# CHAPTER 1.2: APPLICATIONS OF MATRICES AND DETERMINANTS:(Text)

$\color {purple} {Minor\ of\ an\ element\ of\ a\ Matrix}\ \hspace{20cm}$
$Minor\ of\ an\ element\ is\ a\ determinant\ obtained\ by\ deleting\ the\ row\ and\ column\ in\ which\ the\ element\ occurs$
$\color{purple}{Example:\ 1}\ \color{red}{Find\ the\ Minor\ of\ a_1\ in\ the\ matrix}\ \hspace{15cm}$
$A \ = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ \hspace{22cm}$
$Minor\ of\ a_1 =\begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \\ \end{vmatrix}\ = b_2c_3\ -\ b_3c_2\ \hspace{15cm}$
$\color {purple}{Example\ 2:}\ \color{red}{Find\ the\ Minor\ of\ 2\ in\ the\ following\ matrix}\ \hspace{15cm}$
$A \ = \begin{bmatrix} 1 & 0 & -1 \\ 2 & 3 & 4 \\ 7 & 8 & -2 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ \hspace{22cm}$
$\therefore\ Minor\ of\ 2 =\begin{vmatrix} 0 & -1 \\ 8 & -2 \\ \end{vmatrix}\ \hspace{15cm}$
$= 0 + 8\ \hspace{15cm}$
$\boxed{Minor\ of\ 2= 8}\ \hspace{15cm}$
$\color {purple} {Cofactor\ of\ an\ element of\ a\ Matrix}\ \hspace{20cm}$
$Cofactor\ of\ an\ element\ is\ a\ signed\ minor\ of\ that\ element$
$\therefore\ cofactor\ of\ a_{ij} = (-1)^{i\ +\ j}\ minor\ of\ a_{ij}$
$\color {purple}{Example\ 3 :}\ \color {red}{Find\ the\ cofactor\ of\ 4\ in\ the\ following\ matrix}\ \hspace{20cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ = \begin{vmatrix} 1 & 0 & -1 \\ 2 & 3 & 4 \\ 7 & 8 & -2 \\ \end{vmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ \hspace{22cm}$
$cofactor\ of\ 4 = (-1)^{2\ +\ 3}\ \begin{vmatrix} 1 & 0 \\ 7 & 8 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (8 -0)\ \hspace{15cm}$
$= (-1) (8)\ \hspace{15cm}$
$\boxed{cofactor\ of\ 4 = – 8}\ \hspace{15cm}$
$\color {purple} {Example\ 4:} \color {red}{Find\ the\ cofactor\ matrix\ of}\ \begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ A\ =\begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ \begin{vmatrix} 2 & 3 \\ 1 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^2 (4 -3)\ \hspace{15cm}$
$= (1) (1)\ \hspace{15cm}$
$cofactor\ of\ 2 = 1\ \hspace{15cm}$
$cofactor\ of\ 3 = (-1)^{1\ +\ 2}\ \begin{vmatrix} 1 & 3 \\ -1 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (2 + 3)\ \hspace{15cm}$
$= (-1) (5)\ \hspace{15cm}$
$cofactor\ of\ 3 = -5\ \hspace{15cm}$
$cofactor\ of\ 4 = (-1)^{1\ +\ 3}\ \begin{vmatrix} 1 & 2 \\ -1 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (1 + 2)\ \hspace{15cm}$
$= (1) (3)\ \hspace{15cm}$
$cofactor\ of\ 4 = 3\ \hspace{15cm}$
$cofactor\ of\ 1 = (-1)^{2\ +\ 1}\ \begin{vmatrix} 3 & 4 \\ 1 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (6 – 4)\ \hspace{15cm}$
$= (-1) (2)\ \hspace{15cm}$
$cofactor\ of\ 1 = -2\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{2\ +\ 2}\ \begin{vmatrix} 2 & 4 \\ -1 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (4 + 4)\ \hspace{15cm}$
$= (1) (8)\ \hspace{15cm}$
$cofactor\ of\ 2 = 8\ \hspace{15cm}$
$cofactor\ of\ 3 = (-1)^{2\ +\ 3}\ \begin{vmatrix} 2 & 3 \\ -1 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (2 + 3)\ \hspace{15cm}$
$= (-1) (5)\ \hspace{15cm}$
$cofactor\ of\ 3 = -5\ \hspace{15cm}$
$cofactor\ of\ -1 = (-1)^{3\ +\ 1}\ \begin{vmatrix} 3 & 4 \\ 2 & 3 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (9 – 8)\ \hspace{15cm}$
$= (1) (1)\ \hspace{15cm}$
$cofactor\ of\ -1 = 1\ \hspace{15cm}$
$cofactor\ of\ 1 = (-1)^{3\ +\ 2}\ \begin{vmatrix} 2 & 4 \\ 1 & 3 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (6 – 4)\ \hspace{15cm}$
$= (-1) (2)\ \hspace{15cm}$
$cofactor\ of\ 1 = -2\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{3\ +\ 3}\ \begin{vmatrix} 2 & 3 \\ 1 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^6 (4 – 3)\ \hspace{15cm}$
$= (1) (1)\ \hspace{15cm}$
$cofactor\ of\ 2 = 1\ \hspace{15cm}$
