# SOLUTIONS TO SEMINAR – 1 FOR ENGINEERING MATHEMATICS – I

$\color {black} {1.}\ Solve\ by\ using\ Cramers\ Rule\ \hspace{20cm}$
$3x – y + 2z=8,\ x – y + z = 2\ and\ 2x + y – z = 1\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{20cm}$
$3x – y + 2z=8\ ——————-(1)\ \hspace{6cm}$
$x – y + z = 2\ \hspace{15cm}$
$2x + y – z = 1\ \hspace{15cm}$
$\Delta = \begin{vmatrix} 3 & -1 & 2 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta =3\begin{vmatrix} -1 & 1 \\ 1 & -1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 1 & 1 \\ 2 & -1 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 1 & -1\\ 2 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta =3(1\ -\ 1)\ + 1 (-1\ -\ 2) + 2(1\ -\ (-2))\ \hspace{9cm}$
$\Delta =3(0)\ + 1 (-3) + 2(3)\ \hspace{13cm}$
$\Delta =0\ – 3 + 6\ \hspace{14cm}$
$\Delta =3\ \hspace{17cm}$
$\Delta_x = \begin{vmatrix} 8 & -1 & 2 \\ 2 & -1 & 1 \\ 1 & 1 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_x =8\begin{vmatrix} -1 & 1 \\ 1 & -1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & 1 \\ 1 & -1 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 2 & -1\\ 1 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_x =8(1\ -\ 1) + 1 (-2\ -\ 1) + 2(2\ -\ (-1))\ \hspace{9cm}$
$\Delta_x =8(0)\ + 1 (-3) + 2(3)\ \hspace{13cm}$
$\Delta_x = 0\ – 3 + 6\ \hspace{14cm}$
$\Delta_x =3\ \hspace{17cm}$
$\Delta_y = \begin{vmatrix} 3 & 8 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_y =3\begin{vmatrix} 2 & 1 \\ 1 & -1 \\ \end{vmatrix}\ -\ 8\begin{vmatrix} 1 & 1 \\ 2 & -1 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 1 & 2\\ 2 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_y =3(-2\ -\ 1)\ – 8 (-1\ -\ 2) + 2(1\ -\ (4))\ \hspace{9cm}$
$\Delta_y =3(- 3)\ – 8 (-3) + 2(-3)\ \hspace{13cm}$
$\Delta_y = -9\ + 24\ – 6\ \hspace{14cm}$
$\Delta_y = 9\ \hspace{17cm}$
$\Delta_z = \begin{vmatrix} 3 & -1 & 8 \\ 1 & -1 & 2 \\ 2 & 1 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_z =3\begin{vmatrix} -1 & 2 \\ 1 & 1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 1 & 2 \\ 2 & 1 \\ \end{vmatrix}\ +\ 8\begin{vmatrix} 1 & – 1\\ 2 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_z =3(-1\ -\ 2)\ + 1 (1\ -\ 4) + 8(1\ -\ (-2))\ \hspace{9cm}$
$\Delta_z =3(-3)\ + 1 (-3) + 8(3)\ \hspace{13cm}$
$\Delta_z =-9\ – 3 + 24\ \hspace{14cm}$
$\Delta_z =12\ \hspace{17cm}$
$The\ Solution\ is\ \hspace{20cm}$
$x=\ \frac{\Delta_x}{\Delta} =\ \frac{3}{3} =\ 1\ \hspace{20cm}$
$y=\ \frac{\Delta_y}{\Delta} =\ \frac{8}{4} =\ 2\ \hspace{20cm}$
$z=\ \frac{\Delta_z}{\Delta} =\ \frac{12}{4} =\ 3\ \hspace{20cm}$
$For\ cross\ verification\ \hspace{20cm}$
$Put\ x =1\ y = 2\ z = 3\ in\ equation (1)\ \hspace{18cm}$
$LHS = 3(1) – 3 + 2(4)$$= 3 – 3 + 8 = 8$$= RHS$
$2.\ Find\ the\ inverse\ of\ \begin{bmatrix} 3 & 1 & -1\\ 2 & -2 & 0 \\ 1 & 2 & -1 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {black}{Solution:}\ A\ =\begin{bmatrix} 3 & 1 & -1 \\ 2 & -2 & 0 \\ 1 & 2 & -1 \\ \end{bmatrix}\ \hspace{15cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =\ 3\begin{vmatrix} -2 & 0 \\ 2 & -1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 0 \\ 1 & -1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & -2\\ 1 & 2 \\ \end{vmatrix}\ \hspace{10cm}$
