APPLICATIONS OF MATRICES AND DETERMINANTS:(Exercise Problems)

\[\LARGE{\color {purple} {PART- A}}\]
\[1.\ Find\ the\ cofactor\ of\ 3\ in\ the\ matrix\ \begin{bmatrix} 1 & 2 & 0 \\ -1 & 3 & 4 \\ 5 & 6 & 7 \\ \end{bmatrix}\ \hspace{10cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 1 & 2 & 0 \\ -1 & 3 & 4 \\ 5 & 6 & 7 \\ \end{bmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{1\ +\ 1}\ \begin{vmatrix} 1 & 0 \\ 5 & 7 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 [1(7)\ -\ 5(0)]\ \hspace{15cm}\]
\[= 1 (7\ -\ 0)\ \hspace{15cm}\]
\[=\ 1 (7)\ \hspace{15cm}\]
\[cofactor\ of\ 3 =\ 7\ \hspace{15cm}\]
\[2.\ Find\ the\ Adjoint\ matrix\ of\ \begin{bmatrix} 2 & 2 \\ 3 & -5 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 2 & 2 \\ 3 & -5 \\ \end{bmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ (-5)\ \hspace{15cm}\]
\[=\ {(-1)}^2 (-5)\ \hspace{15cm}\]
\[= (1) (-5)\ \hspace{15cm}\]
\[cofactor\ of\ 2 =\ -5\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 2}\ (3)\ \hspace{15cm}\]
\[=\ {(-1)}^3 (3)\ \hspace{15cm}\]
\[= (-1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 2\ =\ -\ 3\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{2\ +\ 1}\ (2)\ \hspace{15cm}\]
\[=\ {(-1)}^3 (2)\ \hspace{15cm}\]
\[= (-1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = -2\ \hspace{15cm}\]
\[cofactor\ of\ -5 = (-1)^{2\ +\ 2}\ (2)\ \hspace{15cm}\]
\[=\ {(-1)}^4 (2)\ \hspace{15cm}\]
\[= (1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ -5 = 2\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{bmatrix} -5 & -3 \\ -2 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{bmatrix} -5 & -2 \\ -3 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[3.\ Find\ the\ Adjoint\ matrix\ of\ \begin{bmatrix} 3 & -4 \\ 2 & 5 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 3 & -4 \\ 2 & 5 \\ \end{bmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 3\ =\ (-1)^{1\ +\ 1}\ (5)\ \hspace{15cm}\]
\[=\ {(-1)}^2 (5)\ \hspace{15cm}\]
\[= (1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 3 =\ 5\ \hspace{15cm}\]
\[cofactor\ of\ -4 = (-1)^{1\ +\ 2}\ (2)\ \hspace{15cm}\]
\[=\ {(-1)}^3 (2)\ \hspace{15cm}\]
\[= (-1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ -4\ =\ -\ 2\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 1}\ (-4)\ \hspace{15cm}\]
\[=\ {(-1)}^3 (-4)\ \hspace{15cm}\]
\[= (-1) (-4)\ \hspace{15cm}\]
\[cofactor\ of\ 2 =\ 4\ \hspace{15cm}\]
\[cofactor\ of\ 5 = (-1)^{2\ +\ 2}\ (3)\ \hspace{15cm}\]
\[=\ {(-1)}^4 (3)\ \hspace{15cm}\]
\[= (1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 5\ =\ 3\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{bmatrix} 5 & -2 \\ 4 & 3 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{bmatrix} 5 & 4 \\ -2 & 3 \\ \end{bmatrix}\ \hspace{15cm}\]
\[4.\ Find\ the\ Adjoint\ matrix\ of\ \begin{bmatrix} 1 & 4 \\ 5 & -2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 1 & 4 \\ 5 & -2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ (-2)\ \hspace{15cm}\]
\[=\ {(-1)}^2 (-2)\ \hspace{15cm}\]
\[= (1) (-2)\ \hspace{15cm}\]
\[cofactor\ of\ 1 =\ -2\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{1\ +\ 2}\ (5)\ \hspace{15cm}\]
\[=\ {(-1)}^3 (5)\ \hspace{15cm}\]
\[= (-1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 4 =\ -\ 5\ \hspace{15cm}\]
\[cofactor\ of\ 5 = (-1)^{2\ +\ 1}\ (4)\ \hspace{15cm}\]
\[=\ {(-1)}^3 (4)\ \hspace{15cm}\]
\[= (-1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = -4\ \hspace{15cm}\]
\[cofactor\ of\ -2\ = (-1)^{2\ +\ 2}\ (1)\ \hspace{15cm}\]
\[=\ {(-1)}^4 (1)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ -2 = 1\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{bmatrix} -2 & -5 \\ -4 & 1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{bmatrix} -2 & -4 \\ -5 & 1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[5.\ Find\ the\ rank\ of\ \begin{bmatrix} 3 & -4 \\ -6 & 8 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 3 & -4 \\ -6 & 8 \\ \end{bmatrix}\ \hspace{15cm}\]
