# APPLICATIONS OF MATRICES AND DETERMINANTS:(Exercise Problems)

$\LARGE{\color {purple} {PART- A}}$
$1.\ \color {red}{Find\ the\ cofactor\ of\ 3\ in\ the\ matrix\ \begin{bmatrix} 1 & 2 & 0 \\ -1 & 3 & 4 \\ 5 & 6 & 7 \\ \end{bmatrix}}\ \hspace{15cm}$
$\color {black}{Solution:}\ A\ =\begin{bmatrix} 1 & 2 & 0 \\ -1 & 3 & 4 \\ 5 & 6 & 7 \\ \end{bmatrix}\ \hspace{15cm}$
$cofactor\ of\ 3 = (-1)^{1\ +\ 1}\ \begin{vmatrix} 1 & 0 \\ 5 & 7 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^2 [1(7)\ -\ 5(0)]\ \hspace{15cm}$
$= 1 (7\ -\ 0)\ \hspace{15cm}$
$=\ 1 (7)\ \hspace{15cm}$
$cofactor\ of\ 3 =\ 7\ \hspace{15cm}$
$2.\ \color {red}{Find\ the\ Adjoint\ matrix\ of\ \begin{bmatrix} 2 & 2 \\ 3 & -5 \\ \end{bmatrix}}\ \hspace{15cm}$
$\color {black}{Solution:}\ A\ =\begin{bmatrix} 2 & 2 \\ 3 & -5 \\ \end{bmatrix}\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ (-5)\ \hspace{15cm}$
$=\ {(-1)}^2 (-5)\ \hspace{15cm}$
$= (1) (-5)\ \hspace{15cm}$
$cofactor\ of\ 2 =\ -5\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{1\ +\ 2}\ (3)\ \hspace{15cm}$
$=\ {(-1)}^3 (3)\ \hspace{15cm}$
$= (-1) (3)\ \hspace{15cm}$
$cofactor\ of\ 2\ =\ -\ 3\ \hspace{15cm}$
$cofactor\ of\ 3 = (-1)^{2\ +\ 1}\ (2)\ \hspace{15cm}$
$=\ {(-1)}^3 (2)\ \hspace{15cm}$
$= (-1) (2)\ \hspace{15cm}$
$cofactor\ of\ 3 = -2\ \hspace{15cm}$
$cofactor\ of\ -5 = (-1)^{2\ +\ 2}\ (2)\ \hspace{15cm}$
$=\ {(-1)}^4 (2)\ \hspace{15cm}$
$= (1) (2)\ \hspace{15cm}$
$cofactor\ of\ -5 = 2\ \hspace{15cm}$
$\therefore\ cofactor\ matrix\ = \begin{bmatrix} -5 & -3 \\ -2 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A = \begin{bmatrix} -5 & -2 \\ -3 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$3.\ \color {red}{Find\ the\ Adjoint\ matrix\ of\ \begin{bmatrix} 3 & -4 \\ 2 & 5 \\ \end{bmatrix}}\ \hspace{15cm}$
$\color {black}{Solution:}\ A\ =\begin{bmatrix} 3 & -4 \\ 2 & 5 \\ \end{bmatrix}\ \hspace{15cm}$
$cofactor\ of\ 3\ =\ (-1)^{1\ +\ 1}\ (5)\ \hspace{15cm}$
$=\ {(-1)}^2 (5)\ \hspace{15cm}$
$= (1) (5)\ \hspace{15cm}$
$cofactor\ of\ 3 =\ 5\ \hspace{15cm}$
$cofactor\ of\ -4 = (-1)^{1\ +\ 2}\ (2)\ \hspace{15cm}$
$=\ {(-1)}^3 (2)\ \hspace{15cm}$
$= (-1) (2)\ \hspace{15cm}$
$cofactor\ of\ -4\ =\ -\ 2\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{2\ +\ 1}\ (-4)\ \hspace{15cm}$
$=\ {(-1)}^3 (-4)\ \hspace{15cm}$
$= (-1) (-4)\ \hspace{15cm}$
$cofactor\ of\ 2 =\ 4\ \hspace{15cm}$
$cofactor\ of\ 5 = (-1)^{2\ +\ 2}\ (3)\ \hspace{15cm}$
$=\ {(-1)}^4 (3)\ \hspace{15cm}$
$= (1) (3)\ \hspace{15cm}$
$cofactor\ of\ 5\ =\ 3\ \hspace{15cm}$
$\therefore\ cofactor\ matrix\ = \begin{bmatrix} 5 & -2 \\ 4 & 3 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A = \begin{bmatrix} 5 & 4 \\ -2 & 3 \\ \end{bmatrix}\ \hspace{15cm}$
$4.\ \color {red}{Find\ the\ Adjoint\ matrix\ of\ \begin{bmatrix} 1 & 4 \\ 5 & -2 \\ \end{bmatrix}}\ \hspace{15cm}$
$\color {black}{Solution:}\ A\ =\begin{bmatrix} 1 & 4 \\ 5 & -2 \\ \end{bmatrix}\ \hspace{15cm}$
$cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ (-2)\ \hspace{15cm}$
$=\ {(-1)}^2 (-2)\ \hspace{15cm}$
$= (1) (-2)\ \hspace{15cm}$
$cofactor\ of\ 1 =\ -2\ \hspace{15cm}$
$cofactor\ of\ 4 = (-1)^{1\ +\ 2}\ (5)\ \hspace{15cm}$
$=\ {(-1)}^3 (5)\ \hspace{15cm}$
$= (-1) (5)\ \hspace{15cm}$