$Cofactor\ matrix=\begin{bmatrix} 1 & -5 & 3 \\ -2 & 8 & -5 \\ 1 & -2 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {purple} {Adjoint\ of\ Matrix}\ \hspace{20cm}$
$The\ Adjoint\ of\ a\ square\ matrix\ A\ is\ the\ transpose\ of\ the\ matrix\ which\ is\ formed\ by$$the\ elements\ which\ are\ the\ cofactors\ of\ the\ corresponding\ elements\ of\ the\ determinant\ of\ the\ matrix\ A$
$\color {purple}{Method\ for\ to\ find\ adjoint\ of\ Matrix\ of\ order\ 3\ (order 2)}\ \hspace{20cm}$
$i)\ A\ is\ square\ Matrix\ of\ order\ 3\ (order 2)\ \hspace{5cm}$
$ii)\ Find\ the\ co-factor\ of\ all\ the\ elements\ of\ det\ A\ \hspace{8cm}$
$iii)\ Form\ the\ matrix\ by\ replacing\ all\ the\ elements\ of\ A\ by\ the\ corresponding\ cofactor\ in\ \begin{vmatrix} A \\ \end{vmatrix}$
$iv)\ Then\ take\ the\ Transpose\ of\ that\ matrix,\ then\ we\ get\ adj. A.\ \hspace{8cm}$
$\color {purple} {Example\ 5:}\ \color{red}{Find\ the\ Adjoint\ Matrix\ of}\ \begin{bmatrix} 2 & -1 \\ -5 & 3 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ A\ =\begin{bmatrix} 2 & -1 \\ -5 & 3 \\ \end{bmatrix}\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ (3)\ \hspace{15cm}$
$= (1) (3)\ \hspace{15cm}$
$cofactor\ of\ 2 = \ 3\ \hspace{15cm}$
$cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ (-5)\ \hspace{15cm}$
$= (-1) (-5)\ \hspace{15cm}$
$cofactor\ of\ -1 = 5\ \hspace{15cm}$
$cofactor\ of\ -5 = (-1)^{2\ +\ 1}\ (-1)\ \hspace{15cm}$
$= (-1) (-1)\ \hspace{15cm}$
$cofactor\ of\ -5 = 1\ \hspace{15cm}$
$cofactor\ of\ 3 = (-1)^{2\ +\ 2}\ (2)\ \hspace{15cm}$
$= (1) (2)\ \hspace{15cm}$
$cofactor\ of\ 3 = 2\ \hspace{15cm}$
$\therefore\ cofactor\ matrix\ = \begin{bmatrix} 3 & 5 \\ 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {purple} {Example\ 6:}\ \color{red}{Find\ the\ Adjoint\ of}\ A\ = \begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ A\ =\begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$Cofactor\ matrix=\begin{bmatrix} 1 & -5 & 3 \\ -2 & 8 & -5 \\ 1 & -2 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A=\ (Cofactor Matrix)^T\ =\ \begin{bmatrix} 1 & -2 & 1 \\ -5 & 8 & -2 \\ 3 & -5 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {purple} {Inverse\ of\ Matrix}\ \hspace{20cm}$
$Let\ A\ be\ a\ non\ singular\ matrix\ if\ there\ exists\ a\ square\ matrix\ B\, such\ that\ AB\ =\ BA\ = I$$Where\ I\ is\ the\ unit\ matrix\ of\ same\ order\ then\ B\ is\ called\ the\ the\ Inverse\ of\ A\ and\ it\ is\ denoted\ by\ A^{-1}.$
$\color {black}{Note:}\ \hspace{12cm}$
$i)\ inverse\ of\ matrix\ is\ unique\ \hspace{6cm}$
$ii)\ AB=\ BA\ =\ I\ \hspace{8cm}$
$iii)\ (AB)^{-1} =\ B^{-1}A^{-1}\ \hspace{8cm}$
$\color {black}{Formula\ for\ Inverse\ of\ Matrix:}\ \hspace{12cm}$
$\color {green} {\boxed {A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A}}$
$\color {purple} {Example\ 7:}\ \color{red}{Find\ the\ inverse\ of}\ \begin{bmatrix} 1 & -1 \\ -2 & 0 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Let\ A\ =\begin{bmatrix} 1 & -1 \\ -2 & 0 \\ \end{bmatrix}\ \hspace{15cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ = 1(0)\ -\ (-2) \hspace{13cm}$
$= 0 – 2 \hspace{12cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ = – 2\ \neq {0}\ \hspace{13cm}$
$\therefore\ A^{-1}\ exist\ \hspace{10cm}$
$\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}$
$cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ (0)\ \hspace{15cm}$
$= (-1)^2 (0)\ \hspace{15cm}$
$cofactor\ of\ 1 = 0\ \hspace{15cm}$
$cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ (-2)\ \hspace{15cm}$
$= (-1)^3 (-2)\ \hspace{15cm}$
$cofactor\ of\ – 1 = 2\ \hspace{15cm}$
$cofactor\ of\ -2 = (-1)^{2\ +\ 1}\ (-1)\ \hspace{15cm}$
$= (-1)^3 (-1)\ \hspace{15cm}$
$cofactor\ of\ – 2 = 1\ \hspace{15cm}$
$cofactor\ of\ 0 = (-1)^{2\ +\ 2}\ (1)\ \hspace{15cm}$
$= (-1)^4 (1)\ \hspace{15cm}$
$cofactor\ of\ 0 = 1\ \hspace{15cm}$