\[The\ minor\ is\ \begin{vmatrix} 1 & 2 \\ 1 & 2 \\ \end{vmatrix}\ =\ 24\ -\ 24\ =\ 0\ \hspace{5cm}\]
\[\therefore \ \rho(A)\ \neq 2 \hspace{15cm}\]
\[To\ find\ at\ least\ one\ non\ zero\ first\ order\ minor\ i.e.\ to\ find\ at\ least\ one\ non\ zero\ element\]\[of\ A\ ,\ non\ zero\ element\ exist\ in\ A\ .\]
\[Rank\ =\ 1\ \hspace{15cm}\]
\[\therefore \ \rho(A)\ =\ 1 \hspace{15cm}\]
\[\LARGE{\color {purple} {PART- B}}\]
\[6.\ Find\ the\ inverse\ of\ \begin{bmatrix} 2 & -1 \\ 4 & 5 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ Let\ A\ =\begin{bmatrix} 2 & -1 \\ 4 & 5 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = 2(5)\ -\ 4(-1) \hspace{13cm}\]
\[=\ 10\ +\ 4 \hspace{12cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = 14\ \neq {0}\ \hspace{13cm}\]
\[\therefore\ A^{-1}\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ (5)\ \hspace{15cm}\]
\[= (-1)^2 (5)\ \hspace{15cm}\]
\[= (1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 5\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ (4)\ \hspace{15cm}\]
\[= (-1)^3 (4)\ \hspace{15cm}\]
\[= (-1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ -1 =\ -4\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{2\ +\ 1}\ (-1)\ \hspace{15cm}\]
\[= (-1)^3 (-1)\ \hspace{15cm}\]
\[= (-1) (-1)\ \hspace{15cm}\]
\[cofactor\ of\ 4 =\ 1\ \hspace{15cm}\]
\[cofactor\ of\ 5 = (-1)^{2\ +\ 2}\ (2)\ \hspace{15cm}\]
\[= (-1)^4 (2)\ \hspace{15cm}\]
\[= (1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 5 =\ 2\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{bmatrix} 5 & -4 \\ 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{bmatrix} 5 & 1 \\ -4 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} =\ \frac{1}{14}\ \begin{bmatrix} 5 & 1 \\ -4 & 2 \\ \end{bmatrix}\ \hspace{2cm}\]
\[7.\ Find\ the\ inverse\ of\ \begin{bmatrix} 2 & 1 \\ 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ Let\ A\ =\begin{bmatrix} 2 & 1 \\ 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ =\ 2(2)\ -\ 1(1) \hspace{13cm}\]
\[=\ 4 -\ 1 \hspace{12cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ =\ 3\ \neq {0}\ \hspace{13cm}\]
\[\therefore\ A^{-1}\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ (2)\ \hspace{15cm}\]
\[= (-1)^2 (2)\ \hspace{15cm}\]
\[= (1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 2\ =\ 2\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 2}\ (1)\ \hspace{15cm}\]
\[= (-1)^3 (1)\ \hspace{15cm}\]
\[= (-1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 1\ =\ -\ 1\ \hspace{15cm}\]
\[cofactor\ of\ 1 =\ (-1)^{2\ +\ 1}\ (1)\ \hspace{15cm}\]
\[= (-1)^3 (1)\ \hspace{15cm}\]
\[= (-1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 1\ =\ -\ 1\ \hspace{15cm}\]
\[cofactor\ of\ 2\ =\ (-1)^{2\ +\ 2}\ (2)\ \hspace{15cm}\]
\[= (-1)^4 (2)\ \hspace{15cm}\]
\[= (1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 2\ =\ 2\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{bmatrix} 2 & -1 \\ -1 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{3}\ \begin{bmatrix} 2 & -1 \\ -1 & 2 \\ \end{bmatrix}\ \hspace{2cm}\]
\[\LARGE{\color {purple} {PART- C}}\]
\[\color {black} {8.}\ By\ using\ Cramers\ Rule\,\ Solve\ the\ equations\ \hspace{20cm}\]
\[x\ +\ y\ +\ z\ =\ 2,\ 2x\ -\ y\ -\ 2z\ =\ -1\ and\ x\ -\ 2y\ -\ z\ =\ 1\ \hspace{15cm}\]
\[\color {black}{Solution:}\ \hspace{20cm}\]
\[x\ +\ y\ +\ z\ =\ 2\ ————— (1) \hspace{10cm}\]
\[2x\ -\ y\ -\ 2z\ =\ -1\ \hspace{15cm}\]
