# APPLICATIONS OF MATRICES AND DETERMINANTS:(Revision)

$\color {black} {1.}\ Solve\ the\ following\ equations\ using\ Cramers\ Rule\ \hspace{20cm}$
$x\ +\ y\ +\ z\ =\ 3,\ 2\ x\ -\ y\ +\ z\ =\ 2\ and\ 3\ x\ +\ 2\ y\ -\ 2\ z\ =\ 3\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{20cm}$
$x\ +\ y\ +\ z=3\ ——————-(1)\ \hspace{6cm}$
$2\ x\ -\ y\ +\ z\ =\ 2\ \hspace{15cm}$
$3\ x\ +\ 2\ y\ -\ 2\ z\ =\ 3\ \hspace{15cm}$
$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 3 & 2 & -2 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta =1\begin{vmatrix} -1 & 1 \\ 2 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 1 \\ 3 & -2 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & -1\\ 3 & 2 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta =1(2\ -\ 2)\ – 1 (-4\ -\ 3)\ +\ 1(4\ +\ 3)\ \hspace{9cm}$
$\Delta\ =\ 1(0)\ – 1 (-\ 7)\ +\ 1(7)\ \hspace{13cm}$
$\Delta =0\ +\ 7\ +\ 7\ \hspace{14cm}$
$\Delta\ =\ 14\ \hspace{17cm}$
$\Delta_x = \begin{vmatrix} 3 & 1 & 1 \\ 2 & -1 & 1 \\ 3 & 2 & -2 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_x =3\begin{vmatrix} -1 & 1 \\ 2 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 1 \\ 3 & -2 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & -1\\ 3 & 2 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_x\ =\ 3(2\ -\ 2) – 1 (-\ 4\ -\ (3))\ +\ 1(4\ +\ 3)\ \hspace{9cm}$
$\Delta_x\ =\ 3(0)\ – 1 (-\ 7)\ +\ 1(7)\ \hspace{13cm}$
$\Delta_x = 0\ +\ 7\ +\ 7\ \hspace{14cm}$
$\Delta_x\ =\ 14\ \hspace{17cm}$
$\Delta_y = \begin{vmatrix} 1 & 3 & 1 \\ 2 & 2 & 1 \\ 3 & 3 & -2 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_y\ =\ 1\begin{vmatrix} 2 & 1 \\ 3 & -2 \\ \end{vmatrix}\ -\ 3\begin{vmatrix} 2 & 1 \\ 3 & -2 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & 2\\ 3 & 3 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_y\ =\ 1(-\ 4\ -\ 3)\ -\ 3 (-4\ -\ 3)\ +\ 1(6\ -\ 6)\ \hspace{9cm}$
$\Delta_y\ =\ 1(- \ 7)\ -\ 3 (-\ 7)\ +\ 1(0)\ \hspace{13cm}$
$\Delta_y\ =\ -\ 7\ +\ 21\ +\ 0\ \hspace{14cm}$
$\Delta_y\ =\ 14\ \hspace{17cm}$
$\Delta_z = \begin{vmatrix} 1 & 1 & 3 \\ 2 & -1 & 2 \\ 3 & 2 & 3 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_z\ =\ 1\begin{vmatrix} -1 & 2 \\ 2 & 3 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 2 \\ 3 & 3 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} 2 & – 1\\ 3 & 2 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_z\ =\ 1(-\ 3\ -\ 4)\ -\ 1 (6\ -\ 6)\ +\ 3(4\ +\ 3)\ \hspace{9cm}$
$\Delta_z\ =\ 1(-\ 7)\ -\ 1 (0)\ +\ 3(7)\ \hspace{13cm}$
$\Delta_z\ =\ -\ 7\ -\ 0\ +\ 21\ \hspace{14cm}$
$\Delta_z\ =\ 14\ \hspace{17cm}$
$The\ Solution\ is\ \hspace{20cm}$
$x=\ \frac{\Delta_x}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}$
$y=\ \frac{\Delta_y}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}$
$z=\ \frac{\Delta_z}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}$
$For\ cross\ verification\ \hspace{20cm}$
$Put\ x\ =\ 1,\ y\ =\ 2\ and\ z = 3\ in\ equation (1)\ \hspace{18cm}$
$LHS\ =\ 1 + 2 – 3$$= 3 + 2 – 3 = 2$$= RHS$