$cofactor\ of\ 4 =\ -\ 5\ \hspace{15cm}$
$cofactor\ of\ 5 = (-1)^{2\ +\ 1}\ (4)\ \hspace{15cm}$
$=\ {(-1)}^3 (4)\ \hspace{15cm}$
$= (-1) (4)\ \hspace{15cm}$
$cofactor\ of\ 3 = -4\ \hspace{15cm}$
$cofactor\ of\ -2\ = (-1)^{2\ +\ 2}\ (1)\ \hspace{15cm}$
$=\ {(-1)}^4 (1)\ \hspace{15cm}$
$= (1) (1)\ \hspace{15cm}$
$cofactor\ of\ -2 = 1\ \hspace{15cm}$
$\therefore\ cofactor\ matrix\ = \begin{bmatrix} -2 & -5 \\ -4 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A = \begin{bmatrix} -2 & -4 \\ -5 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$5.\ \color {red}{Find\ the\ Adjoint\ matrix\ of\ \begin{bmatrix} 5 & -6 \\ 3 & 2 \\ \end{bmatrix}}\ \hspace{15cm}$
$\color {black}{Solution:}\ A\ =\begin{bmatrix} 5 & -6 \\ 3 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$cofactor\ of\ 5 = (-1)^{1\ +\ 1}\ (2)\ \hspace{15cm}$
$= (1) (2)\ \hspace{15cm}$
$cofactor\ of\ 5 = \ 2\ \hspace{15cm}$
$cofactor\ of\ -6 = (-1)^{1\ +\ 2}\ (3)\ \hspace{15cm}$
$= (-1) (3)\ \hspace{15cm}$
$cofactor\ of\ -6 = -3\ \hspace{15cm}$
$cofactor\ of\ 3 = (-1)^{2\ +\ 1}\ (-6)\ \hspace{15cm}$
$= (-1) (-6)\ \hspace{15cm}$
$cofactor\ of\ 3 = 6\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{2\ +\ 2}\ (5)\ \hspace{15cm}$
$= (1) (5)\ \hspace{15cm}$
$cofactor\ of\ 2 = 5\ \hspace{15cm}$
$\therefore\ cofactor\ matrix\ = \begin{bmatrix} 2 & -3 \\ 6 & 5 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A = \begin{bmatrix} 2 & 6 \\ -3 & 5 \\ \end{bmatrix}\ \hspace{15cm}$
$6.\ \color {red}{Find\ the\ rank\ of\ \begin{bmatrix} 3 & -4 \\ -6 & 8 \\ \end{bmatrix}}\ \hspace{15cm}$
$\color {black}{Solution:}\ A\ =\begin{bmatrix} 3 & -4 \\ -6 & 8 \\ \end{bmatrix}\ \hspace{15cm}$
$The\ minor\ is\ \begin{vmatrix} 1 & 2 \\ 1 & 2 \\ \end{vmatrix}\ =\ 24\ -\ 24\ =\ 0\ \hspace{5cm}$
$\therefore \ \rho(A)\ \neq 2 \hspace{15cm}$
$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}$
$\therefore \ \rho(A)\ =\ 1 \hspace{15cm}$
$\LARGE{\color {purple} {PART- B}}$
$7.\ \color {red}{Find\ the\ inverse\ of\ \begin{bmatrix} 2 & -1 \\ 4 & 5 \\ \end{bmatrix}}\ \hspace{18cm}$
$\color {black}{Solution:}\ Let\ A\ =\begin{bmatrix} 2 & -1 \\ 4 & 5 \\ \end{bmatrix}\ \hspace{15cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ = 2(5)\ -\ 4(-1) \hspace{13cm}$
$=\ 10\ +\ 4 \hspace{12cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ = 14\ \neq {0}\ \hspace{13cm}$
$\therefore\ A^{-1}\ exist\ \hspace{10cm}$
$\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}$
$cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ (5)\ \hspace{15cm}$
$= (-1)^2 (5)\ \hspace{15cm}$
$= (1) (5)\ \hspace{15cm}$
$cofactor\ of\ 2 = 5\ \hspace{15cm}$
$cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ (4)\ \hspace{15cm}$
$= (-1)^3 (4)\ \hspace{15cm}$
$= (-1) (4)\ \hspace{15cm}$
$cofactor\ of\ -1 =\ -4\ \hspace{15cm}$
$cofactor\ of\ 4 = (-1)^{2\ +\ 1}\ (-1)\ \hspace{15cm}$
$= (-1)^3 (-1)\ \hspace{15cm}$
$= (-1) (-1)\ \hspace{15cm}$
$cofactor\ of\ 4 =\ 1\ \hspace{15cm}$
$cofactor\ of\ 5 = (-1)^{2\ +\ 2}\ (2)\ \hspace{15cm}$
$= (-1)^4 (2)\ \hspace{15cm}$
$= (1) (2)\ \hspace{15cm}$
$cofactor\ of\ 5 =\ 2\ \hspace{15cm}$
$\therefore\ cofactor\ matrix\ = \begin{bmatrix} 5 & -4 \\ 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A = \begin{bmatrix} 5 & 1 \\ -4 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}$
