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

\[\color {purple} {Minor\ of\ an\ element\ of\ a\ Matrix}\ \hspace{20cm}\]
\[Minor\ of\ an\ element\ is\ a\ determinant\ obtained\ by\ deleting\ the\ row\ and\ column\ in\ which\ the\ element\ occurs\]
\[\color {black}{For\ EX:\ 1}\ \hspace{20cm}\]
\[Find\ the\ Minor\ a_1\ in\ the\ matrix\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix}\ \hspace{15cm}\]
\[Minor\ of\ a_1 =\begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \\ \end{vmatrix}\ = b_2c_3\ -\ b_3c_2\ \hspace{15cm}\]
\[\color {black}{Example\ 2:}\ \hspace{20cm}\]
\[Find\ the\ Minor\ of\ 2\ in\ the\ following\ matrix\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = \begin{vmatrix} 1 & 0 & -1 \\ 2 & 3 & 4 \\ 7 & 8 & -2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\therefore\ Minor\ of\ 2 =\begin{vmatrix} 0 & -1 \\ 8 & -2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= 0 + 8\ \hspace{15cm}\]
\[Minor\ of\ 2= 8\ \hspace{15cm}\]
\[\color {purple} {Cofactor\ of\ an\ element of\ a\ Matrix}\ \hspace{20cm}\]
\[Cofactor\ of\ an\ element\ is\ a\ signed\ minor\ of\ that\ element\]
\[\therefore\ cofactor\ of\ a_{ij} = (-1)^{i\ +\ j}\ minor\ of\ a_{ij}\]
\[\color {black}{Example\ :}\ \hspace{20cm}\]
\[Find\ the\ cofactor\ of\ 4\ in\ the\ following\ matrix\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = \begin{vmatrix} 1 & 0 & -1 \\ 2 & 3 & 4 \\ 7 & 8 & -2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{2\ +\ 3}\ \begin{vmatrix} 1 & 0 \\ 7 & 8 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (8 -0)\ \hspace{15cm}\]
\[= (-1) (8)\ \hspace{15cm}\]
\[cofactor\ of\ 4 = – 8\ \hspace{15cm}\]
\[\color {purple} {Adjoint\ of\ Matrix}\ \hspace{20cm}\]
\[The\ Adjoint\ of\ a\ square\ matrix\ A\ is\ the\ transpose\ of\ the\ matrix\ which\ is\ formed\ by\]\[the\ elements\ which\ are\ the\ cofactors\ of\ the\ corresponding\ elements\ of\ the\ determinant\ of\ the\ matrix\ A\]
\[\color {purple}{Method\ for\ to\ find\ adjoint\ of\ Matrix\ of\ order\ 3\ (order 2)}\ \hspace{20cm}\]
\[i)\ A\ is\ square\ Matrix\ of\ order\ 3\ (order 2)\ \hspace{5cm}\]
\[ii)\ Find\ the\ co-factor\ of\ all\ the\ elements\ of\ det\ A\ \hspace{8cm}\]
\[iii)\ Form\ the\ matrix\ by\ replacing\ all\ the\ elements\ of\ A\ by\ the\ corresponding\ cofactor\ in\ \begin{vmatrix} A \\ \end{vmatrix}\]
\[iv)\ Then\ take\ the\ Transpose\ of\ that\ matrix,\ then\ we\ get\ adj. A.\ \hspace{8cm}\]
\[\color {black}{Example\ 1:}\ \hspace{20cm}\]
\[Find\ the\ Adjoint\ Matrix\ of\ \begin{bmatrix} 2 & -1 \\ -5 & 3 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 2 & -1 \\ -5 & 3 \\ \end{bmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ (3)\ \hspace{15cm}\]
\[= (1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = \ 3\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ (-5)\ \hspace{15cm}\]
\[= (-1) (-5)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = 5\ \hspace{15cm}\]
\[cofactor\ of\ -5 = (-1)^{2\ +\ 1}\ (-1)\ \hspace{15cm}\]
\[= (-1) (-1)\ \hspace{15cm}\]
\[cofactor\ of\ -5 = 1\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{2\ +\ 2}\ (2)\ \hspace{15cm}\]
