SOLUTIONS TO ASSIGNMENT – 3 FOR ENGINEERING MATHEMATICS – I

\[\color {black} {1.}\ By\ using\ Cramers\ Rule\,\ Solve\ the\ equations\ \hspace{20cm}\]
\[x + y + z=3,\ 2x – y + z = 2\ and\ 3x + 2y – 2z = 3\ \hspace{15cm}\]
\[\color {black}{Solution:}\ \hspace{20cm}\]
\[x + y + z=3\ ——————-(1)\ \hspace{6cm}\]
\[2x – y + z = 2\ \hspace{15cm}\]
\[3x + 2y – 2z = 3\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 3 & 2 & -2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta =1\begin{vmatrix} -1 & 1 \\ 2 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 1 \\ 3 & -2 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & -1\\ 3 & 2 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta =1(2\ -\ 2)\ – 1 (-4\ -\ 3) + 1(4\ -\ (-3))\ \hspace{9cm}\]
\[\Delta =1(0)\ – 1 (-7) + 1(7)\ \hspace{13cm}\]
\[\Delta =0\ + 7 + 7\ \hspace{14cm}\]
\[\Delta =14\ \hspace{17cm}\]
\[\Delta_x = \begin{vmatrix} 3 & 1 & 1 \\ 2 & -1 & 1 \\ 3 & 2 & -2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =3\begin{vmatrix} -1 & 1 \\ 2 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 1 \\ 3 & -2 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & -1\\ 3 & 2 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x =3(2\ -\ 2) – 1 (-4\ -\ 3) + 1(4\ -\ (-3))\ \hspace{9cm}\]
\[\Delta_x = 0\ + 7 + 7\ \hspace{14cm}\]
\[\Delta_x =14\ \hspace{17cm}\]
\[\Delta_y = \begin{vmatrix} 1 & 3 & 1 \\ 2 & 2 & 1 \\ 3 & 3 & -2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y =1\begin{vmatrix} 2 & 1 \\ 3 & -2 \\ \end{vmatrix}\ -\ 3\begin{vmatrix} 2 & 1 \\ 3 & -2 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & 2\\ 3 & 3 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y =1(-4\ -\ 3)\ – 3 (-4\ -\ 3) + 1(6\ -\ 6)\ \hspace{9cm}\]
\[\Delta_y = 1(-7)\ – 3 (-7) + 1(0)\ \hspace{13cm}\]
\[\Delta_y = -7\ + 21\ + 0\ \hspace{14cm}\]
\[\Delta_y = 14\ \hspace{17cm}\]
\[\Delta_z = \begin{vmatrix} 1 & 1 & 3 \\ 2 & -1 & 2 \\ 3 & 2 & 3 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z =1\begin{vmatrix} -1 & 2 \\ 2 & 3 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 2 \\ 3 & 3 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} 2 & – 1\\ 3 & 2 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z =1(-7)\ – 1 (0) + 3(7)\ \hspace{13cm}\]
\[\Delta_z =-7\ – 0 + 21\ \hspace{14cm}\]
\[\Delta_z =14\ \hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x =1\ y = 1\ z = 1\ in\ equation (1)\ \hspace{18cm}\]
\[LHS = 1 + 1 + 1\]\[ = 3\]\[ = RHS\]
\[2\ Solve\ x^6 + 1\ =\ 0\ \hspace{18cm}\]
\[\color {black}{Solution:}\ x^6\ +\ 1\ =\ 0 \hspace{18cm}\]
\[ x^6\ =\ -\ 1\ \hspace{10cm}\]
\[\ x\ =\ (-\ 1)^\frac{1}{6}\ \hspace{10cm}\]
\[ =\ (cos\ π\ +\ i\ sin\ π)^\frac{1}{6}\ \hspace{10cm}\]
\[ =\ cos\ (\frac{π\ + 2kπ}{6})\ +\ i\ sin\ (\frac{π\ + 2kπ}{6})\ where\ k\ =\ 0,\ 1,\ 2,\ 3,\ 4,\ 5\ \hspace{5cm}\]
\[The\ roots\ are\]
\[When\ k = 0,\ \hspace{2cm}\ x\ =\ cos\ \frac{π}{6}\ +\ i\ sin\ \frac{π}{6}\]
\[When\ k = 1,\ \hspace{2cm}\ x\ =\ cos\ \frac{3π}{6}\ +\ i\ sin\ \frac{3π}{6}\]
\[When\ k = 2,\ \hspace{2cm}\ x\ =\ cos\ \frac{5π}{6}\ +\ i\ sin\ \frac{5π}{6}\]
\[When\ k = 3,\ \hspace{2cm}\ x\ =\ cos\ \frac{7π}{6}\ +\ i\ sin\ \frac{7π}{6}\]
\[When\ k = 4,\ \hspace{2cm}\ x\ =\ cos\ \frac{9π}{6}\ +\ i\ sin\ \frac{9π}{6}\]
\[When\ k = 5,\ \hspace{2cm}\ x\ =\ cos\ \frac{11π}{6}\ +\ i\ sin\ \frac{11π}{6}\]
\[3.\ Prove\ that\ \frac{Cos^3\ A\ -\ Cos\ 3\ A}{Cos\ A}\ +\ \frac{Sin^3\ A\ +\ Sin\ 3\ A}{Sin\ A}\ =\ 3\ \hspace{15cm}\]
\[\color {black}{Solution:}\ L.\ H.\ S\ =\ \frac{Cos^3\ A\ -\ Cos\ 3\ A}{Cos\ A}\ +\ \frac{Sin^3\ A\ +\ Sin\ 3\ A}{Sin\ A}\ \hspace{18cm}\]
\[=\ \frac{Cos^3\ A\ -\ (4\ Cos^3\ A\ -\ 3\ Cos\ A)}{Cos\ A}\ +\ \frac{Sin^3\ A\ +\ 3\ Sin\ A\ -\ 4\ Sin^3\ A}{Sin\ A}\ \hspace{15cm}\]
\[=\ \frac{-\ 3Cos^3\ A\ -\ 3\ Cos\ A}{Cos\ A} +\ \frac{-\ 3\ Sin^3\ A\ +\ 3\ Sin\ A}{Sin\ A}\ \hspace{15cm}\]
\[=\ \frac{Cos\ A(-\ 3Cos^2\ A\ +\ 3)}{Cos\ A} +\ \frac{Sin\ A(-\ 3\ Sin^2\ A\ +\ 3)}{Sin\ A}\ \hspace{15cm}\]
\[=\ -\ 3\ Cos^2\ A\ +\ 3\ -\ 3\ Sin^2\ A\ +\ 3\ \hspace{10cm}\]
\[=\ 6\ -\ 3\ (Sin^2\ A\ +\ Cos^2\ A)\ \hspace{10cm}\]
\[=\ 6\ -\ 3\ (1)\ \hspace{10cm}\]
\[=\ 6\ -\ 3\ \hspace{10cm}\]
\[=\ 3\ \hspace{10cm}\]

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