$A^{-1} =\ \frac{1}{14}\ \begin{bmatrix} 5 & 1 \\ -4 & 2 \\ \end{bmatrix}\ \hspace{2cm}$
$8.\ \color {red}{Find\ the\ inverse\ of\ \begin{bmatrix} 2 & 1 \\ 1 & 2 \\ \end{bmatrix}}\ \hspace{15cm}$
$\color {black}{Solution:}\ Let\ A\ =\begin{bmatrix} 2 & 1 \\ 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =\ 2(2)\ -\ 1(1) \hspace{13cm}$
$=\ 4 -\ 1 \hspace{12cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =\ 3\ \neq {0}\ \hspace{13cm}$
$\therefore\ A^{-1}\ exist\ \hspace{10cm}$
$\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}$
$cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ (2)\ \hspace{15cm}$
$= (-1)^2 (2)\ \hspace{15cm}$
$= (1) (2)\ \hspace{15cm}$
$cofactor\ of\ 2\ =\ 2\ \hspace{15cm}$
$cofactor\ of\ 1 = (-1)^{1\ +\ 2}\ (1)\ \hspace{15cm}$
$= (-1)^3 (1)\ \hspace{15cm}$
$= (-1) (1)\ \hspace{15cm}$
$cofactor\ of\ 1\ =\ -\ 1\ \hspace{15cm}$
$cofactor\ of\ 1 =\ (-1)^{2\ +\ 1}\ (1)\ \hspace{15cm}$
$= (-1)^3 (1)\ \hspace{15cm}$
$= (-1) (1)\ \hspace{15cm}$
$cofactor\ of\ 1\ =\ -\ 1\ \hspace{15cm}$
$cofactor\ of\ 2\ =\ (-1)^{2\ +\ 2}\ (2)\ \hspace{15cm}$
$= (-1)^4 (2)\ \hspace{15cm}$
$= (1) (2)\ \hspace{15cm}$
$cofactor\ of\ 2\ =\ 2\ \hspace{15cm}$
$\therefore\ cofactor\ matrix\ = \begin{bmatrix} 2 & -1 \\ -1 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}$
$A^{-1} = \frac{1}{3}\ \begin{bmatrix} 2 & -1 \\ -1 & 2 \\ \end{bmatrix}\ \hspace{2cm}$
$\color {purple} { 9\ .}\ \color {red}{Find\ the\ rank\ of}\ \ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & -1 & 4 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ A\ =\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & -1 & 4 \\ \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 \\ 2 & 4 & 6 \\ 3 & -1 & 4 \\ \end{vmatrix}\ \hspace{5cm}$
$=-1\begin{vmatrix} 4 & 6 \\ -1 & 4 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & 6 \\ 3 & 4 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} 2 & 4\\ 3 & -1 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(16\ +\ 6)\ – 2 (8\ -\ 18) + 3(-2\ +\ 12)\ \hspace{9cm}$
$=1(22)\ – 2 (-10) + 3(10)\ \hspace{13cm}$
$= 22\ +\ 20\ +\ 30\ =\ 72\ \neq\ 0\ \hspace{14cm}$
$\boxed{\therefore\ rank\ of\ A=\ \rho(A)\ =\ 3}\ \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 {black}{Solution:}\ \hspace{20cm}$
$x\ +\ y\ +\ z\ =\ 2\ ————— (1) \hspace{10cm}$
$2x\ -\ y\ -\ 2z\ =\ -1\ \hspace{15cm}$
$x\ -\ 2y\ -\ z\ =\ 1\ \hspace{15cm}$
$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & -2 \\ 1 & -2 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta =1\begin{vmatrix} -1 & -2 \\ -2 & -1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & -2 \\ 1 & -1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & -1\\ 1 & -2 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta =1(1\ -\ 4)\ – 1 (-2\ +\ 2) + 1(-4\ +\ 1)\ \hspace{9cm}$
$\Delta =1(-3)\ – 1 (0) + 1(-3)\ \hspace{13cm}$
$\Delta =-3\ – 0 – 3\ \hspace{14cm}$
$\Delta =-\ 6\ \hspace{17cm}$
$\Delta_x = \begin{vmatrix} 2 & 1 & 1 \\ -1 & -1 & -2 \\ 1 & -2 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_x =\ 2\begin{vmatrix} -1 & -2 \\ -2 & -1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} -1 & -2 \\ 1 & -1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} -1 & -1\\ 1 & -2 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_x =\ 2(1\ -\ 4)\ – 1 (1\ +\ 2) + 1(2\ +\ 1)\ \hspace{9cm}$