\[= (1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = 2\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{bmatrix} 3 & 5 \\ 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Example\ 2:}\ \hspace{20cm}\]
\[Find\ the\ Adjoint\ of\ A\ = \begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ \begin{vmatrix} 2 & 3 \\ 1 & 2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 (4 -3)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 1\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{1\ +\ 2}\ \begin{vmatrix} 1 & 3 \\ -1 & 2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (2 + 3)\ \hspace{15cm}\]
\[= (-1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = -5\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{1\ +\ 3}\ \begin{vmatrix} 1 & 2 \\ -1 & 1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (1 + 2)\ \hspace{15cm}\]
\[= (1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 4 = 3\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{2\ +\ 1}\ \begin{vmatrix} 3 & 4 \\ 1 & 2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (6 – 4)\ \hspace{15cm}\]
\[= (-1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -2\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 2}\ \begin{vmatrix} 2 & 4 \\ -1 & 2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (4 + 4)\ \hspace{15cm}\]
\[= (1) (8)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 8\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{2\ +\ 3}\ \begin{vmatrix} 2 & 3 \\ -1 & 1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (2 + 3)\ \hspace{15cm}\]
\[= (-1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = -5\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{3\ +\ 1}\ \begin{vmatrix} 3 & 4 \\ 2 & 3 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (9 – 8)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = 1\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{3\ +\ 2}\ \begin{vmatrix} 2 & 4 \\ 1 & 3 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (6 – 4)\ \hspace{15cm}\]
\[= (-1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -2\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{3\ +\ 3}\ \begin{vmatrix} 2 & 3 \\ 1 & 2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (4 – 3)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 1\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix} 1 & -5 & 3 \\ -2 & 8 & -5 \\ 1 & -2 & 1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A=\begin{bmatrix} 1 & -2 & 1 \\ -5 & 8 & -2 \\ 3 & -5 & 1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {green} {Singular\ and\ Non-Singular\ Matrix}:\ \hspace{20cm}\]
A square matrix A is called a singular matrix
\[if\ \begin{vmatrix} A \\ \end{vmatrix}\ = 0\ and\ non\ –\ singular\ matrix\ if\ \begin{vmatrix} A \\ \end{vmatrix}\ \neq {0}\ \hspace{10cm}\]
\[\color {black} {Eg:}\ a)\ A =\begin{bmatrix} 3 & 6 \\ 2 & 4 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = 3(4) – 6(2) \hspace{13cm}\]
\[= 12 – 12 \hspace{12cm}\]
\[= 0 \hspace{13cm}\]
\[A\ is\ a\ singular\ matrix\ \hspace{10cm}\]
\[ b)\ A =\begin{bmatrix} 2 & 3 \\ 4 & 5 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = 2(5) – 3(4) \hspace{13cm}\]
