Question 1: The trace and determinant of a \(2 \times 2\) matrix are known to be \(-2\) and \(-35\) respectively. Its eigenvalues are
\(-30\) and \(-5\)
\(-37\) and \(-1\)
\(-7\) and \(5\)
\(17.5\) and \(-2\)
Solution: Let the matrix is \[\mathbf{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\] Trace of a square matrix is sum of its diagonal entries \[\begin{aligned} \text{Trace } \mathbf{A} = a+d &= -2 \\ \text{Determinant } ad-bc &= -35 \\ \text{Eigenvalue } |\mathbf{A} - \lambda \mathbf{I}| &= 0 \end{aligned}\]
\[\begin{vmatrix} a-\lambda & b \\ c & d-\lambda \end{vmatrix} = 0\]
\[\begin{aligned} (a-\lambda)(d-\lambda) - bc &= 0 \\ \lambda^2 - (a+d)\lambda + (ad - bc) &= 0 \\ \lambda^2 - (-2)\lambda + (-35) &= 0 \\ \lambda^2 + 2\lambda - 35 &= 0 \\ (\lambda - 5)(\lambda + 7) &= 0 \\ \lambda_1, \lambda_2 &= 5, -7 \end{aligned}\] The correct option is C.
Question 2: \(f(x,y)\) is a continuous function defined over \((x,y) \in [0,1] \times [0,1]\). Given the two constraints, \(x > y^2\) and \(y > x^2\), the volume under \(f(x,y)\) is
\(\displaystyle \int_{y=0}^{y=1} \int_{x=y^2}^{x=\sqrt{y}} f(x,y)\,dxdy\)
\(\displaystyle \int_{y=\sqrt{x}}^{y=x^2} \int_{x=0}^{x=1} f(x,y)\,dxdy\)
\(\displaystyle \int_{y=0}^{y=1} \int_{x=0}^{x=1} f(x,y)\,dxdy\)
\(\displaystyle \int_{y=0}^{y=\sqrt{x}} \int_{x=0}^{x=\sqrt{y}} f(x,y)\,dxdy\)
Solution: Given constraints \(x > y^2\) and \(y > x^2\)
Limit of \(y\) : \(y=0\) to \(y=1\)
Limit of \(x\) : \(x=y^2\) to \(x=
\sqrt{y}\) (since \(y=x^2 \implies
x=\sqrt{y}\))
So volume under \(f(x,y)\) \[V = \int_{y=0}^{y=1} \int_{x=y^2}^{x=\sqrt{y}}
f(x,y)\, dx dy\] The correct option is A.
Question 3: The state-transition matrix of the above system is
\(\begin{bmatrix} e^{-3t} & 0 \\ e^{-2t} + e^{-3t} & e^{-2t} \end{bmatrix}\)
\(\begin{bmatrix} e^{-3t} & e^{-2t} - e^{-3t} \\ 0 & e^{-2t} \end{bmatrix}\)
\(\begin{bmatrix} e^{3t} & e^{2t} + e^{-3t} \\ 0 & e^{-2t} \end{bmatrix}\)
\(\begin{bmatrix} e^{3t} & e^{2t} - e^{-3t} \\ 0 & e^{-2t} \end{bmatrix}\)
Solution: Given state equations in matrix form can be written as, \[\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} -3 & 1 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 2 \\ 1 \end{bmatrix} u(t)\] \[\dfrac{d\mathbf{X}(t)}{dt} = \mathbf{A}\mathbf{X}(t) + \mathbf{B}u(t)\]
State transition matrix is given by \[\mathbf{\Phi}(t) = \mathcal{L}^{-1}\left[ \mathbf{\Phi}(s) \right]\] \[\mathbf{\Phi}(s) = (s\mathbf{I} - \mathbf{A})^{-1}\] \[(s\mathbf{I} - \mathbf{A}) = \begin{bmatrix} s & 0 \\ 0 & s \end{bmatrix} - \begin{bmatrix} -3 & 1 \\ 0 & -2 \end{bmatrix}\] \[(s\mathbf{I} - \mathbf{A}) = \begin{bmatrix} s+3 & -1 \\ 0 & s+2 \end{bmatrix}\] \[(s\mathbf{I} - \mathbf{A})^{-1} = \dfrac{1}{(s+3)(s+2)} \begin{bmatrix} s+2 & 1 \\ 0 & s+3 \end{bmatrix}\] So \[\mathbf{\Phi}(s) = (s\mathbf{I} - \mathbf{A})^{-1} = \begin{bmatrix} \dfrac{1}{s+3} & \dfrac{1}{(s+3)(s+2)} \\ 0 & \dfrac{1}{s+2} \end{bmatrix}\] \[\mathbf{\Phi}(t) = \mathcal{L}^{-1}\left[\mathbf{\Phi}(s)\right] = \begin{bmatrix} e^{-3t} & e^{-2t} - e^{-3t} \\ 0 & e^{-2t} \end{bmatrix}\] The correct option is B.