$\Delta_x =\ 2(-3)\ – 1 (3) + 1(3)\ \hspace{13cm}$
$\Delta_x =-6\ -3 + 3\ \hspace{14cm}$
$\Delta_x =\ -6\ \hspace{17cm}$
$\Delta_y = \begin{vmatrix} 1 & 2 & 1 \\ 2 & -1 & -2 \\ 1 & 1 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_y =1\begin{vmatrix} -1 & -2 \\ 1 & -1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & -2 \\ 1 & -1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & -1\\ 1 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_y =1(1\ +\ 2)\ – 2 (-2\ +\ 2) + 1(2\ +\ 1)\ \hspace{9cm}$
$\Delta_y =1(3)\ – 2 (0) + 1(3)\ \hspace{13cm}$
$\Delta_y = 3\ -\ 0\ +\ 3\ \hspace{14cm}$
$\Delta_y =\ 6\ \hspace{17cm}$
$\Delta_z = \begin{vmatrix} 1 & 1 & 2 \\ 2 & -1 & -1 \\ 1 & -2 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_z =1\begin{vmatrix} -1 & -1 \\ -2 & 1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & -1 \\ 1 & 1 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 2 & -1\\ 1 & -2 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_z =1(-1\ -\ 2)\ – 1 (2\ +\ 1) + 2(-4\ +\ 1)\ \hspace{9cm}$
$\Delta_z =1(-3)\ – 1 (3) + 2(-3)\ \hspace{13cm}$
$\Delta_z =-3\ -3 – 6\ \hspace{14cm}$
$\Delta_z =\ -12\ \hspace{17cm}$
$The\ Solution\ is\ \hspace{20cm}$
$x=\ \frac{\Delta_x}{\Delta} =\ \frac{-6}{-6} =\ 1\ \hspace{20cm}$
$y=\ \frac{\Delta_y}{\Delta} =\ \frac{6}{-6} =\ -1\ \hspace{20cm}$
$z=\ \frac{\Delta_z}{\Delta} =\ \frac{-12}{-6} =\ 2\ \hspace{20cm}$
$For\ cross\ verification\ \hspace{20cm}$
$Put\ x =1\ y = -1\ z = 2\ in\ equation (1)\ \hspace{18cm}$
$LHS\ =\ 1 – 1 + 2 = 2$$= RHS$
$11.\ \color {red}{Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}$
$x + 2y + 5z=4,\ 3x + y + 4z = 6\ and\ -x + y +z = -1\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{20cm}$
$x + 2y + 5z=4\ ————— (1) \hspace{10cm}$
$3x + y + 4z = 6\ \hspace{15cm}$
$-x + y +z = -1\ \hspace{15cm}$
$\Delta = \begin{vmatrix} 1 & 2 & 5 \\ 3 & 1 & 4 \\ -1 & 1 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta =1\begin{vmatrix} 1 & 4 \\ 1 & 1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 3 & 4 \\ -1 & 1 \\ \end{vmatrix}\ +\ 5\begin{vmatrix} 3 & 1\\ -1 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta =1(1\ -\ 4)\ – 2 (3\ -\ (-4)) + 5(3\ -\ (-1))\ \hspace{9cm}$
$\Delta =1(-3)\ – 2 (7) + 5(4)\ \hspace{13cm}$
$\Delta =-3\ -14 + 20\ \hspace{14cm}$
$\Delta =3\ \hspace{17cm}$
$\Delta_x = \begin{vmatrix} 4 & 2 & 5 \\ 6 & 1 & 4 \\ -1 & 1 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_x =4\begin{vmatrix} 1 & 4 \\ 1 & 1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 6 & 4 \\ -1 & 1 \\ \end{vmatrix}\ +\ 5\begin{vmatrix} 6 & 1\\ -1 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_x =4(1\ -\ 4)\ – 2 (6\ -\ (-4)) + 5(6\ -\ (-1))\ \hspace{9cm}$
$\Delta_x =4(-3)\ – 2 (10) + 5(7)\ \hspace{13cm}$
$\Delta_x =-12\ -20 + 35\ \hspace{14cm}$
$\Delta_x =3\ \hspace{17cm}$
$\Delta_y = \begin{vmatrix} 1 & 4 & 5 \\ 3 & 6 & 4 \\ -1 & -1 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_y =1\begin{vmatrix} 6 & 4 \\ -1 & 1 \\ \end{vmatrix}\ -\ 4\begin{vmatrix} 3 & 4 \\ -1 & 1 \\ \end{vmatrix}\ +\ 5\begin{vmatrix} 3 & 6\\ -1 & -1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_y =1(6\ -\ (-4))\ – 4 (3\ -\ (-4)) + 5(-3\ -\ (-6))\ \hspace{9cm}$
