# What is an elegant way to find the third eigenvalue of $A=\left[\begin{smallmatrix} 51&-12 & -21\\ 60 & -40&28\\ 57&-68&1 \end{smallmatrix}\right]$?

We had this in an exam today: Given the matrix $$A=\begin{bmatrix} 51&-12 & -21\\ 60 & -40&28\\ 57&-68&1 \end{bmatrix}$$ Here is the precise question as asked in our exam:

Someone tells you (accurately) that $-48$ and $24$ are eigenvalues of the matrix $A$. Without using a computer, calculator or writing anything down find the third eigenvalue of $A$.

I have used the classical method (through the characteristic polynomial of A) but this leaded to a some tremendous computation and finally got the answer.

But the question intentionally asked to do not make any use of such computation.

How can one elegantly find the third eigenvalue of $A$?

• Excellent!${}{}{}$ – Dave Nov 23 '17 at 20:17