\[= 10 – 12 \hspace{12cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = -2\ \neq {0}\ \hspace{13cm}\]
\[A\ is\ a\ Non\ – \ Singular\ matrix\ \hspace{10cm}\]
\[\color {black} {c)\,Prove\ that}\ \begin{bmatrix} 1 & -1 & 2 \\ 2 & -2 & 4 \\ 3 & -3 & 6 \\ \end{bmatrix}\ is\ a\ singular\ matrix\ \hspace{15cm}\]
\[A\ =\begin{bmatrix} 1 & -1 & 2 \\ 2 & -2 & 4 \\ 3 & -3 & 6 \\ \end{bmatrix}\ \hspace{12cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ =1\begin{vmatrix} -2 & 4 \\ -3 & 6 \\ \end{vmatrix}\ – \ 1\begin{vmatrix} 2 & 2 \\ 2 & -2 \\ \end{vmatrix}\ -\ 1 \begin{vmatrix} 2 & -1 \\ 2 & 1 \\ \end{vmatrix}\ \hspace{4cm}\]
\[ =1(-12\ +\ 12)\ + 1 (12\ -\ 12) + 2(-6\ +\ 6)\ \hspace{10cm}\]
\[= 10 – 12 \hspace{12cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = 0\ \neq {0}\ \hspace{13cm}\]
\[A\ is\ a\ Singular\ matrix\ \hspace{10cm}\]
\[\color {purple} {Inverse\ of\ Matrix}\ \hspace{20cm}\]
\[Let\ A\ be\ a\ non\ singular\ matrix\ if\ there\ exists\ a\ square\ matrix\ B\, such\ that\ AB\ =\ BA\ = I\]\[Where\ I\ is\ the\ unit\ matrix\ of\ same\ order\ then\ B\ is\ called\ the\ the\ Inverse\ of\ A\ and\ it\ is\ denoted\ by\ A^{-1}.\]
\[\color {black}{Note:}\ \hspace{12cm}\]
\[i)\ inverse\ of\ matrix\ is\ unique\ \hspace{6cm}\]
\[ii)\ AB=\ BA\ =\ I\ \hspace{8cm}\]
\[iii)\ (AB)^{-1} =\ B^{-1}A^{-1}\ \hspace{8cm}\]
\[\color {black}{Formula\ for\ Inverse\ of\ Matrix:}\ \hspace{12cm}\]
\[\color {green} {\boxed {A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A}}\]
\[\color {black}{Example\ 1\:}\ \hspace{20cm}\]
\[ Find\ the\ inverse\ of\ \begin{bmatrix} 1 & -1 \\ -2 & 0 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ Let\ A\ =\begin{bmatrix} 1 & -1 \\ -2 & 0 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = 1(0) – -(-2) \hspace{13cm}\]
\[= 0 – 2 \hspace{12cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = – 2\ \neq {0}\ \hspace{13cm}\]
\[\therefore\ A^{-1}\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ (0)\ \hspace{15cm}\]
\[= (-1)^2 (0)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = 0\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ (-2)\ \hspace{15cm}\]
\[= (-1)^3 (-2)\ \hspace{15cm}\]
\[cofactor\ of\ – 1 = 2\ \hspace{15cm}\]
\[cofactor\ of\ -2 = (-1)^{2\ +\ 1}\ (-1)\ \hspace{15cm}\]
\[= (-1)^3 (-1)\ \hspace{15cm}\]
\[cofactor\ of\ – 2 = 1\ \hspace{15cm}\]
\[cofactor\ of\ 0 = (-1)^{2\ +\ 2}\ (1)\ \hspace{15cm}\]
\[= (-1)^4 (1)\ \hspace{15cm}\]
\[cofactor\ of\ 0 = 1\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{bmatrix} 0 & 2 \\ 1 & 1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{bmatrix} 0 & 1 \\ 2 & 1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{-2}\ \begin{bmatrix} 0 & 1 \\ 2 & 1 \\ \end{bmatrix}\ \hspace{2cm}\]
\[\color {black}{Example\ 2\:}\ \hspace{20cm}\]
\[Find\ the\ inverse\ of\ \begin{bmatrix} 1 & -1 & 1 \\ 2 & -3 & -3 \\ 6 & -2 & -1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ Let\ A\ =\begin{bmatrix} 1 & -1 & 1 \\ 2 & -3 & -3 \\ 6 & -2 & -1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ =1\begin{vmatrix} -3 & -3 \\ -2 & -1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & -3 \\ 6 & -1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & -3\\ 6 & -2 \\ \end{vmatrix}\ \hspace{10cm}\]
