I'll write it again but the users below don't seem to understand ... THIS IS NOT A DUPLICATE . PLEASE READ MY QUESTION THOROUGLY.
I need to find a matrix with characteristic equation given by
λ² − λ − 1 = 0
I know that Since the polynomial p(λ) = λ² − λ − 1 has degree 2,so we look for a 2 × 2 matrix.
I let A be any 2 × 2 matrix, where a, b, c, d ∈ R.
A=$$ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} $$
The characteristic equation of A is:
p(λ) = det(A − λId) = λ² − (a + d)λ + ad − bc
And here is what I don't get . On the answer sheet is says that this polynomial coincides with λ² − λ − 1 if and only if:
−(a + d) = −1 and ad − bc = −1.
I have dyscalculia and even the things that might seem simple are hard for me to understand. Can somebody please explain where they are getting
−(a + d) = −1 and ad − bc = −1
and why , for example, they are not setting λ²=-1? What does that mean?
Please be as simple and clear as possible! I really appreciate the help ! Thanks guys
Edit: please do not mark this as a duplicate. This is a UNIQUE question in which i am trying to understand ONE of the stages to determine the characteristic equation .
You have $$\lambda^2-\lambda-1=\lambda^2\color{red}{-1}\cdot\lambda+\color{blue}{(-1)},$$ you agree? Now the number which is multiplied by $\lambda$ is $\color{red}{-1}$, the number which is just added is $\color{blue}{-1}$.
In $$\lambda^2\color{red}{-(a+d)}\cdot\lambda+ \color{blue}{ad-bc}$$ the number which is multiplied by $\lambda$ is $\color{red}{-(a+d)}$ and the number which is just added is $\color{blue}{ad-bc}$.
From here $$\color{red}{-1}=\color{red}{-(a+d)}\qquad\text{and}\quad\color{blue}{-1}=\color{blue}{ad-bc}. $$