(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 matrizes:
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]
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.

Passo 1

O enunciado nos dá os vetores \boldsymbol{u}=(1,0,-2), \boldsymbol{v}=(1,3,1) e \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}
Precisamos achar x, y e z de modo 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 e impor \tilde{\boldsymbol{u}}\cdot\tilde{\boldsymbol{v}}=0, \tilde{\boldsymbol{u}}\cdot\tilde{\boldsymbol{w}}=0 e \tilde{\boldsymbol{v}}\cdot\tilde{\boldsymbol{w}}=0.

Passo 2

Calculamos \tilde{\boldsymbol{u}} = U\boldsymbol{u}. Multiplicando, obtemos
\tilde{\boldsymbol{u}} =U\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-4\\\ -10\\\ -2\end{pmatrix}.

Passo 3

Calculamos \tilde{\boldsymbol{v}} = V\boldsymbol{v}. Multiplicando, obtemos
\tilde{\boldsymbol{v}} =V\boldsymbol{v} =\begin{pmatrix} 1 & 1 & 2\\\\ 0 & y & 1\\\\ 1 & 3 & 2 \end{pmatrix} \begin{pmatrix}1\\\ 3\\\ 1\end{pmatrix} =\begin{pmatrix}6\\\ 3y+1\\\ 12\end{pmatrix}.

Passo 4

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

Passo 5

Impondo os produtos escalares nulos, montamos o sistema:
  1. \tilde{\boldsymbol{u}}\cdot\tilde{\boldsymbol{v}}=0
(x-4)\cdot6+(-10)\cdot(3y+1)+(-2)\cdot12=0
6x-24-30y-10-24=0
3x-15y-29=0.
  1. \tilde{\boldsymbol{u}}\cdot\tilde{\boldsymbol{w}}=0
(x-4)\cdot1+(-10)\cdot4+(-2)\cdot(1+z)=0
x-4-40-2-2z=0
x-2z-46=0.
  1. \tilde{\boldsymbol{v}}\cdot\tilde{\boldsymbol{w}}=0
6\cdot1+(3y+1)\cdot4+12\cdot(1+z)=0
6+12y+4+12+12z=0
6y+6z+11=0.

Passo 6

Resolvemos o sistema
\begin{cases} 3x-15y-29=0,\\\ x-2z-46=0,\\\\ 6y+6z+11=0. \end{cases}
Da segunda equação, x=2z+46. Da terceira,
6y=-6z-11 \quad\Longrightarrow\quad y=-z-\frac{11}{6}.
Substituindo em 3x-15y-29=0:
3(2z+46)-15\bigl(-z-\tfrac{11}{6}\bigr)-29=0
6z+138+15z+\tfrac{165}{6}-29=0
21z+109+\tfrac{165}{6}=0 \quad\Longrightarrow\quad 126z+654+165=0
126z+819=0 \quad\Longrightarrow\quad z=-\frac{819}{126}=-\frac{13}{2}.
Logo
y=-\bigl(-\tfrac{13}{2}\bigr)-\tfrac{11}{6} =\frac{13}{2}-\frac{11}{6} =\frac{39-11}{6} =\frac{28}{6} =\frac{14}{3},
e
x=2\Bigl(-\tfrac{13}{2}\Bigr)+46=-13+46=33.

Resposta

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

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