$\therefore\ cofactor\ matrix\ = \begin{bmatrix} 0 & 2 \\ 1 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A = \begin{bmatrix} 0 & 1 \\ 2 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}$
$A^{-1} = \frac{1}{-2}\ \begin{bmatrix} 0 & 1 \\ 2 & 1 \\ \end{bmatrix}\ \hspace{2cm}$
$\color {purple} {Example\ 8:}\ \color{red}{Find\ the\ inverse\ of}\ \begin{bmatrix} 1 & -1 & 1 \\ 2 & -3 & -3 \\ 6 & -2 & -1 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Let\ A\ =\begin{bmatrix} 1 & -1 & 1 \\ 2 & -3 & -3 \\ 6 & -2 & -1 \\ \end{bmatrix}\ \hspace{15cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =1\begin{vmatrix} -3 & -3 \\ -2 & -1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & -3 \\ 6 & -1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & -3\\ 6 & -2 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(3\ -\ 6)\ + 1 (-2\ +\ 18) + 1(-4\ +\ 18)\ \hspace{9cm}$
$=1(-3)\ + 1 (16) + 1(14)\ \hspace{13cm}$
$= -3\ + 16 + 14\ \hspace{14cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ = 27\ \neq\ 0\ \hspace{17cm}$
$\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}$
$\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}$
$cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ \begin{vmatrix} -3 & -3 \\ -2 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^2 (3 – 6)\ \hspace{15cm}$
$= (1) (-3)\ \hspace{15cm}$
$cofactor\ of\ 1 = -3\ \hspace{15cm}$
$cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ \begin{vmatrix} 2 & -3 \\ 6 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (-2 + 18)\ \hspace{15cm}$
$= (-1) (16)\ \hspace{15cm}$
$cofactor\ of\ -1 = -16\ \hspace{15cm}$
$cofactor\ of\ 1 = (-1)^{1\ +\ 3}\ \begin{vmatrix} 2 & -3 \\ 6 & -2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (-4 + 18)\ \hspace{15cm}$
$= (1) (14)\ \hspace{15cm}$
$cofactor\ of\ 1 = 14\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{2\ +\ 1}\ \begin{vmatrix} -1 & 1 \\ -2 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (1+ 2)\ \hspace{15cm}$
$= (-1) (3)\ \hspace{15cm}$
$cofactor\ of\ 2 = -3\ \hspace{15cm}$
$cofactor\ of\ -3 = (-1)^{2\ +\ 2}\ \begin{vmatrix} 1 & 1 \\ 6 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (-1- 6)\ \hspace{15cm}$
$= (1) (-7)\ \hspace{15cm}$
$cofactor\ of\ -3 = -7\ \hspace{15cm}$
$cofactor\ of\ -3 = (-1)^{2\ +\ 3}\ \begin{vmatrix} 1 & -1 \\ 6 & -2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (-2+ 6)\ \hspace{15cm}$
$= (-1) (4)\ \hspace{15cm}$
$cofactor\ of\ -3 = -4\ \hspace{15cm}$
$cofactor\ of\ 6 = (-1)^{3\ +\ 1}\ \begin{vmatrix} -1 & 1 \\ -3 & -3 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (3 + 3)\ \hspace{15cm}$
$= (1) (6)\ \hspace{15cm}$
$cofactor\ of\ 6 = 6\ \hspace{15cm}$
$cofactor\ of\ -2 = (-1)^{3\ +\ 2}\ \begin{vmatrix} 1 & 1 \\ 2 & -3 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (-3- 2)\ \hspace{15cm}$
$= (-1) (-5)\ \hspace{15cm}$
$cofactor\ of\ -2 = 5\ \hspace{15cm}$
$cofactor\ of\ -1 = (-1)^{3\ +\ 3}\ \begin{vmatrix} 1 & -1 \\ 2 & -3 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^6 (-3+ 2)\ \hspace{15cm}$
$= (1) (-1)\ \hspace{15cm}$
$cofactor\ of\ -1 = -1\ \hspace{15cm}$
$Cofactor\ matrix=\begin{bmatrix} -3 & -16 & 14 \\ -3 & -7 & -4 \\ 6 & 5 & -1 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A=\begin{bmatrix} -3 & -3 & 6 \\ -16 & -7 & 5 \\ 14 & -4 & -1 \\ \end{bmatrix}\ \hspace{15cm}$
$A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}$
$A^{-1} = \frac{1}{27}\ \begin{bmatrix} -3 & -3 & 6 \\ -16 & -7 & 5 \\ 14 & -4 & -1 \\ \end{bmatrix}\ \hspace{2cm}$
$\color {purple} {Example\ 9:}\ \color{red}{Find\ the\ inverse\ of\ the\ matrix}\ \begin{bmatrix} 3 & 4 & 1 \\ 0 & -1 & 2 \\ 5 & -2 & 6 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Let\ A\ =\begin{bmatrix} 3 & 4 & 1 \\ 0 & -1 & 2 \\ 5 & -2 & 6 \\ \end{bmatrix}\ \hspace{15cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =\ 3\begin{vmatrix} -1 & 2 \\ -2 & 6 \\ \end{vmatrix}\ -\ 4\begin{vmatrix} 0 & 2 \\ 5 & 6 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 0 & -1\\ 5 & -2 \\ \end{vmatrix}\ \hspace{10cm}$