$=\ 3(2\ -\ 0)\ – 1 (-2\ -\ 0) – 1(4\ +\ 2)\ \hspace{9cm}$
$=3(2)\ – 1 (-2) – 1(6)\ \hspace{13cm}$
$=\ 6\ +\ 2\ -\ 6\ \hspace{14cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =\ 2\ \neq\ 0\ \hspace{17cm}$
$\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}$
$\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}$
$cofactor\ of\ 3 = (-1)^{1\ +\ 1}\ \begin{vmatrix} -2 & 0 \\ 2 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^2 (2 – 0)\ \hspace{15cm}$
$= (1) (2)\ \hspace{15cm}$
$cofactor\ of\ 1 =\ 2\ \hspace{15cm}$
$cofactor\ of\ 1\ =\ (-1)^{1\ +\ 2}\ \begin{vmatrix} 2 & 0 \\ 1 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (-\ 2 -\ 0)\ \hspace{15cm}$
$= (-1) (-2)\ \hspace{15cm}$
$cofactor\ of\ 1\ =\ 2\ \hspace{15cm}$
$cofactor\ of\ -1 = (-1)^{1\ +\ 3}\ \begin{vmatrix} 2 & -2 \\ 1 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (4\ +\ 2)\ \hspace{15cm}$
$= (1) (6)\ \hspace{15cm}$
$cofactor\ of\ -1\ =\ 6\ \hspace{15cm}$
$cofactor\ of\ 2\ =\ (-1)^{2\ +\ 1}\ \begin{vmatrix} 1 & -1 \\ 2 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (-1 +\ 2)\ \hspace{15cm}$
$= (-1) (1)\ \hspace{15cm}$
$cofactor\ of\ 2 =\ -1\ \hspace{15cm}$
$cofactor\ of\ -\ 2 = (-1)^{2\ +\ 2}\ \begin{vmatrix} 3 & -1 \\ 1 & -\ 1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (-3 +\ 1)\ \hspace{15cm}$
$= (1) (-2)\ \hspace{15cm}$
$cofactor\ of\ -2 =\ -2\ \hspace{15cm}$
$cofactor\ of\ 0\ =\ (-1)^{2\ +\ 3}\ \begin{vmatrix} 3 & 1 \\ 1 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (6\ -\ 1)\ \hspace{15cm}$
$= (-1) (5)\ \hspace{15cm}$
$cofactor\ of\ 0\ = -5\ \hspace{15cm}$
$cofactor\ of\ 1\ = (-1)^{3\ +\ 1}\ \begin{vmatrix} 1 & -1 \\ -2 & 0 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (0 – 2)\ \hspace{15cm}$
$= (1) (-2)\ \hspace{15cm}$
$cofactor\ of\ 1 =\ -2\ \hspace{15cm}$
$cofactor\ of\ 2\ = (-1)^{3\ +\ 2}\ \begin{vmatrix} 3 & -1 \\ 2 & 0 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (0 + 2)\ \hspace{15cm}$
$= (-1) (2)\ \hspace{15cm}$
$cofactor\ of\ 2 = -2\ \hspace{15cm}$
$cofactor\ of\ -1 = (-1)^{3\ +\ 3}\ \begin{vmatrix} 3 & 1 \\ 2 & -2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^6 (-6 – 2)\ \hspace{15cm}$
$= (1) (-8)\ \hspace{15cm}$
$cofactor\ of\ -1 =\ -8\ \hspace{15cm}$
$Cofactor\ matrix=\begin{bmatrix} 2 & 2 & 6 \\ -1 & -2 & -5 \\ -2 & -2 & -8 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A=\begin{bmatrix} 2 & -1 & -2 \\ 2 & -2 & -2 \\ 6 & -5 & -8 \\ \end{bmatrix}\ \hspace{15cm}$
$A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}$
$A^{-1} = \frac{1}{2}\ \begin{bmatrix} 2 & -1 & -2 \\ 2 & -2 & -2 \\ 6 & -5 & -8 \\ \end{bmatrix}\ \hspace{2cm}$
$3.\ Find\ the\ rank\ of\ the\ matrix\ \begin{bmatrix} 1 & 2 & 3 & 1 \\ 2 & 3 & 4 & 2 \\ 3 & 4 & 5 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {black}{Solution:}\ A\ =\begin{bmatrix} 1 & 2 & 3 & 1 \\ 2 & 3 & 4 & 2 \\ 3 & 4 & 5 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$Order\ of\ A = 3 × 4\ \hspace{15cm}$