\[ =1(3\ -\ 6)\ + 1 (-2\ +\ 18) + 1(-4\ +\ 18)\ \hspace{9cm}\]
\[ =1(-3)\ + 1 (16) + 1(14)\ \hspace{13cm}\]
\[ = -3\ + 16 + 14\ \hspace{14cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = 27\ \neq\ 0\ \hspace{17cm}\]
\[\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ \begin{vmatrix} -3 & -3 \\ -2 & -1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 (3 – 6)\ \hspace{15cm}\]
\[= (1) (-3)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -3\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ \begin{vmatrix} 2 & -3 \\ 6 & -1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (-2 + 18)\ \hspace{15cm}\]
\[= (-1) (16)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = -16\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 3}\ \begin{vmatrix} 2 & -3 \\ 6 & -2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (-4 + 18)\ \hspace{15cm}\]
\[= (1) (14)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = 14\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 1}\ \begin{vmatrix} -1 & 1 \\ -2 & -1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (1+ 2)\ \hspace{15cm}\]
\[= (-1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = -3\ \hspace{15cm}\]
\[cofactor\ of\ -3 = (-1)^{2\ +\ 2}\ \begin{vmatrix} 1 & 1 \\ 6 & -1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (-1- 6)\ \hspace{15cm}\]
\[= (1) (-7)\ \hspace{15cm}\]
\[cofactor\ of\ -3 = -7\ \hspace{15cm}\]
\[cofactor\ of\ -3 = (-1)^{2\ +\ 3}\ \begin{vmatrix} 1 & -1 \\ 6 & -2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-2+ 6)\ \hspace{15cm}\]
\[= (-1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ -3 = -4\ \hspace{15cm}\]
\[cofactor\ of\ 6 = (-1)^{3\ +\ 1}\ \begin{vmatrix} -1 & 1 \\ -3 & -3 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (3 + 3)\ \hspace{15cm}\]
\[= (1) (6)\ \hspace{15cm}\]
\[cofactor\ of\ 6 = 6\ \hspace{15cm}\]
\[cofactor\ of\ -2 = (-1)^{3\ +\ 2}\ \begin{vmatrix} 1 & 1 \\ 2 & -3 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-3- 2)\ \hspace{15cm}\]
\[= (-1) (-5)\ \hspace{15cm}\]
\[cofactor\ of\ -2 = 5\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{3\ +\ 3}\ \begin{vmatrix} 1 & -1 \\ 2 & -3 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (-3+ 2)\ \hspace{15cm}\]
\[= (1) (-1)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = -1\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix} -3 & -16 & 14 \\ -3 & -7 & -4 \\ 6 & 5 & -1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A=\begin{bmatrix} -3 & -3 & 6 \\ -16 & -7 & 5 \\ 14 & -4 & -1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{27}\ \begin{bmatrix} -3 & -3 & 6 \\ -16 & -7 & 5 \\ 14 & -4 & -1 \\ \end{bmatrix}\ \hspace{2cm}\]
\[\color {purple} {Rank\ of\ Matrix:}\ \hspace{20cm}\]
\[Let\ A\ be\ any\ m×n\ matrix.\ The\ order\ of\ the\ largest\ square\ sub\ matrix\ of\ A\ whose\ determinant\]\[ has\ a\ non\ -\ zero\ value\ is\ known\ as\ the\ rank\ of\ the\ matrix\ A\]\[and\ is\ denoted\ by\ \rho(A)\]
\[\color {black}{Example\ 1:}\ \hspace{20cm}\]