$\Delta_y =1(10)\ – 4 (7) + 5(3)\ \hspace{13cm}$
$\Delta_y = 10\ -28\ + 15\ \hspace{14cm}$
$\Delta_y =-3\ \hspace{17cm}$
$\Delta_z = \begin{vmatrix} 1 & 2 & 4 \\ 3 & 1 & 6 \\ -1 & 1 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_z =1\begin{vmatrix} 1 & 6 \\ 1 & -1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 3 & 6 \\ -1 & -1 \\ \end{vmatrix}\ +\ 4\begin{vmatrix} 3 & 1\\ -1 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_z =1(-1\ -\ 6)\ – 2 (-3\ -\ (-6)) + 4(3\ -\ (-1))\ \hspace{9cm}$
$\Delta_z =1(-7)\ – 2 (3) + 4(4)\ \hspace{13cm}$
$\Delta_z =-7\ -6 + 16\ \hspace{14cm}$
$\Delta_z =3\ \hspace{17cm}$
$The\ Solution\ is\ \hspace{20cm}$
$x=\ \frac{\Delta_x}{\Delta} =\ \frac{3}{3} =\ 1\ \hspace{20cm}$
$y=\ \frac{\Delta_y}{\Delta} =\ \frac{-3}{3} =\ -1\ \hspace{20cm}$
$z=\ \frac{\Delta_z}{\Delta} =\ \frac{3}{3} =\ 1\ \hspace{20cm}$
$For\ cross\ verification\ \hspace{20cm}$
$Put\ x =1\ y = -1\ z = 1\ in\ equation (1)\ \hspace{18cm}$
$LHS = 1 + 2(-1) + 5(1)$$= 1 – 2 + 5 = 4$$= RHS$

$\color{red}{12.\ Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}$
$3x\ -\ y\ +\ 2z\ =\ 8,\ x\ +\ y\ +\ z\ =\ 2\ and\ 2x\ +\ y\ -\ z\ =\ -\ 1\ \hspace{15cm}$
$\color {blue}{Solution:}\ \hspace{20cm}$
$3x\ -\ y\ +\ 2z\ =\ 8\ ————— (1) \hspace{10cm}$
$x\ +\ y\ +\ z\ =\ 2\ \hspace{15cm}$
$2x\ +\ y\ -\ z\ =\ -\ 1\ \hspace{15cm}$
$\Delta = \begin{vmatrix} 3 & -1 & 2 \\ 1 & 1 & 1 \\ 2 & 1 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta =\ 3\begin{vmatrix} 1 & 1 \\ 1 & – 1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 1 & 1 \\ 2 & -1 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 1 & 1\\ 2 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta =\ 3(-\ 1\ -\ 1)\ +\ 1 (-1\ -\ 2)\ +\ 2(1\ -\ 2)\ \hspace{9cm}$
$\Delta =\ 3(-2)\ +\ 1 (-3)\ +\ 2(-1)\ \hspace{13cm}$
$\Delta =\ -6\ -\ 3\ -\ 2\ \hspace{14cm}$
$\Delta =\ -\ 11\ \hspace{17cm}$
$\Delta_x = \begin{vmatrix} 8 & -1 & 2 \\ 2 & 1 & 1 \\ -1 & 1 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_x =\ 8\begin{vmatrix} 1 & 1 \\ 1 & -1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & 1 \\ -1 & -1 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 2 & 1\\ -1 & 1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_x =\ 8(-1\ -\ 1)\ +\ 1 (-2\ +\ 1) + 2(2\ +\ 1)\ \hspace{9cm}$
$\Delta_x =\ 8(-2)\ + 1 (-1)\ +\ 2 (3)\ \hspace{13cm}$
$\Delta_x =\ -\ 16\ -\ 1\ +\ 6\ \hspace{14cm}$
$\Delta_x = -\ 11\ \hspace{17cm}$
$\Delta_y = \begin{vmatrix} 3 & 8 & 2 \\ 1 & 2 & 1 \\ 2 & -1 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_y =\ 3\begin{vmatrix} 2 & 1 \\ -1 & -1 \\ \end{vmatrix}\ -\ 8\begin{vmatrix} 1 & 1 \\ 2 & -1 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 1 & 2\\ 2 & -1 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_y =\ 3(-2\ +\ 1)\ – 8 (-1\ -\ 2) +\ 2(-1\ -\ 4)\ \hspace{9cm}$
$\Delta_y =\ 3(-1)\ -\ 8(-3)\ + 2(-5)\ \hspace{13cm}$
$\Delta_y =\ -3\ +\ 24\ -\ 10\ \hspace{14cm}$
$\Delta_y =\ 11\ \hspace{17cm}$
$\Delta_z = \begin{vmatrix} 3 & -1 & 8 \\ 1 & 1 & 2 \\ 2 & 1 & -1 \\ \end{vmatrix}\ \hspace{15cm}$