$=\ 3(- 6\ +\ 4)\ -\ 4 (0\ -\ 10)\ +\ 1(0\ +\ 5)\ \hspace{9cm}$
$=\ 3(-2)\ -\ 4 (-10)\ +\ 1(5)\ \hspace{13cm}$
$=\ -6\ +\ 40\ + 5\ \hspace{14cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =\ 39\ \neq\ 0\ \hspace{17cm}$
$\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}$
$\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}$
$cofactor\ of\ 3\ = (-1)^{1\ +\ 1}\ \begin{vmatrix} -1 & 2 \\ -2 & 6 \\ \end{vmatrix}\ \hspace{15cm}$
$cofactor\ of\ 3\ = (-1)^2\ (-6\ +\ 4)\ \hspace{15cm}$
$= (1) (-\ 2)\ \hspace{15cm}$
$cofactor\ of\ 3\ =\ -\ 2\ \hspace{15cm}$
$cofactor\ of\ 4\ = (-1)^{1\ +\ 2}\ \begin{vmatrix} 0 & 2 \\ 5 & 6 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (0\ -\ 10)\ \hspace{15cm}$
$= (-1) (-\ 10)\ \hspace{15cm}$
$cofactor\ of\ 4\ =\ 10\ \hspace{15cm}$
$cofactor\ of\ 1 = (-1)^{1\ +\ 3}\ \begin{vmatrix} 0 & -1 \\ 5 & -2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (0\ +\ 5)\ \hspace{15cm}$
$= (1) (5)\ \hspace{15cm}$
$cofactor\ of\ 1 =\ 5\ \hspace{15cm}$
$cofactor\ of\ 0\ =\ (-1)^{2\ +\ 1}\ \begin{vmatrix} 4 & 1 \\ -2 & 6 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (24\ +\ 2)\ \hspace{15cm}$
$= (-1) (26)\ \hspace{15cm}$
$cofactor\ of\ 0\ =\ -\ 26\ \hspace{15cm}$
$cofactor\ of\ -\ 1\ =\ (-1)^{2\ +\ 2}\ \begin{vmatrix} 3 & 1 \\ 5 & 6 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (18\ -\ 5)\ \hspace{15cm}$
$= (1) (13)\ \hspace{13cm}$
$cofactor\ of\ -\ 1\ =\ 13\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{2\ +\ 3}\ \begin{vmatrix} 3 & 4 \\ 5 & -2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (-\ 6 -\ 20)\ \hspace{15cm}$
$= (-1) (-\ 26)\ \hspace{15cm}$
$cofactor\ of\ 2 =\ 26\ \hspace{15cm}$
$cofactor\ of\ 5\ =\ (-1)^{3\ +\ 1}\ \begin{vmatrix} 4 & 1 \\ -1 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (8\ +\ 1)\ \hspace{15cm}$
$= (1) (9)\ \hspace{15cm}$
$cofactor\ of\ 5\ =\ 9\ \hspace{15cm}$
$cofactor\ of\ -\ 2\ =\ (-1)^{3\ +\ 2}\ \begin{vmatrix} 3 & 1 \\ 0 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (6\ -\ 0)\ \hspace{15cm}$
$= (-1) (6)\ \hspace{15cm}$
$cofactor\ of\ -\ 2\ =\ -\ 6\ \hspace{15cm}$
$cofactor\ of\ 6\ =\ (-1)^{3\ +\ 3}\ \begin{vmatrix} 3 & 4 \\ 0 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^6 (-3\ -\ 0)\ \hspace{15cm}$
$= (1) (-3)\ \hspace{15cm}$
$cofactor\ of\ 6\ =\ -\ 3\ \hspace{15cm}$
$Cofactor\ matrix=\begin{bmatrix} -2 & 10 & 5 \\ -26 & 13 & 26 \\ 9 & -6 & -3 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A=\begin{bmatrix} -2 & -26 & 9 \\ 10 & 13 & – 6 \\ 5 & 26 & -3 \\ \end{bmatrix}\ \hspace{15cm}$
$A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}$
$A^{-1} = \frac{1}{39}\ \begin{bmatrix} -2 & -26 & 9 \\ 10 & 13 & -6 \\ 5 & 26 & -3 \\ \end{bmatrix}\ \hspace{2cm}$
$\color {purple} {Rank\ of\ Matrix:}\ \hspace{20cm}$
$Let\ A\ be\ any\ m×n\ matrix.\ The\ order\ of\ the\ largest\ square\ sub\ matrix\ of\ A\ whose\ determinant$$has\ a\ non\ -\ zero\ value\ is\ known\ as\ the\ rank\ of\ the\ matrix\ A$$and\ is\ denoted\ by\ \rho(A)$
$\color {purple} {Example\ 10:}\ \color{red}{Find\ the\ rank\ of\ the\ matrix}\ \begin{bmatrix} 5 & 2 \\ 6 & 3 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ A\ =\begin{bmatrix} 5 & 2 \\ 6 & 3 \\ \end{bmatrix}\ \hspace{17cm}$
$Order\ of\ A = 2 × 2\ \hspace{15cm}$
$\therefore \ \rho(A)\ \leq 2 \hspace{15cm}$
$The\ higher\ order\ of\ minor\ of\ A = 2\ \hspace{10cm}$
$The\ minor\ is\ \begin{vmatrix} 5 & 2 \\ 6 & 3 \\ \end{vmatrix}\ =\ 15 – 12\ =\ 3\ \neq\ 0$
$\boxed{\therefore\ Rank\ of\ A\ =\ 2}\ \hspace{15cm}$
