Em cada caso determine se a matriz A dada é ortogonal, nesse caso determine sua inversa.
A=\left[\begin{array}{ccc}-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\\\ 0 & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}}\end{array}\right]

Passo 1

O enunciado apresenta a matriz
A = \begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}}\\[6pt] 0 & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}}\\[6pt] \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{pmatrix}
e pede para verificar se A é ortogonal. Lembramos que uma matriz real é ortogonal se suas colunas formam um conjunto ortonormal em \mathbb{R}^3, ou seja, cada coluna tem norma 1 e colunas diferentes são ortogonais entre si. Quando isso vale, temos ainda \,A^{-1}=A^T.]

Passo 2

Denotemos as colunas de A por
c_1 = \begin{pmatrix}-\tfrac{1}{\sqrt{2}}\\[3pt]0\\[3pt]\tfrac{1}{\sqrt{2}}\end{pmatrix},\quad c_2 = \begin{pmatrix}\tfrac{1}{\sqrt{6}}\\[3pt]-\tfrac{2}{\sqrt{6}}\\[3pt]\tfrac{1}{\sqrt{6}}\end{pmatrix},\quad c_3 = \begin{pmatrix}\tfrac{1}{\sqrt{3}}\\[3pt]\tfrac{1}{\sqrt{3}}\\[3pt]\tfrac{1}{\sqrt{3}}\end{pmatrix}.
Calculamos as normas:
\|c_1\|^2 = \Bigl(-\tfrac{1}{\sqrt{2}}\Bigr)^2 + 0^2 + \Bigl(\tfrac{1}{\sqrt{2}}\Bigr)^2 = \tfrac12 + 0 + \tfrac12 = 1,
\|c_2\|^2 = \Bigl(\tfrac{1}{\sqrt{6}}\Bigr)^2 + \Bigl(-\tfrac{2}{\sqrt{6}}\Bigr)^2 + \Bigl(\tfrac{1}{\sqrt{6}}\Bigr)^2 = \tfrac16 + \tfrac46 + \tfrac16 = 1,
\|c_3\|^2 = 3 \cdot \Bigl(\tfrac{1}{\sqrt{3}}\Bigr)^2 = 3\cdot\tfrac13 = 1.]

Passo 3

Agora verificamos os produtos internos entre as colunas:
c_1\!\cdot\!c_2 = \bigl(-\tfrac{1}{\sqrt{2}}\bigr)\bigl(\tfrac{1}{\sqrt{6}}\bigr) +0\cdot\bigl(-\tfrac{2}{\sqrt{6}}\bigr) +\bigl(\tfrac{1}{\sqrt{2}}\bigr)\bigl(\tfrac{1}{\sqrt{6}}\bigr) =0,
c_1\!\cdot\!c_3 = \bigl(-\tfrac{1}{\sqrt{2}}\bigr)\bigl(\tfrac{1}{\sqrt{3}}\bigr) +0\cdot\bigl(\tfrac{1}{\sqrt{3}}\bigr) +\bigl(\tfrac{1}{\sqrt{2}}\bigr)\bigl(\tfrac{1}{\sqrt{3}}\bigr) =0,
c_2\!\cdot\!c_3 = \bigl(\tfrac{1}{\sqrt{6}}\bigr)\bigl(\tfrac{1}{\sqrt{3}}\bigr) +\bigl(-\tfrac{2}{\sqrt{6}}\bigr)\bigl(\tfrac{1}{\sqrt{3}}\bigr) +\bigl(\tfrac{1}{\sqrt{6}}\bigr)\bigl(\tfrac{1}{\sqrt{3}}\bigr) =0.
Portanto, as colunas são mutuamente ortogonais e têm norma 1.

Passo 4

Como as colunas de A formam um conjunto ortonormal, concluímos que A é ortogonal. Logo
A^{-1} = A^T = \begin{pmatrix} -\tfrac{1}{\sqrt{2}} & 0 & \tfrac{1}{\sqrt{2}}\\[6pt] \tfrac{1}{\sqrt{6}} & -\tfrac{2}{\sqrt{6}} & \tfrac{1}{\sqrt{6}}\\[6pt] \tfrac{1}{\sqrt{3}} & \tfrac{1}{\sqrt{3}} & \tfrac{1}{\sqrt{3}} \end{pmatrix}.]

Resposta

A matriz A é ortogonal e
A^{-1} = A^T = \begin{pmatrix} -\tfrac{1}{\sqrt{2}} & 0 & \tfrac{1}{\sqrt{2}}\\[6pt] \tfrac{1}{\sqrt{6}} & -\tfrac{2}{\sqrt{6}} & \tfrac{1}{\sqrt{6}}\\[6pt] \tfrac{1}{\sqrt{3}} & \tfrac{1}{\sqrt{3}} & \tfrac{1}{\sqrt{3}} \end{pmatrix}.]

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