Given Information
We are given with following predator-prey matrix: \[ A = \left[ \begin{array} { c c } { 0.4 } & { 0.3 } \\ { - p } & { 1.2 } \end{array} \right] \] We have to estimate the long-term growth rate and the eventual ratio of owls to flying squirrels.

Step-1:
For p=0.5
\[ A = \left[ \begin{array} { c c } { 0.4 } & { 0.3 } \\ { - p } & { 1.2 } \end{array} \right] \]

Step-2:
The Equation of characteristic polynomial \[ \begin{aligned} \operatorname { det } ( A - \lambda I ) & = \left| \begin{array} { c c } { 0.4 - \lambda } & { 0.3 } \\ { - 0.5 } & { 1.2 - \lambda } \end{array} \right| \\ & = ( 0.4 - \lambda ) ( 1.2 - \lambda ) - ( 0.3 ) ( - 0.5 ) \\ & = \lambda ^ { 2 } - 1.6 \lambda + 0.48 + 0.15 \\ = & \lambda ^ { 2 } - 1.6 \lambda + 0.63 \\ = & \lambda ^ { 2 } - 0.9 \lambda - 0.7 \lambda + 0.63 \\ = & \lambda ( \lambda - 0.9 ) - 0.7 \lambda + 0.63 \\ = & \lambda ( \lambda - 0.9 ) ( \lambda - 0.7 ) \\ = & ( \lambda - 0.9 ) ( \lambda - 0.7 ) \end{aligned} \] From the equation, we can see that eigenvalues are 0.7 and 0.9

Step-3:
We have to find value of $p$ for which the populations of both the owls and squirrels tends towards constant levels. Write the characteristic polynomial in terms of $p$. \[ \begin{aligned} \operatorname { det } ( A - \lambda I ) & = \left| \begin{array} { c c } { 0.4 - \lambda } & { 0.3 } \\ { - p } & { 1.2 - \lambda } \end{array} \right| \\ & = ( 0.4 - \lambda ) ( 1.2 - \lambda ) + 0.3 p \\ & = \lambda ^ { 2 } - 1.6 \lambda + 0.48 + 0.3 p \end{aligned} \]

The eigenvalues are: 0.55 and 1.05



Step-4:
Solve for eigenvalues: \[\begin{array}{l} \lambda = \dfrac{{1.6 \pm \sqrt {{{( - 1.6)}^2} - 4(0.48 + 0.3p)} }}{2}\\ = \dfrac{{1.6 \pm \sqrt {2.56 - 1.92 - 1.2p} }}{2}\\ = \dfrac{{1.6 \pm \sqrt {0.64 - 1.2p} }}{2} \end{array}\]

Step-5:
Equate eigenvalue with 1 \[ \begin{aligned} \dfrac { 1.6 \pm \sqrt { 0.64 - 1.2 p } } { 2 } & = 1 \\ 1.6 \pm \sqrt { 0.64 - 1.2 p } & = 2 \\ \pm \sqrt { 0.64 - 1.2 p } & = 0.4 \\ 0.64 - 1.2 p & = 0.16 \\ 0.64 - 0.16 & = 1.2 p \\ 0.48 & = 1.2 p \\ 0.4 & = p \end{aligned} \] Hence,

The population of both the owls and squirrels tends towards constant level whe



Step-6:
Eigenvalues when p=0.4

\[ \begin{aligned} \lambda ^ { 2 } - 1.6 \lambda + 0.48 + 0.3 p & = 0 \\ \lambda ^ { 2 } - 1.6 \lambda + 0.48 + 0.3 ( 0.4 ) & = 0 \\ \lambda ^ { 2 } - 1.6 \lambda + 0.48 + 0.12 & = 0 \\ \lambda ^ { 2 } - 1.6 \lambda + 0.64 & = 0 \\ \lambda ^ { 2 } - 1.0 \lambda - 0.6 \lambda + 0.6 & = 0 \\ \lambda ( \lambda - 1.0 ) - 0.6 ( \lambda - 1.0 ) & = 0 \\ ( \lambda - 1.0 ) ( \lambda - 0.6 ) & = 0 \end{aligned} \] Therefore,

The eigenvalues are 1 and 0.6



Step-7:
The eigenvector for $\lambda = 1$:

\[\begin{array}{l} [A - (1.0)I]{{\bf{v}}_1} = {\bf{0}}\\ \left[ {\begin{array}{*{20}{c}} {0.4 - 1.0}&{0.3}\\ { - 0.4}&{1.2 - 1.0} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}} \end{array}} \right] = 0\\ \left[ {\begin{array}{*{20}{c}} { - 0.6}&{0.3}\\ { - 0.4}&{0.2} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}} \end{array}} \right] = 0 \end{array}\] The system of equations are \[\begin{array}{l} - 0.6{x_1} + 0.3{x_2} = 0\\ - 0.4{x_1} + 0.2{x_2} = 0 \end{array}\] Thus, general solution is: \[ \begin{aligned} \mathbf { v } _ { 1 } & = \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] \\ & = \left[ \begin{array} { l } { x _ { 1 } } \\ { 2 x _ { 1 } } \end{array} \right] \\ & = \left[ \begin{array} { l } { 1 } \\ { 2 } \end{array} \right] x _ { 1 } \end{aligned} \] The eigenvector is: $ \mathbf { v } _ { 1 } = \left[ \begin{array} { l } { 1 } \\ { 2 } \end{array} \right]$. Hence,

The constant level of owls and squirrels is 1 owl for every 2 thousand squirrels. The constant level of owls and squirrels is 1 owl for every 2 thousand squirrels.