$\color {purple} {Example\ 11:}\ \color {red}{Find\ the\ rank\ of\ the\ matrix}\ \begin{bmatrix} 1 & 2 \\ 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ A\ =\begin{bmatrix} 1 & 2 \\ 1 & 2 \\ \end{bmatrix}\ \hspace{17cm}$
$Order\ of\ A = 2 × 2\ \hspace{15cm}$
$\therefore \ \rho(A)\ \leq 2 \hspace{15cm}$
$The\ higher\ order\ of\ minor\ of\ A = 2\ \hspace{10cm}$
$The\ minor\ is\ \begin{vmatrix} 1 & 2 \\ 1 & 2 \\ \end{vmatrix}\ =\ 6- 6\ =\ 0\ \hspace{5cm}$
$\therefore \ \rho(A)\ \neq 2 \hspace{15cm}$
$To\ find\ at\ least\ one\ non\ zero\ first\ order\ minor\ i.e.\ to\ find\ at\ least\ one\ non\ zero\ element$$of\ A\ ,\ non\ zero\ element\ exist\ in\ A\ .$
$Rank\ =\ 1\ \hspace{15cm}$
$\boxed{\therefore \ \rho(A)\ =\ 1} \hspace{15cm}$
$\color {purple} {Example\ 12:}\ \color {red}{Find\ the\ rank\ of\ the\ matrix}\ \begin{bmatrix} -1 & 2 & 3 \\ 0 & 3 & 1 \\ 4 & 5 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ A\ =\begin{bmatrix} -1 & 2 & 3 \\ 0 & 3 & 1 \\ 4 & 5 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$Order\ of\ A = 3 × 3\ \hspace{15cm}$
$\therefore \ \rho(A)\ \leq 3 \hspace{15cm}$
$The\ higher\ order\ of\ minor\ of\ A = 3\ \hspace{10cm}$
$The\ minor\ is\ =\ \begin{vmatrix} -1 & 2 & 3 \\ 0 & 3 & 1 \\ 4 & 5 & 2 \\ \end{vmatrix}\ \hspace{5cm}$
$=-1\begin{vmatrix} 3 & 1 \\ 5 & 2 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 0 & 1 \\ 4 & 2 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} 0 & 3\\ 4 & 5 \\ \end{vmatrix}\ \hspace{10cm}$
$=-1(6\ -\ 5)\ – 2 (0\ -\ 4) + 3(0\ -\ 12)\ \hspace{9cm}$
$=-1(1)\ – 2 (-4) + 3(-12)\ \hspace{13cm}$
$= -1\ + 8 – 36\ =\ -29\ \neq\ 0\ \hspace{14cm}$
$\boxed{\therefore\ rank\ of\ A=\ \rho(A)\ =\ 3}\ \hspace{15cm}$
$\color {purple} {Example\ 13:}\ \color {red}{Find\ the\ rank\ of\ the\ matrix}\ \begin{pmatrix} 1 & 2 & -1 & 3 \\ 2 & 4 & -4 & 7 \\ -1 & -2 & -2 & -2 \\ \end{pmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ A\ =\begin{pmatrix} 1 & 2 & -1 & 3 \\ 2 & 4 & -4 & 7 \\ -1 & -2 & -2 & -2 \\ \end{pmatrix}\ \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 & -1 \\ 2 & 4 & -4 \\ -1 & -2 & -2 \\ \end{vmatrix}\ \hspace{15cm}$
$=1\begin{vmatrix} 4 & -4 \\ -2 & -2 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & -4 \\ -1 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 4\\ -1 & -2 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(-8\ -\ 8)\ – 2 (-4\ -\ 4) – 1(-4\ +\ 4)\ \hspace{9cm}$
$=1(-16)\ – 2 (-8) -1(0)\ \hspace{13cm}$
$= -16\ +\ 16\ -\ 0\ =\ 0\ \hspace{14cm}$
$A_2\ =\begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 7 \\ -1 & -2 & -2 \\ \end{vmatrix}\ \hspace{15cm}$
$=1\begin{vmatrix} 4 & 7 \\ -2 & -2 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & 7 \\ -1 & -2 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 2 & 3\\ 1 & 3 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(-8\ +\ 14)\ – 2 (-4\ +\ 7) +\ 3(-4\ +\ 4)\ \hspace{9cm}$
$=1(6)\ – 2 (3)\ +\ 3(0)\ \hspace{13cm}$
$=\ 6\ -\ 6\ + 0\ =\ 0\ \hspace{14cm}$
$A_3\ =\begin{bmatrix} 1 & -1 & 3 \\ 2 & -4 & 7 \\ -1 & -2 & -2 \\ \end{bmatrix}\ \hspace{15cm}$
$=\ 1\begin{vmatrix} -4 & 7 \\ -2 & -2 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & 7 \\ -1 & -2 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} 2 & -4\\ -1 & -2 \\ \end{vmatrix}\ \hspace{10cm}$
$=\ 1(8\ +\ 14)\ +\ 1 (-4\ +\ 7)\ + 3(-4\ -\ 4)\ \hspace{9cm}$
$=\ 1(22)\ +\ 1 (3)\ +\ 3(-8)\ \hspace{13cm}$
$= 22\ +\ 3\ -\ 24\ =\ 1\ \neq\ 0\ \hspace{14cm}$
$A_4\ =\begin{bmatrix} 2 & -1 & 3 \\ 4 & -4 & 7 \\ -2 & -2 & -2 \\ \end{bmatrix}\ \hspace{15cm}$
$=\ 2\begin{vmatrix} -4 & 7 \\ -2 & -2 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 4 & 7 \\ -2 & -2 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} 4 & -4\\ -2 & -2 \\ \end{vmatrix}\ \hspace{10cm}$
$=\ 2(8\ +\ 14)\ + 1 (-8\ +\ 14)\ +\ 3(-8\ -\ 8)\ \hspace{9cm}$
$=\ 2(22)\ +\ 1 (6)\ +\ 3(-16)\ \hspace{13cm}$