$\Delta_z\ =\ 3\begin{vmatrix} 1 & 2 \\ 1 & -1\\ \end{vmatrix}\ +\ 1\begin{vmatrix} 1 & 2 \\ 2 & -1 \\ \end{vmatrix}\ +\ 8\begin{vmatrix} 1 & 1\\ 2 & 1\\ \end{vmatrix}\ \hspace{10cm}$
$\Delta_z \=\ 3(-\ 1\ -\ 2)\ +\ 1 (-1\ -\ 4)\ +\ 8(1\ -\ 2)\ \hspace{9cm}$
$\Delta_z =\ 3(-3)\ +\ 1 (-5)\ +\ 8(-1)\ \hspace{13cm}$
$\Delta_z =\ -\ 9\ -\ 5\ -\ 8\ \hspace{14cm}$
$\Delta_z =\ -\ 22\ \hspace{17cm}$
$The\ Solution\ is\ \hspace{20cm}$
$x=\ \frac{\Delta_x}{\Delta}\ =\ \frac{-11}{-11} =\ 1\ \hspace{20cm}$
$y=\ \frac{\Delta_y}{\Delta} =\ \frac{11}{-11} =\ -\ 1\ \hspace{20cm}$
$z=\ \frac{\Delta_z}{\Delta}\ =\ \frac{-22}{-11} =\ 2\ \hspace{20cm}$
$For\ cross\ verification\ \hspace{20cm}$
$Put\ x =\ 1\ y\ =\ -\ 1\ z\ =\ 2\ in\ equation (1)\ \hspace{18cm}$
$LHS =\ 3(1)\ -\ (-1)\ +\ 2(2)$$=\ 3\ +\ 1\ +\ 4\ =\ 8$$= RHS$
$13.\ \color {red}{Find\ the\ inverse\ of\ \begin{bmatrix} 2 & 3 & 4 \\ 4 & 3 & 1 \\ 1 & 2 & 4 \\ \end{bmatrix}}\ \hspace{15cm}$
$\color {black}{Solution:}\ Part\ -\ C\ of Question\ 2\ \hspace{20cm}$

https://yanamtakshashila.com/2021/12/14/applications-of-matrices-and-determinantsrevision/

$13.\ \color {red}{Find\ the\ inverse\ of\ \begin{bmatrix} 1 & 2 & -1\\ 3 & 8 & 2 \\ 4 & 9 & 1 \\ \end{bmatrix}}\ \hspace{15cm}$
$\color {black}{Solution:}\ A\ =\begin{bmatrix} 1 & 2 & -1 \\ 3 & 8 & 2 \\ 4 & 9 & 1 \\ \end{bmatrix}\ \hspace{15cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =1\begin{vmatrix} 8 & 2 \\ 9 & 1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 3 & 2 \\ 4 & 1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 3 & 8\\ 4 & 9 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(8\ -\ 18)\ – 2 (3\ -\ 8) – 1(27\ -\ 32)\ \hspace{9cm}$
$=1(-10)\ – 2 (-5) – 1(-5)\ \hspace{13cm}$
$=-10\ +10+ 5\ \hspace{14cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =5\ \neq\ 0\ \hspace{17cm}$
$\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}$
$\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}$
$cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ \begin{vmatrix} 8 & 2 \\ 9 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$cofactor\ of\ 1 = (-1)^2\ (8\ -18)\ \hspace{15cm}$
$= (1) (-10)\ \hspace{15cm}$
$cofactor\ of\ 1 = -10\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{1\ +\ 2}\ \begin{vmatrix} 3 & 2 \\ 4 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (3 – 8)\ \hspace{15cm}$
$= (-1) (-5)\ \hspace{15cm}$
$cofactor\ of\ 2 = 5\ \hspace{15cm}$
$cofactor\ of\ -1 = (-1)^{1\ +\ 3}\ \begin{vmatrix} 3 & 8 \\ 4 & 9 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (27 – 32)\ \hspace{15cm}$
$= (1) (-5)\ \hspace{15cm}$
$cofactor\ of\ -1 = -5\ \hspace{15cm}$
$cofactor\ of\ 3 = (-1)^{2\ +\ 1}\ \begin{vmatrix} 2 & -1 \\ 9 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (2 + 9)\ \hspace{15cm}$
$= (-1) (11)\ \hspace{15cm}$
$cofactor\ of\ 3 = -11\ \hspace{15cm}$
$cofactor\ of\ 8 = (-1)^{2\ +\ 2}\ \begin{vmatrix} 1 & -1 \\ 4 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (1 + 4)\ \hspace{15cm}$
$= (1) (5)\ \hspace{15cm}$
$cofactor\ of\ 8 = 5\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{2\ +\ 3}\ \begin{vmatrix} 1 & 2 \\ 4 & 9 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (9 – 8)\ \hspace{15cm}$
$= (-1) (1)\ \hspace{15cm}$
$cofactor\ of\ 2 = -1\ \hspace{15cm}$