\[Find\ the\ rank\ of\ the\ matrix\ \begin{bmatrix} 5 & 2 \\ 6 & 3 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 5 & 2 \\ 6 & 3 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Order\ of\ A = 2 × 2\ \hspace{15cm}\]
\[\therefore \ \rho(A)\ \leq 2 \hspace{15cm}\]
\[The\ higher\ order\ of\ minor\ of\ A = 2\ \hspace{10cm}\]
\[The\ minor\ is\ \begin{vmatrix} 5 & 2 \\ 6 & 3 \\ \end{vmatrix}\ =\ 15 – 12\ =\ -3\ \neq\ 0\]
\[\therefore\ Rank\ of\ A\ =\ 2\ \hspace{15cm}\]
\[\color {black}{Example\ 2:}\ \hspace{20cm}\]
\[Find\ the\ rank\ of\ the\ matrix\ \begin{bmatrix} 1 & 2 \\ 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 1 & 2 \\ 1 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Order\ of\ A = 2 × 2\ \hspace{15cm}\]
\[\therefore \ \rho(A)\ \leq 2 \hspace{15cm}\]
\[The\ higher\ order\ of\ minor\ of\ A = 2\ \hspace{10cm}\]
\[The\ minor\ is\ \begin{vmatrix} 1 & 2 \\ 1 & 2 \\ \end{vmatrix}\ =\ 6- 6\ =\ 0\ \hspace{5cm}\]
\[\therefore \ \rho(A)\ \neq 2 \hspace{15cm}\]
\[To\ find\ at\ least\ one\ non\ zero\ first\ order\ minor\ i.e.\ to\ find\ at\ least\ one\ non\ zero\ element\]\[of\ A\ ,\ non\ zero\ element\ exist\ in\ A\ .\]
\[Rank\ =\ 1\ \hspace{15cm}\]
\[\therefore \ \rho(A)\ =\ 1 \hspace{15cm}\]
\[\color {black}{Example\ 3:}\ \hspace{20cm}\]
\[Find\ the\ rank\ of\ the\ matrix\ \begin{bmatrix} -1 & 2 & 3 \\ 0 & 3 & 1 \\ 4 & 5 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} -1 & 2 & 3 \\ 0 & 3 & 1 \\ 4 & 5 & 2 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Order\ of\ A = 3 × 3\ \hspace{15cm}\]
\[\therefore \ \rho(A)\ \leq 3 \hspace{15cm}\]
\[The\ higher\ order\ of\ minor\ of\ A = 3\ \hspace{10cm}\]
\[The\ minor\ is\ =\ \begin{vmatrix} -1 & 2 & 3 \\ 0 & 3 & 1 \\ 4 & 5 & 2 \\ \end{vmatrix}\ \hspace{5cm}\]
\[=-1\begin{vmatrix} 3 & 1 \\ 5 & 2 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 0 & 1 \\ 4 & 2 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} 0 & 3\\ 4 & 5 \\ \end{vmatrix}\ \hspace{10cm}\]
\[ =-1(6\ -\ 5)\ – 2 (0\ -\ 4) + 3(0\ -\ 12)\ \hspace{9cm}\]
\[ =-1(1)\ – 2 (-4) + 3(-12)\ \hspace{13cm}\]
\[ = -1\ + 8 – 36\ =\ -29\ \neq\ 0\ \hspace{14cm}\]
\[\therefore\ rank\ of\ A=\ \rho(A)\ =\ 3\ \hspace{15cm}\]
\[\color {black}{Example\ 4:}\ \hspace{20cm}\]
\[Find\ the\ rank\ of\ the\ matrix\ \begin{bmatrix} 1 & 2 & 3 & 2 \\ 2 & 3 & 5 & 1 \\ 1 & 3 & 4 & 5 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 1 & 2 & 3 & 2 \\ 2 & 3 & 5 & 1 \\ 1 & 3 & 4 & 5 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Order\ of\ A = 3 × 4\ \hspace{15cm}\]
\[\therefore \ rank\ of\ A=\rho(A)\ \leq\ Min{3,4} =3\ . \hspace{15cm}\]
\[The\ highest\ order\ of\ minors\ of\ A = 3.\ \hspace{15cm}\]
\[A\ has\ the\ following\ minors\ of\ order 3.\ \hspace{15cm}\]
\[ A_1\ =\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 5 \\ 1 & 3 & 4 \\ \end{bmatrix}\ \hspace{15cm}\]
\[=1\begin{vmatrix} 3 & 5 \\ 3 & 4 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & 5 \\ 1 & 4 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} 2 & 3\\ 1 & 3 \\ \end{vmatrix}\ \hspace{10cm}\]
\[ =1(12\ -\ 15)\ – 2 (8\ -\ 5) + 3(6\ -\ 3)\ \hspace{9cm}\]
\[ =1(-3)\ – 2 (3) + 3(3)\ \hspace{13cm}\]
\[ = -3\ – 6 + 9\ =\ 0\ \hspace{14cm}\]