$=\ 44\ +\ 6\ -\ 48\ =\ 2\ \neq\ 0\ \hspace{14cm}$
$\boxed{\therefore\ rank\ of\ A=\ \rho(A)\ =\ 3}\ \hspace{15cm}$
$\color {purple} {Example\ 14:}\ \color {red}{Find\ the\ rank\ of\ the\ matrix}\ \begin{bmatrix} 1 & 2 & 3 & 2 \\ 2 & 3 & 5 & 1 \\ 1 & 3 & 4 & 5 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ A\ =\begin{bmatrix} 1 & 2 & 3 & 2 \\ 2 & 3 & 5 & 1 \\ 1 & 3 & 4 & 5 \\ \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{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 5 \\ 1 & 3 & 4 \\ \end{bmatrix}\ \hspace{15cm}$
$=1\begin{vmatrix} 3 & 5 \\ 3 & 4 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & 5 \\ 1 & 4 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} 2 & 3\\ 1 & 3 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(12\ -\ 15)\ – 2 (8\ -\ 5) + 3(6\ -\ 3)\ \hspace{9cm}$
$=1(-3)\ – 2 (3) + 3(3)\ \hspace{13cm}$
$= -3\ – 6 + 9\ =\ 0\ \hspace{14cm}$
$A_2\ =\begin{bmatrix} 1 & 2 & 2 \\ 2 & 3 & 1 \\ 1 & 3 & 5 \\ \end{bmatrix}\ \hspace{15cm}$
$=1\begin{vmatrix} 3 & 1 \\ 3 & 5 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & 1 \\ 1 & 5 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 2 & 3\\ 1 & 3 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(15\ -\ 3)\ – 2 (10\ -\ 1) + 2(6\ -\ 3)\ \hspace{9cm}$
$=1(12)\ – 2 (9) + 2(3)\ \hspace{13cm}$
$= 12\ – 18 + 6\ =\ 0\ \hspace{14cm}$
$A_3\ =\begin{bmatrix} 2 & 3 & 2 \\ 3 & 5 & 1 \\ 3 & 4 & 5 \\ \end{bmatrix}\ \hspace{15cm}$
$=2\begin{vmatrix} 5 & 1 \\ 4 & 5 \\ \end{vmatrix}\ -\ 3\begin{vmatrix} 3 & 1 \\ 3 & 5 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 3 & 5\\ 3 & 4 \\ \end{vmatrix}\ \hspace{10cm}$
$=2(25\ -\ 4)\ – 3 (15\ -\ 3) + 2(12\ -\ 15)\ \hspace{9cm}$
$=2(21)\ – 3 (12) + 2(-3)\ \hspace{13cm}$
$= 42\ – 36 – 6\ =\ 0\ \hspace{14cm}$
$A_4\ =\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 1 & 4 & 5 \\ \end{bmatrix}\ \hspace{15cm}$
$=1\begin{vmatrix} 5 & 1 \\ 4 & 5 \\ \end{vmatrix}\ -\ 3\begin{vmatrix} 2 & 1 \\ 1 & 5 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 2 & 5\\ 1 & 4 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(25\ -\ 4)\ – 3 (10\ -\ 1) + 2(8\ -\ 5)\ \hspace{9cm}$
$=1(21)\ – 3 (9) + 2(3)\ \hspace{13cm}$
$= 21\ – 27 + 6\ =\ 0\ \hspace{14cm}$
$All\ third\ order\ minors\ vanish\ \hspace{8cm}$
$\therefore\ \rho(A)\ \lt 3\ \hspace{15cm}$
$To\ find\ at\ least\ a\ non\ zero\ of\ order\ 2 × 2\ \ \hspace{10cm}$
$\begin{vmatrix} 1 & 3 \\ 2 & 5\\ \end{vmatrix}\ =\ 5- 6\ =\ -1\ \neq\ 0\ \hspace{5cm}$
$\therefore\ A\ has\ at\ least\ one\ non\ zero\ minor\ of\ order\ 2\ \hspace{5cm}$
$\boxed{\therefore\ \rho(A)\ =2}\ \hspace{15cm}$
$\color {purple} {SOLUTION\ OF\ SIMULTANEOUS\ EQUATIONS\ USING\ CRAMERS\ RULE}\ \hspace{20cm}$
$a_1x\ + b_1y\ +\ c_1z\ = d_1\ —– (1)\ \hspace{20cm}$
$a_2x\ + b_2y\ +\ c_2z\ = d_2\ \hspace{20cm}$
$a_3x\ + b_3y\ +\ c_3z\ = d_3\ \hspace{20cm}$
$\color {black}{Solution:}\ to\ find\ x,\ y,\ z\ \hspace{20cm}$
$Step\ 1:\ \Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix}\ \hspace{20cm}$
$Step\ 2:\ \Delta_x = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \\ \end{vmatrix}\ \hspace{20cm}$
$Step\ 3:\ \Delta_y= \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \\ \end{vmatrix}\ \hspace{20cm}$
$Step\ 4:\ \Delta_z = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \\ \end{vmatrix}\ \hspace{20cm}$
$Solution\ is\ x=\ \frac{\Delta_x}{\Delta}.\ y=\ \frac{\Delta_y}{\Delta},\ z=\ \frac{\Delta_z}{\Delta}\ \hspace{15cm}$
$\color {purple}{Example\ 15:}\ \color{red}{Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}$
$x + 2y – z=-1,\ 3x + 8y + 2z = 28\ and\ 4x + 9y – z = 14\ \hspace{15cm}$
$\color {blue}{Solution:}\ \hspace{22cm}$
$x + 2y – z = – 1\ ————— (1) \hspace{10cm}$
$3x + 8y + 2z = 28\ \hspace{15cm}$
$4x + 9y – z = 14\ \hspace{15cm}$
