Find the eigenvalue of $\begin{bmatrix}6&-3\\-3&6\end{bmatrix}x=\frac{\lambda}{18}\begin{bmatrix}4&1\\1&4\end{bmatrix}x$

Question

Find the eigenvalue of $\begin{bmatrix}6&-3\\-3&6\end{bmatrix}x=\frac{\lambda}{18}\begin{bmatrix}4&1\\1&4\end{bmatrix}x$

$\begin{bmatrix}6&-3\\-3&6\end{bmatrix}x-\frac{\lambda}{18}\begin{bmatrix}4&1\\1&4\end{bmatrix}x=0$

$=\begin{bmatrix} 6-\frac{4\lambda}{18}&-3-\frac{\lambda}{18}\\-3-\frac{\lambda}{18}&6-\frac{4\lambda}{18}\end{bmatrix}x=0$

Now I am stuck. The solution suggests to take the determinant of the matrix $\begin{bmatrix} 6-\frac{4\lambda}{18}&-3-\frac{\lambda}{18}\\-3-\frac{\lambda}{18}&6-\frac{4\lambda}{18}\end{bmatrix}$, but why? Also the solutions wrote that $\det(A)=(6-\frac{4\lambda}{18}-3-\frac{\lambda}{18})(6-\frac{4\lambda}{18}+3+\frac{\lambda}{18})$, where did this come from? I thought the determinant would be $\left(6-\frac{4\lambda}{18}\right)^{2}-(-3-\frac{\lambda}{18})^{2}$?

Answer 1

We want the values of $\lambda$ for which the matrix equation $$ \begin{bmatrix} 6-\frac{4\lambda}{18}&-3-\frac{\lambda}{18}\\-3-\frac{\lambda}{18}&6-\frac{4\lambda}{18}\end{bmatrix}x=0 $$ Has some non-zero solution $x$. This is the case exactly when the matrix $$ A = \begin{bmatrix} 6-\frac{4\lambda}{18}&-3-\frac{\lambda}{18}\\-3-\frac{\lambda}{18}&6-\frac{4\lambda}{18}\end{bmatrix} $$ is singular. $A$ will be singular if and only if $\det A = 0$.

So, set $\det A$ equal to zero, and solve for $\lambda$.