How to calculate $a,b,c,d$ given the eigenvectors

Question

Given the matrix $$A=\begin{bmatrix} a & b & 2 \\ c & d & 6 \\ 3 & 4 & -3 \end{bmatrix}$$ with eigenvectors $$v_1=\begin{bmatrix} 5 \\ 1 \\ 3 \end{bmatrix} \quad\text{and}\quad v_2=\begin{bmatrix} 7 \\ 4 \\ 3 \end{bmatrix}$$ find $a,b,c,d$.

After this i know i should compute $Av_1$ and $Av_2$. $$Av_1=\begin{bmatrix} 5a+b+6 \\ 5c+d+18 \\ 10 \end{bmatrix}$$ So what next in order to find value for $a,b,c,d$?

Answer 1

$A*V_1 = \lambda*V_1$

$A*V_1$ is \begin{bmatrix} 5a+b+6 \\ 5c+d+18 \\ 10 \\ \end{bmatrix}

$\lambda*V_1$ is \begin{bmatrix} 5*\lambda \\ 1*\lambda \\ 3*\lambda \\ \end{bmatrix} On equating, we see that $\lambda = \frac{10}{3}$

Similarly solve for other eigen value. Then with equations formed, get the desired values

Answer 2

Hint. Note that by the definitions of eigenvalue and eigenvector, $$\begin{bmatrix} 5a+b+6 \\ 5c+d+18 \\ 10 \end{bmatrix}=Av_1=\lambda_1 v_1=\begin{bmatrix} 5\lambda_1 \\ \lambda_1 \\ 3\lambda_1 \end{bmatrix}$$ and looking at the third component wefind that $10=3\lambda_1$, that is $\lambda_1=10/3$. Hence $$\begin{cases} 5a+b+6=5\lambda_1 \\ 5c+d+18=\lambda_1 \\ \end{cases}\implies \begin{cases} 5a+b=32/3 &\text{(1)}\\ 5c+d=-44/3 &\text{(2)}\\ \end{cases}.$$ Now do the same thing with the other eigenvector $v_2$ and you will obtain another couple of equations (3) and (4). Finally solve the linear system of (1) and (3) with respect to $a$, $b$ and the linear system of (2) and (4) with respect to $c$, $d$.