$\Delta = \begin{vmatrix} 1 & 2 & -1 \\ 3 & 8 & 2 \\ 4 & 9 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta =1\begin{vmatrix} 8 & 2 \\ 9 & – 1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 3 & 2 \\ 4 & -1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 3 & 8\\ 4 & 9 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta =1(-8\ -\ 18)\ – 2 (-3\ -\ 8) – 1(27\ -\ 32)\ \hspace{9cm}$
$\Delta =1(-26)\ – 2 (-11) – 1(-5)\ \hspace{13cm}$
$\Delta =-26\ +22 + 5\ \hspace{14cm}$
$\boxed{\Delta =1}\ \hspace{17cm}$
$\Delta_x = \begin{vmatrix} -1 & 2 & -1 \\ 28 & 8 & 2 \\ 14 & 9 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_x =-1\begin{vmatrix} 8 & 2 \\ 9 & -1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 28 & 2 \\ 14 & -1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 28 & 8\\ 14 & 9 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_x =-1(-8\ -\ 18)\ – 2 (-28\ -\ 28) – 1(252\ -\ 112)\ \hspace{9cm}$
$\Delta_x =-1(-26)\ – 2 (-56) -1 (140)\ \hspace{13cm}$
$\Delta_x =26\ + 112 – 140\ \hspace{14cm}$
$\boxed{\Delta_x = -2}\ \hspace{17cm}$
$\Delta_y = \begin{vmatrix} 1 & -1 & -1 \\ 3 & 28 & 2 \\ 4 & 14 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_y =1\begin{vmatrix} 28 & 2 \\ 14 & -1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 3 & 2 \\ 4 & -1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 3 & 28\\ 4 & 14 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_y =1(-28\ -\ 28)\ + 1 (-3\ -\ 8) – 1(42\ -\ 112)\ \hspace{9cm}$
$\Delta_y =1(-56)\ + 1 (-11) – 1(-70)\ \hspace{13cm}$
$\Delta_y = -56\ -11\ + 70\ \hspace{14cm}$
$\boxed{\Delta_y =3}\ \hspace{17cm}$
$\Delta_z = \begin{vmatrix} 1 & 2 & – 1 \\ 3 & 8 & 28 \\ 4 & 9 & 14 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_z =1\begin{vmatrix} 8 & 28 \\ 9 & 14 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 3 & 28 \\ 4 & 14 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 3 & 8\\ 4 & 9 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_z =1(112\ -\ 252)\ – 2 (42\ -\ 112) – 1(27\ -\ 32)\ \hspace{9cm}$
$\Delta_z =1(-140)\ – 2 (-70) – 1(-5)\ \hspace{13cm}$
$\Delta_z =-140\ + 140 + 5\ \hspace{14cm}$
$\boxed{\Delta_z =5}\ \hspace{17cm}$
$The\ Solution\ is\ \hspace{20cm}$
$x=\ \frac{\Delta_x}{\Delta} =\ \frac{-2}{1} =\ -2\ \hspace{20cm}$
$y=\ \frac{\Delta_y}{\Delta} =\ \frac{3}{1} =\ 3\ \hspace{20cm}$
$z=\ \frac{\Delta_z}{\Delta} =\ \frac{5}{1} =\ 5\ \hspace{20cm}$
$For\ cross\ verification\ \hspace{20cm}$
$Put\ x =-2\ y = 3\ z = 5\ in\ equation (1)\ \hspace{18cm}$
$LHS = -2 + 2(3) – 5$$= -2 + 6 – 5 = -1$$= RHS$
$\color {violet}{Example\ 16:}\ \color {red} {Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}$
$4\ x\ +\ y\ +\ z\ =\ 6,\ 2\ x\ -\ y\ -\ 2\ z\ =\ -\ 6\ and\ x\ +\ y\ +\ z\ =\ 3\ \hspace{15cm}$
$\color {blue}{Solution:}\ \hspace{22cm}$
$4\ x\ +\ y\ +\ z\ =\ 6\ ——————-(1)\ \hspace{6cm}$
$2\ x\ -\ y\ -\ 2\ z\ =\ -\ 6\ \hspace{15cm}$
$x\ +\ y\ +\ z\ =\ 3\ \hspace{15cm}$
$\Delta = \begin{vmatrix} 4 & 1 & 1 \\ 2 & -1 & -2\\ 1 & 1 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta =4\begin{vmatrix} -1 & -2 \\ 1 & 1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & -2 \\ 1 & 1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & -1\\ 1 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta =4(-1\ +\ 2)\ – 1 (2\ +\ 2)\ +\ 1(2\ +\ 1)\ \hspace{9cm}$
$\Delta\ =\ 4(1)\ – 1 (4)\ +\ 1(7)\ \hspace{13cm}$
$\Delta =4\ -\ 4\ +\ 3\ \hspace{14cm}$
$\boxed{\Delta\ =\ 3}\ \hspace{17cm}$
$\Delta_x = \begin{vmatrix} 6 & 1 & 1 \\ -6 & -1 & -2 \\ 3 & 1 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_x =6\begin{vmatrix} -1 & -2 \\ 1 & 1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} -6 & -2 \\ 3 & 1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} -6 & -1\\ 3 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_x\ =\ 6(-1\ +\ 2) – 1 (\ -6\ +\ 6)\ +\ 1(-6\ +\ 3)\ \hspace{9cm}$