$cofactor\ of\ 4 = (-1)^{3\ +\ 1}\ \begin{vmatrix} 2 & -1 \\ 8 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (4 + 8)\ \hspace{15cm}$
$= (1) (12)\ \hspace{15cm}$
$cofactor\ of\ 4 = 12\ \hspace{15cm}$
$cofactor\ of\ 9 = (-1)^{3\ +\ 2}\ \begin{vmatrix} 1 & -1 \\ 3 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (2 + 3)\ \hspace{15cm}$
$= (-1) (5)\ \hspace{15cm}$
$cofactor\ of\ 9 = -5\ \hspace{15cm}$
$cofactor\ of\ 1 = (-1)^{3\ +\ 3}\ \begin{vmatrix} 1 & 2 \\ 3 & 8 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^6 (8 – 6)\ \hspace{15cm}$
$= (1) (2)\ \hspace{15cm}$
$cofactor\ of\ 1 = 2\ \hspace{15cm}$
$Cofactor\ matrix=\begin{bmatrix} -10 & 5 & -5 \\ -11 & 5 & -1 \\ 12 & -5 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A=\begin{bmatrix} -10 & -11 & 12 \\ 5 & 5 & -5 \\ -5 & -1 & 2 \\ \end{bmatrix}\ \hspace{15cm}$
$A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}$
$A^{-1} = \frac{1}{5}\ \begin{bmatrix} -10 & -11 & 12 \\ 5 & 5 & -5 \\ -5 & -1 & 2 \\ \end{bmatrix}\ \hspace{2cm}$
$15.\ \color{red}{Find\ the\ inverse\ of}\ \begin{bmatrix} 1 & 1 & -\ 1 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Let\ A\ =\begin{bmatrix} 1 & 1 & -\ 1 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \\ \end{bmatrix}\ \hspace{15cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =1\begin{vmatrix} 1 & 0 \\ 2 & 3 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 0 \\ -1 & 3 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 1\\ -1 & 2 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(3\ -\ 0)\ – 1 (6\ -\ 0)\ -\ 1(4\ +\ 1)\ \hspace{9cm}$
$=1(3)\ -\ 1 (6)\ -\ 1(5)\ \hspace{13cm}$
$=\ 3\ -\ 6\ -\ 5\ \hspace{14cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =\ -\ 8\ \neq\ 0\ \hspace{17cm}$
$\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}$
$\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}$
$cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ \begin{vmatrix} 1 & 0 \\ 2 & 3 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^2 (3 – 0)\ \hspace{15cm}$
$= (1) (3)\ \hspace{15cm}$
$cofactor\ of\ 1 = 3\ \hspace{15cm}$
$cofactor\ of\ 1 = (-1)^{1\ +\ 2}\ \begin{vmatrix} 2 & 0 \\ -1 & 3 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (6 + 0)\ \hspace{15cm}$
$= (-1) (6)\ \hspace{15cm}$
$cofactor\ of\ 1 = -6\ \hspace{15cm}$
$cofactor\ of\ -1 = (-1)^{1\ +\ 3}\ \begin{vmatrix} 2 & 1 \\ -1 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (4+ 1)\ \hspace{15cm}$
$= (1) (5)\ \hspace{15cm}$
$cofactor\ of\ -1 = 5\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{2\ +\ 1}\ \begin{vmatrix} 1 & -1 \\ 2 & 3 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^3 (3 + 2)\ \hspace{15cm}$
$= (-1) (5)\ \hspace{15cm}$
$cofactor\ of\ 2 = -\ 5\ \hspace{15cm}$
$cofactor\ of\ 1 = (-1)^{2\ +\ 2}\ \begin{vmatrix} 1 & -1 \\ -1 & 3 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (3- 1)\ \hspace{15cm}$
$= (1) (2)\ \hspace{15cm}$
$cofactor\ of\ 1 = 2\ \hspace{15cm}$
$cofactor\ of\ 0 = (-1)^{2\ +\ 3}\ \begin{vmatrix} 1 & 1 \\ -1 & 2 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (2+ 1)\ \hspace{15cm}$
$= (-1) (3)\ \hspace{15cm}$
$cofactor\ of\ 0 = -3\ \hspace{15cm}$
$cofactor\ of\ -1 = (-1)^{3\ +\ 1}\ \begin{vmatrix} 1 & -1 \\ 1 & 0 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^4 (0+ 1)\ \hspace{15cm}$
