&-2\\\ -11/6&1&1\end{pmatrix},\;\text{etc.}
Calcula-se
D=7,\quad D_x=231,\quad D_y=\tfrac{98}{3},\quad D_z=-\tfrac{91}{2}.

Passo 7

Pela regra de Cramer,
x=\frac{D_x}{D}=33,\quad y=\frac{D_y}{D}=\frac{14}{3},\quad z=\frac{D_z}{D}=-\frac{13}{2}.

Resposta

x=33
y=\tfrac{14}{3}
z=-\tfrac{13}{2}

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