$\Delta_x\ =\ 6(1)\ – 1 (0)\ +\ 1(-3)\ \hspace{13cm}$
$\Delta_x = 6\ +\ 0\ -3\ \hspace{14cm}$
$\boxed{\Delta_x\ =\ 3}\ \hspace{17cm}$
$\Delta_y = \begin{vmatrix} 4 & 6 & 1 \\ 2 & -6 & -2 \\ 1 & 3 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_y\ =\ 4\begin{vmatrix} -6 & -2 \\ 3 & 1\\ \end{vmatrix}\ -\ 6\begin{vmatrix} 2 & -2 \\ 1 & 1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & -6\\ 1 & 3 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_y\ =\ 4(-\ 6\ +\ 6)\ -\ 6 (2\ +\ 2)\ +\ 1(6\ +\ 6)\ \hspace{9cm}$
$\Delta_y\ =\ 4(0)\ -\ 6 (4)\ +\ 1(12)\ \hspace{13cm}$
$\Delta_y\ =\ -\ 0\ -\ 24\ +\ 12\ \hspace{14cm}$
$\boxed{\Delta_y\ =\ -12}\ \hspace{17cm}$
$\Delta_z = \begin{vmatrix} 4 & 1 & 6 \\ 2 & -1 & -6 \\ 1 & 1 & 3 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_z\ =\ 4\begin{vmatrix} -1 & -6 \\ 1 & 3 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & -6 \\ 1 & 3 \\ \end{vmatrix}\ +\ 6\begin{vmatrix} 2 & – 1\\ 1 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_z\ =\ 4(-\ 3\ +\ 6)\ -\ 1 (6\ +\ 6)\ +\ 6(2\ +\ 1)\ \hspace{9cm}$
$\Delta_z\ =\ 4(3)\ -\ 1 (12)\ +\ 6(3)\ \hspace{13cm}$
$\Delta_z\ =\ 12\ -\ 12\ +\ 18\ \hspace{14cm}$
$\boxed{\Delta_z\ =\ 18}\ \hspace{17cm}$
$The\ Solution\ is\ \hspace{20cm}$
$x=\ \frac{\Delta_x}{\Delta} =\ \frac{3}{3} =\ 1\ \hspace{20cm}$
$y=\ \frac{\Delta_y}{\Delta} =\ \frac{-12}{3} =\ -4\ \hspace{20cm}$
$z=\ \frac{\Delta_z}{\Delta} =\ \frac{18}{3} =\ 6\ \hspace{20cm}$
$For\ cross\ verification\ \hspace{20cm}$
$Put\ x\ =\ 1,\ y\ =\ -4\ and\ z = 6\ in\ equation (1)\ \hspace{18cm}$
$LHS\ =\ 4(1) – 4 + 6$$= 4 – 4 + 6 = 6$$= RHS$

### Exercise Problems

$\LARGE{\color {purple} {PART- A}}$
$1.\ \color {red}{Find\ the\ cofactor\ of\ 3\ in\ the\ matrix}\ \begin{bmatrix} 1 & 2 & 0 \\ -1 & 3 & 4 \\ 5 & 6 & 7 \\ \end{bmatrix}\ \hspace{15cm}$
$2.\ \color {red}{Find\ the\ Adjoint\ matrix\ of}\ \begin{bmatrix} 2 & 2 \\ 3 & -5 \\ \end{bmatrix}\ \hspace{15cm}$
$3.\ \color {red}{Find\ the\ Adjoint\ matrix\ of}\ \begin{bmatrix} 3 & -4 \\ 2 & 5 \\ \end{bmatrix}\ \hspace{15cm}$
$4.\ \color {red}{Find\ the\ Adjoint\ matrix\ of}\ \begin{bmatrix} 1 & 4 \\ 5 & -2 \\ \end{bmatrix}\ \hspace{15cm}$
$5.\ \color {red}{Find\ the\ Adjoint\ matrix\ of}\ \begin{bmatrix} 5 & -6 \\ 3 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$6.\ \color {red}{Find\ the\ rank\ of}\ \begin{bmatrix} 3 & -4 \\ -6 & 8 \\ \end{bmatrix}\ \hspace{18cm}$
$\LARGE{\color {purple} {PART- B}}$
$7.\ \color {red}{Find\ the\ inverse\ of}\ \begin{bmatrix} 2 & -1 \\ 4 & 5 \\ \end{bmatrix}\ \hspace{18cm}$
$8.\ \color {red}{Find\ the\ inverse\ of}\ \begin{bmatrix} 2 & 1 \\ 1 & 2 \\ \end{bmatrix}\ \hspace{18cm}$
$\color {purple} { 9\ .}\ \color {red}{Find\ the\ rank\ of}\ \ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & -1 & 4 \\ \end{bmatrix}\ \hspace{15cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {red} {10.\ By\ using\ Cramers\ Rule\,\ Solve\ the\ equations}\ \hspace{20cm}$
$x\ +\ y\ +\ z\ =\ 2,\ 2x\ -\ y\ -\ 2z\ =\ -1\ and\ x\ -\ 2y\ -\ z\ =\ 1\ \hspace{15cm}$
$\color{red}{11.\ Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}$
$3x\ -\ y\ +\ 2z\ =\ 8,\ x\ +\ y\ +\ z\ =\ 2\ and\ 2x\ +\ y\ -\ z\ =\ -\ 1\ \hspace{15cm}$
$12.\ \color {red}{Find\ the\ inverse\ of}\ \begin{bmatrix} 2 & 3 & 4 \\ 4 & 3 & 1 \\ 1 & 2 & 4 \\ \end{bmatrix}\ \hspace{15cm}$
$13.\ \color {red}{Find\ the\ inverse\ of}\ \begin{bmatrix} 1 & 2 & -1\\ 3 & 8 & 2 \\ 4 & 9 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$14.\ \color{red}{Find\ the\ inverse\ of}\ \begin{bmatrix} 1 & 1 & -\ 1 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {purple} {15}\ \color {red}{Find\ the\ rank\ of\ the\ matrix}\ \begin{pmatrix} 1 & 2 & 3 & 1 \\ 2 & 3 & 4 & 2 \\ 3 & 4 & 5 & 1 \\ \end{pmatrix}\ \hspace{15cm}$