Using Power method, obtain the largest eigen value and corresponding eigen vector correct upto 2d of the following matrices:

1. \(\left[\begin{array}{ccc} 1&2&-1 \\ 2&1&2 \\ -1&2&1\end{array}\right]\)   
2. \(\left[\begin{array}{ccc} 4&1&1&1 \\ 3&-1&1&1\\ -1&1&2&0\\1&1&0&2\end{array}\right]\)

       3. \(\left[\begin{array}{ccc} 0&11&-5\\ -2&17&-7 \\ -4&26&-10\end{array}\right]\)

Find the smallest eigen value and corresponding eigen vector of the following matrices using Power method:

1. \(\left[\begin{array}{ccc} 3&2&1 \\ 2&3&2 \\ 1&2&2\end{array}\right]\)
2. \(\left[\begin{array}{ccc} 2&2&-1 \\ -5&9&-3 \\ -4&4&1\end{array}\right]\)
3. \(\left[\begin{array}{ccc} 2&-1&0 \\ -1&2&-1 \\ 0&-1&2\end{array}\right]\)

Find all the eigen values and corresponding eigen vectors using Power method:

1. \(\left[\begin{array}{ccc} 1&2&3 \\ 3&1&0 \\ -2&0&1\end{array}\right]\)
2. \(\left[\begin{array}{ccc} 4&1&0\\ 1&20&1 \\ 0&1&4\end{array}\right]\)
3. \(\left[\begin{array}{ccc} 2&1&1&0 \\ 1&1&0&1 \\ 1&0&0&1\\1&1&1&2\end{array}\right]\)

If the largest eigen value of the matrix\(\left[\begin{array}{ccc} 5&-2&0 \\ 1&2&-3 \\ 1&-2&4\end{array}\right]\)is 6 and corresponding eigen vector is\(\left[\begin{array}{ccc} 1 \\ -1/2 \\ 1\end{array}\right]\),then find the subdominant eigen value.

Find the eigen values of following matrix:

\(\left[\begin{array}{ccc} 1&-1&0&0&0&0 \\ 1&1&-1&0&0&0 \\ 0&1&1&-1&0&0\\0&0&1&1&-1&0\\0&0&0&1&1&-1\\0&0&0&0&1&1\end{array}\right]\)

Let y(x) be a continuous solution of the initial value problem

y' + 2y = f(x), y(0) = 0, where \( f(x) = \begin{cases} 1, & \quad \text{if } 0 \le x\le1,\\ 0, & \quad \text{if } x>1 \end{cases} \)

Then y\(( \frac{3}{2})\) is equal to

(a)  \(( \frac{sinh(1)}{e^3})\)  (b)   \(( \frac{cosh(1)}{e^3})\)   (c)  \(( \frac{sinh(1)}{e^2})\)  (d)   \(( \frac{cosh(1)}{e^2})\)

The singular integral of the ODE 

\((xy'-y)^2 =x^2(x^2-y^2)\)

is

(a) \(y=xsinx\)       (b)  \(y=xsin(x+\frac{\pi}{4})\)   (c)  \(y=x\)     (d)  \(y=x+ \frac{\pi}{4}\)

The initial value problem

\(y'=2\sqrt{y}, y(0)=a,\)

has

(a) a unique solution if a < 0,

(b) no solution if a > 0,

(c) infinitely many solutions if a = 0

(d) a unique solution if a ≥ 0

For the initial value problem

\(\begin {cases} \frac{dy}{dx}=y^2+cos^2x, x>0,\\ y(0)=0 \end{cases}\)

The largest interval of existence of the solution predicted by Picard’s theorem is:

(a) [0, 1] (b) [0, \(\frac{1}{2}\) ] (c) [0, \(\frac{1}{3}\) ] (d) [0, \(\frac{1}{4} \) ]

Consider the ODE \(y'=f(y(x))\). If f is an even function and y is an odd function, then

(a) \(-y(-x)\)is also a solution.

(b)\(y(-x)\)is also a solution.

(c) \(-y(x)\) is also a solution.

(d) \(y(x)y(-x)\) is also a solution.

Let \(y:[o,\infty) \rightarrow [0,\infty) \)be a continuously differentiable function satisfying 

\(y(t)=y(0) +\int_0^t y(s)ds,\) for \(t\ge0.\)

Then

(a)   \(y^2(t)=y^2(0)=\int_0^t y^2(s)ds.\)

(b)   \(y^2(t)=y^2(0)+2\int_0^t y^2(s)ds.\)

(c)   \(y^2(t)=y^2(0)+\int_0^t y(s)ds.\)

(d)   \(y^2(t)=y^2(0)+\left(\int_0^ty(s)ds\right)^2 +2y(0)\int_0^ty(s)ds.\)