Let $A$ be a $3\times 3$ matrix such that $A^{T}A=I$ and $\det (A)=1$. Then $\det (A-I)$ is equal to:

Solution

1. Given $A^{T}A=I$, $A$ is an orthogonal matrix, which implies that its eigenvalues $\lambda$ satisfy $|\lambda |=1$.
2. The characteristic polynomial is defined as $$\begin{array}{rl}p(\lambda )& =\det (A-\lambda I)\\ & =0\end{array}$$.
3. For any $3\times 3$ orthogonal matrix with $\det (A)=1$, the product of the eigenvalues is 1, and it is a known property that $\lambda =1$ must be an eigenvalue.
4. Substituting $\lambda =1$ into the characteristic equation gives $\det (A-I)=0$.
5. Hence the answer is (A).