Find $a_{11}$ and $a_{22}$ given two eigen-values?

Question

This is one question on my homework, and I can't seem to find how to solve it online (not sure how to word it in a search) or in the textbook.

Determine $a_{11}$ and $a_{22}$ for $$\begin{pmatrix} a_{11} & -2.4 \\ 1.6 & a_{22} \end{pmatrix}$$ if $\lambda_1 =2$, $\lambda_2 =5$ and $a_{11}>a_{22}$.

Thank you in advance for any help.

Answer 1

Since 4 and 5 are eigenvalues, we know that the characteristic polynomial is ($\lambda$ - 4)($\lambda$-5) = $\lambda^2 - 9 \lambda+20$. But if we compute the characteristic polynomial of the given matrix we get $\lambda^2 - (a_{11}+a_{22})\lambda + 3.84+a_{11}a_{22}$. So we have

$$ a_{11}+a_{22} = 9\\a_{11}a_{22} = 16.16$$

You can now solve for $a_{11}, a_{22}$.

Answer 2

Here $$A=\begin{pmatrix} a_{11} & -2.4 \\ 1.6 & a_{22} \end{pmatrix}$$

Given that the two eigen-value of $~A~$ are $~\lambda_1 =2\quad \text{and}\quad \lambda_2 =5~$ and also $~a_{11}>a_{22}$ .

Now $~Trace~ (A)= a_{11}+a_{22}=2+5=7$

and $~Determinant ~(A)=a_{11}~a_{22}~-~(-2.4)~(1.6)=10$

$\implies a_{11}~a_{22}~=6.16$

So we have two equations $$a_{11}+a_{22}= 7$$and $$a_{11}~a_{22}~=6.16$$

Solving we have $$a_{11}=\frac{7}{2}\pm\frac{\sqrt {609}}{10},\quad a_{22}=\frac{7}{2}\mp\frac{\sqrt {609}}{10}$$ Since $~a_{11}>a_{22}~$,so $$a_{11}=\frac{7}{2}+\frac{\sqrt {609}}{10},\quad a_{22}=\frac{7}{2}-\frac{\sqrt {609}}{10}$$