Calcule $A^{-1}$ , sendo $A=\left(\begin{array}{ccc}\frac{1}{\sqrt{3}} & \frac{-2}{\sqrt{6}} & 0 \\\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{-1}{\sqrt{2}} \\\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}}\end{array}\right)$ . Dica: note que $\operatorname{det}(A)=1$ !

O enunciado pede a inversa da matriz

A=\begin{pmatrix} \tfrac{1}{\sqrt{3}} & \tfrac{-2}{\sqrt{6}} & 0\\[6pt] \\ \tfrac{1}{\sqrt{3}} & \tfrac{1}{\sqrt{6}} & \tfrac{-1}{\sqrt{2}}\\[6pt] \\ \tfrac{1}{\sqrt{3}} & \tfrac{1}{\sqrt{6}} & \tfrac{1}{\sqrt{2}} \end{pmatrix}

sabendo que

\det(A)=1

. Vamos usar a ideia de matrizes ortogonais: se as colunas de

A

forem vetores unitários e mutuamente ortogonais, então

A^{-1}=A^{T}

Verificamos as normas e produtos internos das colunas de

A

. Denotando

\mathbf{c}_1=(\tfrac1{\sqrt3},\tfrac1{\sqrt3},\tfrac1{\sqrt3})

\mathbf{c}_2=(-\tfrac2{\sqrt6},\tfrac1{\sqrt6},\tfrac1{\sqrt6})

\mathbf{c}_3=(0,-\tfrac1{\sqrt2},\tfrac1{\sqrt2})

, temos

\|\mathbf{c}_1\|^2=\frac13+\frac13+\frac13=1,\quad \|\mathbf{c}_2\|^2=\frac{4}{6}+\frac{1}{6}+\frac{1}{6}=1,\quad \|\mathbf{c}_3\|^2=\frac12+\frac12=1

\mathbf{c}_i\cdot\mathbf{c}_j=0\quad(\forall i\neq j).

Logo, as colunas são ortonormais.

Como

A

é ortogonal e

\det(A)=1

, vale

A^{-1}=A^{T}.

Transpondo

A

, obtemos

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

]

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

]

Calcule A^{-1}, sendo A=\left(\begin{array}{ccc}\frac{1}{\sqrt{3}} & \frac{-2}{\sqrt{6}} & 0 \\\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{-1}{\sqrt{2}} \\\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}}\end{array}\right). Dica: note que \operatorname{det}(A)=1 !