$2.\ Find\ the\ inverse\ of\ \begin{bmatrix} 2 & 3 & 4 \\ 4 & 3 & 1 \\ 1 & 2 & 4 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {black}{Solution:}\ A\ =\begin{bmatrix} 2 & 3 & 4 \\ 4 & 3 & 1 \\ 1 & 2 & 4 \\ \end{bmatrix}\ \hspace{15cm}$
$Let\ A=\begin{bmatrix} 2 & 3 & 4 \\ 4 & 3 & 1 \\ 1 & 2 & 4 \\ \end{bmatrix}\ \hspace{15cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =2\begin{vmatrix} 3 & 1 \\ 2 & 4 \\ \end{vmatrix}\ -\ 3\begin{vmatrix} 4 & 1 \\ 1 & 4 \\ \end{vmatrix}\ +\ 4\begin{vmatrix} 4 & 3\\ 1 & 2 \\ \end{vmatrix}\ \hspace{10cm}$
$=2(12\ -\ 2)\ – 3 (16\ -\ 1) + 4(8\ -\ 3)\ \hspace{9cm}$
$=2(10)\ – 3 (15) + 4(5)\ \hspace{13cm}$
$=20\ -45 + 20\ \hspace{14cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =-5\ \neq\ 0\ \hspace{17cm}$
$\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}$
$\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}$
$cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ \begin{vmatrix} 3 & 1 \\ 2 & 4 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^2 (12 – 2)\ \hspace{15cm}$
$= (1) (10)\ \hspace{15cm}$
$cofactor\ of\ 2 = 10\ \hspace{15cm}$
$cofactor\ of\ 3 = (-1)^{1\ +\ 2}\ \begin{vmatrix} 4 & 1 \\ 1 & 4 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (16 – 1)\ \hspace{15cm}$
$= (-1) (15)\ \hspace{15cm}$
$cofactor\ of\ 3 = -15\ \hspace{15cm}$
$cofactor\ of\ 4 = (-1)^{1\ +\ 3}\ \begin{vmatrix} 4 & 3 \\ 1 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (8- 3)\ \hspace{15cm}$
$= (1) (5)\ \hspace{15cm}$
$cofactor\ of\ 4 = 5\ \hspace{15cm}$
$cofactor\ of\ 4 = (-1)^{2\ +\ 1}\ \begin{vmatrix} 3 & 4 \\ 2 & 4 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (12- 8)\ \hspace{15cm}$
$= (-1) (4)\ \hspace{15cm}$
$cofactor\ of\ 4 = -4\ \hspace{15cm}$
$cofactor\ of\ 3 = (-1)^{2\ +\ 2}\ \begin{vmatrix} 2 & 4 \\ 1 & 4 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (8- 4)\ \hspace{15cm}$
$= (1) (4)\ \hspace{15cm}$
$cofactor\ of\ 3 = 4\ \hspace{15cm}$
$cofactor\ of\ 1 = (-1)^{2\ +\ 3}\ \begin{vmatrix} 2 & 3 \\ 1 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (4- 3)\ \hspace{15cm}$
$= (-1) (1)\ \hspace{15cm}$
$cofactor\ of\ 1 = -1\ \hspace{15cm}$
$cofactor\ of\ 1 = (-1)^{3\ +\ 1}\ \begin{vmatrix} 3 & 4 \\ 3 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (3- 12)\ \hspace{15cm}$
$= (1) (-9)\ \hspace{15cm}$
$cofactor\ of\ 1 = -9\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{3\ +\ 2}\ \begin{vmatrix} 2 & 4 \\ 4 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (2- 16)\ \hspace{15cm}$
$= (-1) (-14)\ \hspace{15cm}$
$cofactor\ of\ 2 = 14\ \hspace{15cm}$
$cofactor\ of\ 4 = (-1)^{3\ +\ 3}\ \begin{vmatrix} 2 & 3 \\ 4 & 3 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^6 (6- 12)\ \hspace{15cm}$
$= (1) (-6)\ \hspace{15cm}$
$cofactor\ of\ 4 = -6\ \hspace{15cm}$
$Cofactor\ matrix=\begin{bmatrix} 10 & -15 & 5 \\ -4 & 4 & -1 \\ -9 & 14 & -6 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A=\begin{bmatrix} 10 & -4 & -9 \\ -15 & 4 & 14 \\ 5 & -1 & -6 \\ \end{bmatrix}\ \hspace{15cm}$
$A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}$
$A^{-1} = \frac{1}{-5}\ \begin{bmatrix} 10 & -4 & -9 \\ -15 & 4 & 14 \\ 5 & -1 & -6 \\ \end{bmatrix}\ \hspace{2cm}$