Practicing Success
If $A=\frac{1}{3}\begin{bmatrix}1&2&2\\2&1&-2\\a&2&b\end{bmatrix}$ is an orthogonal matrix, then |
$a = 2,b=1$ $a=-2,b=-1$ $a = 2, b = -1$ $a = -2,b=1$ |
$a=-2,b=-1$ |
Since A is an orthogonal matrix. $∴AA^T = I$ $⇒\frac{1}{3}\begin{bmatrix}1&2&2\\2&1&-2\\a&2&b\end{bmatrix}.\frac{1}{3}\begin{bmatrix}1&2&a\\2&1&2\\2&-2&b\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$ $⇒\frac{1}{9}\begin{bmatrix}1&2&2\\2&1&-2\\a&2&b\end{bmatrix}\begin{bmatrix}1&2&a\\2&1&2\\2&-2&b\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$ $⇒\begin{bmatrix}9&0&a+4+2b\\0&9&2a+2-2b\\a+4+2b&2a+2-2b&a^2+4+b^2\end{bmatrix}=\begin{bmatrix}9&0&0\\0&9&0\\0&0&9\end{bmatrix}$ $⇒a+4+2b=0, 2a+2-2b=0$ and $a^2+4+b^2 = 9$ $⇒a+2b+4=0, a-b+1=0$ and $a^2 + b^2=5$ $⇒a=-2, b=-1$ |