Question

Let $A$ be a $3\times3$ real matrix such that $A\begin{pmatrix}1\\0\\1\end{pmatrix}=2\begin{pmatrix}1\\0\\1\end{pmatrix},\ A\begin{pmatrix}-1\\0\\1\end{pmatrix}=4\begin{pmatrix}-1\\0\\1\end{pmatrix},\ A\begin{pmatrix}0\\1\\0\end{pmatrix}=2\begin{pmatrix}0\\1\\0\end{pmatrix}$.
Then, the system $(A-3I)\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}$ has :

• A. exactly two solutions
• B. infinitely many solutions
• C. unique solution
• D. no solution

Hint:

The three given vectors are eigenvectors of $A$; find the eigenvalues of $A-3I$ and its determinant.

Answer 1

The Correct Option is C

The given equations say that $A$ has eigenvalues $2,\,4,\,2$ with the eigenvectors $(1,0,1)^T$, $(-1,0,1)^T$, $(0,1,0)^T$, which are linearly independent.
For $A-3I$ the same vectors are eigenvectors with eigenvalues $2-3=-1,\ 4-3=1,\ 2-3=-1$.
$\det(A-3I)=(-1)(1)(-1)=1\ne0$, so $A-3I$ is invertible and the system has a unique solution. Option (3).