$= (1) (1)\ \hspace{15cm}$
$cofactor\ of\ -1 = 1\ \hspace{15cm}$
$cofactor\ of\ 2 = (-1)^{3\ +\ 2}\ \begin{vmatrix} 1 & -1 \\ 2 & 0 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^5 (0 + 2)\ \hspace{15cm}$
$= (-1) (2)\ \hspace{15cm}$
$cofactor\ of\ 2 = -2\ \hspace{15cm}$
$cofactor\ of\ 3 = (-1)^{3\ +\ 3}\ \begin{vmatrix} 1 & 1 \\ 2 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$= (-1)^6 (1- 2)\ \hspace{15cm}$
$= (1) (-1)\ \hspace{15cm}$
$cofactor\ of\ 3 = -1\ \hspace{15cm}$
$Cofactor\ matrix=\begin{bmatrix} 3 & -6 & 5 \\ -5 & 2 & -3 \\ 1 & -2 & -1 \\ \end{bmatrix}\ \hspace{15cm}$
$Adj.\ A=\begin{bmatrix} 3 & -5 & 1 \\ -6 & 2 & -2 \\ 5 & -3 & -1 \\ \end{bmatrix}\ \hspace{15cm}$
$A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}$
$A^{-1} = \frac{1}{-8}\ \begin{bmatrix} 3 & -5 & 1 \\ -6 & 2 & -2 \\ 5 & -3 & -1 \\ \end{bmatrix}\ \hspace{2cm}$

$\color {purple} {16}\ \color {red}{Find\ the\ rank\ of\ the\ matrix}\ \begin{pmatrix} 1 & 2 & 3 & 1 \\ 2 & 3 & 4 & 2 \\ 3 & 4 & 5 & 1 \\ \end{pmatrix}\ \hspace{15cm}$
$\color {blue}{Solution:}\ A\ =\begin{pmatrix} 1 & 2 & 3 & 1 \\ 2 & 3 & 4 & 2 \\ 3 & 4 & 5 & 1 \\ \end{pmatrix}\ \hspace{15cm}$
$Order\ of\ A = 3 × 4\ \hspace{15cm}$
$\therefore \ \rho(A)\ \leq 3 \hspace{15cm}$
$The\ higher\ order\ of\ minor\ of\ A = 3\ \hspace{10cm}$
$A_1\ =\begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \\ \end{vmatrix}\ \hspace{15cm}$
$=-1\begin{vmatrix} 3 & 4 \\ 4 & 5 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & 4 \\ 3 & 5 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} 2 & 3\\ 3 & 4 \\ \end{vmatrix}\ \hspace{10cm}$
$=-1(15\ -\ 16)\ – 2 (10\ -\ 12) + 3(8\ -\ 9)\ \hspace{9cm}$
$=-1(-1)\ – 2 (-2) + 3(-1)\ \hspace{13cm}$
$= -1\ +\ 4\ -\ 3\ =\ 0\ \hspace{14cm}$
$A_2\ =\begin{vmatrix} 1 & 2 & 1 \\ 2 & 3 & 2 \\ 3 & 4 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$=1\begin{vmatrix} 3 & 2 \\ 4 & 1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & 2 \\ 3 & 1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & 3\\ 3 & 4 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(3\ -\ 8)\ – 2 (2\ -\ 6) +\ 1(8\ -\ 9)\ \hspace{9cm}$
$=1(-5)\ – 2 (-4)\ +\ 1(-1)\ \hspace{13cm}$
$=\ – 5\ +\ 8\ – 1\ =\ -2\ \neq\ 0\ \hspace{14cm}$
$A_3\ =\begin{vmatrix} 1 & 3 & 1 \\ 2 & 4 & 2 \\ 3 & 5 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$=\ 1\begin{vmatrix} 4 & 2 \\ 5 & 1 \\ \end{vmatrix}\ -\ 3\begin{vmatrix} 2 & 2 \\ 3 & 1 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} 2 & -4\\ -1 & -2 \\ \end{vmatrix}\ \hspace{10cm}$
$=\ 1(4\ -\ 10))\ -\ 3 (2\ -\ 6)\ + 1(10\ -\ 12)\ \hspace{9cm}$
$=\ 1(-6)\ -\ 3 (-4)\ +\ 1(-2)\ \hspace{13cm}$
$= -6\ +\ 12\ -\ 2\ =\ -6\ \neq\ 0\ \hspace{14cm}$
$A_4\ =\begin{vmatrix} 2 & 3 & 1 \\ 3 & 4 & 2 \\ 4 & 5 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$=\ 2\begin{vmatrix} 4 & 2 \\ 5 & 1 \\ \end{vmatrix}\ -\ 3\begin{vmatrix} 3 & 2 \\ 4 & 1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 3 & 4\\ 4 & 5 \\ \end{vmatrix}\ \hspace{10cm}$
$=\ 2(4\ -\ 10)\ – 3 (3\ -\ 8)\ +\ 1(15\ -\ 16)\ \hspace{9cm}$
$=\ 2(-6)\ -\ 3 (-5)\ +\ 1(-1)\ \hspace{13cm}$
$=\ -\ 12\ +\ 15\ -\ 1\ =\ 2\ \neq\ 0\ \hspace{14cm}$
$\boxed{\therefore\ rank\ of\ A=\ \rho(A)\ =\ 3}\ \hspace{15cm}$