Summer 2015 - Q.2 - Grad Plus

# Summer 2015 – Q.2

2. (a) Solve $latex x^7+x^4+i\left(x^3+1\right)=0$ [6M]

$latex x^7+x^4+i\left(x^3+1\right)=0$

$latex \begin{array}{l}x^4\left(x^3+1\right)+i\left(x^3+1\right)=0\\\\\left(x^4+i\right)\left(x^3+1\right)=0\\\\x^3=-1\\\\x^4=-i\end{array}$

$latex \begin{array}{l}x^3=-1\\\\\;\;\;\;=\cos\pi+\sin\pi\end{array}$

$latex \begin{array}{l}x^3=cos\left(2n\pi+\pi\right)+isin\left(2n\pi+\pi\right)\\\\x=\left[cos\left(2n+1\right)\pi+isin\left(2n+1\right)\pi\right]^\frac13\\\\x=cos\frac{\left(2n+1\right)}3+isin\frac{\left(2n+1\right)\pi}3\end{array}$

n = 0, 1, 2

Root of the equation are

$latex \left(\cos\frac\pi3+i\sin\frac\pi3\right),\;\left(\cos\pi+i\sin\pi\right)\;and\;\left(\cos\frac{5\pi}3+i\sin\frac{5\pi}3\right)$

(b) Reduce the matrix A to normal form and hence find its rank where

$latex A=\begin{bmatrix}0&1&-3&-1\\1&0&4&3\\3&1&0&2\\1&1&-2&0\end{bmatrix}$ [6M]

$latex A=\begin{bmatrix}0&1&-3&-1\\1&0&4&3\\3&1&0&2\\1&1&-2&0\end{bmatrix}\;\;\;R_1\leftrightarrow R_2$

$latex \begin{array}{l}=\begin{bmatrix}1&0&4&3\\0&1&-3&-1\\3&1&0&2\\1&1&-2&0\end{bmatrix}\;\;\;R_3\rightarrow R_3-3R_1\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;R_4\rightarrow R_4-R_1\end{array}$

$LATEX \begin{array}{l}\\\;=\;\;\begin{bmatrix}1&0&4&3\\0&1&-3&-1\\0&1&-12&-7\\0&1&-6&-3\end{bmatrix}\;\;\;\;\;\;\;\;R_3\rightarrow R_3\;-R_2\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;R_4\rightarrow R_4-R_2\;\;\;\;\;\;\;\;\;\;\;\;\;\end{array}$

(c) State and prove that Euler’s theorem for three variables and hence find $latex x\frac{\partial u}{\partial x}+y\frac{\displaystyle\partial u}{\displaystyle\partial y}+z\frac{\displaystyle\partial u}{\displaystyle\partial x}$ where

$latex u=\frac{x^3\;y^3\;z^3}{x^3+y^3+z^3}$

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