$\therefore \ rank\ of\ A=\rho(A)\ \leq\ Min{3,4} =3\ . \hspace{15cm}$
$The\ highest\ order\ of\ minors\ of\ A = 3.\ \hspace{15cm}$
$A\ has\ the\ following\ minors\ of\ order 3.\ \hspace{15cm}$
$A_1\ =\begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \\ \end{vmatrix}\ \hspace{15cm}$
$=1\begin{vmatrix} 3 & 4 \\ 4 & 5 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & 4 \\ 3 & 5 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} 2 & 3\\ 3 & 4 \\ \end{vmatrix}\ \hspace{10cm}$
$=\ 1(15\ -\ 16)\ – 2 (10\ -\ 12) +\ 3(8\ -\ 9)\ \hspace{9cm}$
$=\ 1(-1)\ – 2 (-2) + 3(-1)\ \hspace{13cm}$
$=\ -1\ +\ 4 -\ 3\ =\ 12\ =\ 0\ \hspace{14cm}$
$A_2\ =\begin{vmatrix} 1 & 2 & 1 \\ 2 & 3 & 2 \\ 3 & 4 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$=1\begin{vmatrix} 3 & 2 \\ 4 & 1\\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & 2 \\ 3 & 1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & 3\\ 3 & 4 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(3\ -\ 8)\ – 2 (2\ -\ 6) +\ 1(8\ -\ 9)\ \hspace{9cm}$
$=1(-5)\ – 2 (-4) + 1(-1)\ \hspace{13cm}$
$= 5\ +\ 8 – 1\ =\ 2\ \neq 0\ \hspace{14cm}$
$A_3\ =\begin{bmatrix} 1 & 3 & 1 \\ 2 & 4 & 2 \\ 3 & 5 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$=1\begin{vmatrix} 4 & 2 \\ 5 & 1 \\ \end{vmatrix}\ -\ 3\begin{vmatrix} 2 & 2 \\ 3 & 1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & 4\\ 3 & 5 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(4\ -\ 10)\ – 3 (2\ -\ 6) + 1(10\ -\ 12)\ \hspace{9cm}$
$=1(-6)\ – 3 (-4) + 1(-2)\ \hspace{13cm}$
$= -6\ +\ 12 – 2\ =\ -6\ \neq 0\ \hspace{14cm}$
$A_4\ =\begin{bmatrix} 2 & 3 & 1 \\ 3 & 4 & 2 \\ 4 & 5 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$=2\begin{vmatrix} 4 & 2 \\ 5 & 1 \\ \end{vmatrix}\ -\ 3\begin{vmatrix} 3 & 2 \\ 4 & 1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 3 & 4\\ 4 & 5 \\ \end{vmatrix}\ \hspace{10cm}$
$=2(4\ -\ 10)\ – 3 (3\ -\ 8) + 1(15\ -\ 16)\ \hspace{9cm}$
$=2(-6)\ – 3 (-5) + 1(-1)\ \hspace{13cm}$
$= -12\ + 15 – 1\ =\ 2\ \neq 0\ \hspace{14cm}$
$\therefore\ \rho(A)\ =\ 3\ \hspace{15cm}$
$4.\ Find\ the\ coefficient\ of\ x^{30}\ in\ the\ expansion\ of\ (x^4\ +\ \frac{1}{x^6})^{15}\ \hspace{15cm}$
$\color {black}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ x^4,\ a\ =\ \frac{1}{x^6},\ n\ =\ 15,\ r\ =\ r\ \hspace{14cm}$
$T_{r + 1} = 15C_r\ (x^4)^{15-r}\ (\frac{1}{x^6})^r \hspace{15cm}$
$= 15C_r\ x^{60 – 4r}\ \frac{1^r}{x^{6r}} \hspace{15cm}$
$= 15C_r\ x^{60 – 4r – 6r} \hspace{15cm}$
$T_{r + 1} = 15C_r\ x^{60 – 10r}\ ———————— (1)\ \hspace{15cm}$
$We\ assume\ that\ x^{30}\ occurs\ in\ 15C_r\ x^{60 – 10r}\ \hspace{15cm}$
$\therefore\ x^{60 – 10r}\ =\ x^{32}\ \hspace{15cm}$
$Equating\ the\ indices\ 60 – 10r\ =\ 30\ \hspace{15cm}$
$-10r\ =\ 30\ -\ 60\ \hspace{15cm}$
$-10r\ = -\ 30\ \hspace{15cm}$
$r\ = 3\ \hspace{15cm}$
$Put\ r\ =\ 3\ in\ (1)\ \hspace{15cm}$
$T_{3 + 1} = 15C_3\ 1^3\ x^{60\ -\ 30}\ \hspace{15cm}$
$T_4 = 15C_3\ x^{30}\ \hspace{15cm}$
$The\ coefficient\ of\ x^{30}\ =\ 15\ C_3\ \hspace{15cm}$