\[x\ -\ 2y\ -\ z\ =\ 1\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix} 1 & 2 & 5 \\ 3 & 1 & 4 \\ -1 & 1 & 1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta =1\begin{vmatrix} 1 & 4 \\ 1 & 1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 3 & 4 \\ -1 & 1 \\ \end{vmatrix}\ +\ 5\begin{vmatrix} 3 & 1\\ -1 & 1 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta =1(1\ -\ 4)\ – 2 (3\ -\ (-4)) + 5(3\ -\ (-1))\ \hspace{9cm}\]
\[\Delta =1(-3)\ – 2 (7) + 5(4)\ \hspace{13cm}\]
\[\Delta =-3\ -14 + 20\ \hspace{14cm}\]
\[\Delta =3\ \hspace{17cm}\]
\[9.\ Find\ the\ inverse\ of\ \begin{bmatrix} 2 & 3 & 4 \\ 4 & 3 & 1 \\ 1 & 2 & 4 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ Part\ -\ C\ of Question\ 2\ \ https://yanamtakshashila.com/2021/12/14/applications-of-matrices-and-determinantsrevision/ \hspace{20cm}\]
\[10.\ Find\ the\ inverse\ of\ \begin{bmatrix} 1 & 2 & -1\\ 3 & 8 & 2 \\ 4 & 9 & 1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 1 & 2 & -1 \\ 3 & 8 & 2 \\ 4 & 9 & 1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ =1\begin{vmatrix} 8 & 2 \\ 9 & 1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 3 & 2 \\ 4 & 1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 3 & 8\\ 4 & 9 \\ \end{vmatrix}\ \hspace{10cm}\]
\[ =1(8\ -\ 18)\ – 2 (3\ -\ 8) – 1(27\ -\ 32)\ \hspace{9cm}\]
\[ =1(-10)\ – 2 (-5) – 1(-5)\ \hspace{13cm}\]
\[ =-10\ +10+ 5\ \hspace{14cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ =5\ \neq\ 0\ \hspace{17cm}\]
\[\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ \begin{vmatrix} 8 & 2 \\ 9 & 1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 (8 – 18)\ \hspace{15cm}\]
\[= (1) (-10)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -10\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 2}\ \begin{vmatrix} 3 & 2 \\ 4 & 1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (3 – 8)\ \hspace{15cm}\]
\[= (-1) (-5)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 5\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{1\ +\ 3}\ \begin{vmatrix} 3 & 8 \\ 4 & 9 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (27 – 32)\ \hspace{15cm}\]
\[= (1) (-5)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = -5\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{2\ +\ 1}\ \begin{vmatrix} 2 & -1 \\ 9 & 1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (2 + 9)\ \hspace{15cm}\]
\[= (-1) (11)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = -11\ \hspace{15cm}\]
\[cofactor\ of\ 8 = (-1)^{2\ +\ 2}\ \begin{vmatrix} 1 & -1 \\ 4 & 1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (1 + 4)\ \hspace{15cm}\]
\[= (1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 8 = 5\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 3}\ \begin{vmatrix} 1 & 2 \\ 4 & 9 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (9 – 8)\ \hspace{15cm}\]
\[= (-1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = -1\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{3\ +\ 1}\ \begin{vmatrix} 2 & -1 \\ 8 & 2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (4 + 8)\ \hspace{15cm}\]
\[= (1) (12)\ \hspace{15cm}\]
\[cofactor\ of\ 4 = 12\ \hspace{15cm}\]
\[cofactor\ of\ 9 = (-1)^{3\ +\ 2}\ \begin{vmatrix} 1 & -1 \\ 3 & 2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (2 + 3)\ \hspace{15cm}\]
\[= (-1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 9 = -5\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{3\ +\ 3}\ \begin{vmatrix} 1 & 2 \\ 3 & 8 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (8 – 6)\ \hspace{15cm}\]
\[= (1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = 2\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix} -10 & 5 & -5 \\ -11 & 5 & -1 \\ 12 & -5 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A=\begin{bmatrix} -10 & -11 & 12 \\ 5 & 5 & -5 \\ -5 & -1 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{5}\ \begin{bmatrix} -10 & -11 & 12 \\ 5 & 5 & -5 \\ -5 & -1 & 2 \\ \end{bmatrix}\ \hspace{2cm}\]