Find \(a,b,c\) & \(A^{-1}\) if \(A=\frac19\begin{bmatrix}8&-4&a\\1&4&b\\4&7&c\end{bmatrix}\) is orthogonal.

A is orthogonal \(\Rightarrow\) its columns are orthonormal.

Check: \(C_1\cdot C_1=\frac{64+1+16}{81}=1,\ C_2\cdot C_2=\frac{16+16+49}{81}=1,\ C_1\cdot C_2=\frac{-32+4+28}{81}=0\). ✓

\(C_3\perp C_1:\ 8a+b+4c=0\); \(C_3\perp C_2:\ -4a+4b+7c=0\).

Solving by cross-multiplication: \(\frac{a}{1(7)-4(4)}=\frac{b}{4(-4)-8(7)}=\frac{c}{8(4)-1(-4)}\Rightarrow\frac a{-9}=\frac b{-72}=\frac c{36}\Rightarrow\frac a{-1}=\frac b{-8}=\frac c4=k\).

Unit length: \(C_3\cdot C_3=1\Rightarrow a^2+b^2+c^2=81\Rightarrow k^2(1+64+16)=81\Rightarrow k=\pm1\).

\(a=-1,\ b=-8,\ c=4\) (or \(a=1,\ b=8,\ c=-4\)).

For an orthogonal matrix \(A^{-1}=A^T\): \(A^{-1}=\frac19\begin{bmatrix}8&1&4\\-4&4&7\\-1&-8&4\end{bmatrix}\) (with the second choice the last row is \(1,\ 8,\ -4\)).