$5.\ Find\ the\ term\ independent\ of\ x\ in\ the\ expansion\ of\ (3\sqrt{x} – \frac{2}{x^2})^{10}\ \hspace{15cm}$
$\color {black}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ 3\sqrt{x},\ a\ =\ \frac{-2}{x^2},\ n\ =\ 10,\ r\ =\ r\ \hspace{14cm}$
$T_{r + 1} = 10C_r\ (3\sqrt{x})^{10-r}\ (-\frac{2}{x^2})^r \hspace{15cm}$
$= 10C_r\ 3^{10\ -\ r}\ x^(\frac{10-r}{2})\ \frac{-2^r}{x^{2r}} \hspace{15cm}$
$= 10C_r\ x^(\frac{10-r}{2})\ {-2}^r\ \ x^{-2r}\ \hspace{15cm}$
$= 10C_r\ x^(\frac{10-r}{2}\ -\ 2r)\ {-2}^r\ \hspace{15cm}$
$= 10C_r\ x^(\frac{10-r – 4r}{2})\ {-2}^r\ \hspace{15cm}$
$T_{r + 1} = 10C_r\ x^(\frac{10-5r}{2})\ {-2}^r\ ———————— (1)\ \hspace{15cm}$
$To\ find\ the\ independent\ term\ of\ x,\ find\ the\ coefficient\ of\ x^0.\ \hspace{15cm}$
$\therefore\ x^{\frac{10-5r}{2}}\ =\ x^0\ \hspace{15cm}$
$\frac{10-5r}{2}\ =\ 0\ \hspace{15cm}$
$10 – 5r\ =\ 0\ \hspace{15cm}$
$-5r\ = -\ 10\ \hspace{15cm}$
$r\ = 2\ \hspace{15cm}$
$Put\ r\ =\ 2\ in\ (1)\ \hspace{15cm}$
$\therefore\ independent\ term\ of\ x =\ 10C_2\ 3^{10\ -\ 2}\ (-2)^2\ \hspace{15cm}$
$=\ 10C_2\ 3^8\ (4)\ \hspace{15cm}$
$6.\ Show\ that\ the\ complex\ numbers\ 2\ -\ 2i,\ 8\ +\ 4i,\ 5\ +\ 7i\ and\ -1\ +\ i\ form\ a\ rectangle\ \hspace{10cm}$
$\color {black}{Solution:} \hspace{20cm}$
$Let\ A = 2\ -\ 2i\ = (2, – 2)\ \hspace{15cm}$
$B = 8\ +\ 4i\ = (8, 4)\ \hspace{13cm}$
$C = 5\ +\ 7i\ = (5, 7)\ \hspace{13cm}$
$D = -1\ +\ i\ = (-1, 1)\ \hspace{13cm}$
$AB\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (2\ -\ 8)^2 +\ (-2\ -\ 4)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((-6)^2 + (-6)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (36 + 36 )}\ \hspace{8cm}$
$AB = \sqrt {72}\ \hspace{8cm}$
$BC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (8\ -\ 5)^2 +\ (4\ -\ 7)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((3)^2 + (-3)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (9 + 9 )}\ \hspace{8cm}$
$BC = \sqrt{18}\ \hspace{8cm}$
$CD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (5\ +\ 1)^2 +\ (7\ -\ 1)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((6)^2 + (6)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (36 + 36 )}\ \hspace{8cm}$
$CD = \sqrt{72}\ \hspace{8cm}$
$DA\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (-1\ -\ 2)^2 +\ (1\ +\ 2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((-3)^2 + (3)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (9 + 9 )}\ \hspace{8cm}$
$DA = \sqrt{18}\ \hspace{8cm}$
$∴\ AB = CD \ and\ BC = DA$
$AC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (2\ -\ 5)^2 + (-2\ -\ 7)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((-3)^2 + (-9)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (9 + 81 )}\ \hspace{8cm}$
$AC = \sqrt{90}\ \hspace{8cm}$
$BD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (8\ +\ 1)^2 + (4\ – 1)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((9)^2 + (3)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (81 + 9 )}\ \hspace{8cm}$
$BD = \sqrt{90}\ \hspace{8cm}$
$∴\ AC = BD$
$∴\ The\ given\ complex\ numbers\ form\ a\ rectangle$