\[ A_2\ =\begin{bmatrix} 1 & 2 & 2 \\ 2 & 3 & 1 \\ 1 & 3 & 5 \\ \end{bmatrix}\ \hspace{15cm}\]
\[=1\begin{vmatrix} 3 & 1 \\ 3 & 5 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & 1 \\ 1 & 5 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 2 & 3\\ 1 & 3 \\ \end{vmatrix}\ \hspace{10cm}\]
\[ =1(15\ -\ 3)\ – 2 (10\ -\ 1) + 2(6\ -\ 3)\ \hspace{9cm}\]
\[ =1(12)\ – 2 (9) + 2(3)\ \hspace{13cm}\]
\[ = 12\ – 18 + 6\ =\ 0\ \hspace{14cm}\]
\[ A_3\ =\begin{bmatrix} 2 & 3 & 2 \\ 3 & 5 & 1 \\ 3 & 4 & 5 \\ \end{bmatrix}\ \hspace{15cm}\]
\[=2\begin{vmatrix} 5 & 1 \\ 4 & 5 \\ \end{vmatrix}\ -\ 3\begin{vmatrix} 3 & 1 \\ 3 & 5 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 3 & 5\\ 3 & 4 \\ \end{vmatrix}\ \hspace{10cm}\]
\[ =2(25\ -\ 4)\ – 3 (15\ -\ 3) + 2(12\ -\ 15)\ \hspace{9cm}\]
\[ =2(21)\ – 3 (12) + 2(-3)\ \hspace{13cm}\]
\[ = 42\ – 36 – 6\ =\ 0\ \hspace{14cm}\]
\[ A_4\ =\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 1 & 4 & 5 \\ \end{bmatrix}\ \hspace{15cm}\]
\[=1\begin{vmatrix} 5 & 1 \\ 4 & 5 \\ \end{vmatrix}\ -\ 3\begin{vmatrix} 2 & 1 \\ 1 & 5 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 2 & 5\\ 1 & 4 \\ \end{vmatrix}\ \hspace{10cm}\]
\[ =1(25\ -\ 4)\ – 3 (10\ -\ 1) + 2(8\ -\ 5)\ \hspace{9cm}\]
\[ =1(21)\ – 3 (9) + 2(3)\ \hspace{13cm}\]
\[ = 21\ – 27 + 6\ =\ 0\ \hspace{14cm}\]
\[All\ third\ order\ minors\ vanish\ \hspace{8cm}\]
\[\therefore\ \rho(A)\ \lt 3\ \hspace{15cm}\]
\[To\ find\ at\ least\ a\ non\ zero\ of\ order\ 2 × 2\ \ \hspace{10cm}\]
\[\begin{vmatrix} 1 & 3 \\ 2 & 5\\ \end{vmatrix}\ =\ 5- 6\ =\ -1\ \neq\ 0\ \hspace{5cm}\]
\[\therefore\ A\ has\ at\ least\ one\ non\ zero\ minor\ of\ order\ 2\ \hspace{5cm}\]
\[\therefore\ \rho(A)\ =2\ \hspace{15cm}\]
\[\color {purple} {SOLUTION\ OF\ SIMULTANEOUS\ EQUATIONS\ USING\ CRAMERS\ RULE}\ \hspace{20cm}\]
\[a_1x\ + b_1y\ +\ c_1z\ = d_1\ —– (1)\ \hspace{20cm}\]
\[a_2x\ + b_2y\ +\ c_2z\ = d_2\ \hspace{20cm}\]
\[a_3x\ + b_3y\ +\ c_3z\ = d_3\ \hspace{20cm}\]
\[\color {black}{Solution:}\ to\ find\ x,\ y,\ z\ \hspace{20cm}\]
\[Step\ 1:\ \Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix}\ \hspace{20cm}\]
\[Step\ 2:\ \Delta_x = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \\ \end{vmatrix}\ \hspace{20cm}\]
\[Step\ 3:\ \Delta_y= \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \\ \end{vmatrix}\ \hspace{20cm}\]
\[Step\ 4:\ \Delta_z = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \\ \end{vmatrix}\ \hspace{20cm}\]
\[Solution\ is\ x=\ \frac{\Delta_x}{\Delta}.\ y=\ \frac{\Delta_y}{\Delta},\ z=\ \frac{\Delta_z}{\Delta}\ \hspace{15cm}\]
\[Example\ 1:\ Solve\ the\ following\ equations\ using\ Cramers\ Rule\ \hspace{20cm}\]
\[x + 2y + 5z=4,\ 3x + y + 4z = 6\ and\ -x + y +z = -1\ \hspace{15cm}\]
\[\color {black}{Solution:}\ \hspace{20cm}\]
\[x + 2y + 5z=4\ ————— (1) \hspace{10cm}\]
\[3x + y + 4z = 6\ \hspace{15cm}\]
\[-x + y +z = -1\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix} 1 & 2 & 5 \\ 3 & 1 & 4 \\ -1 & 1 & 1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta =1\begin{vmatrix} 1 & 4 \\ 1 & 1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 3 & 4 \\ -1 & 1 \\ \end{vmatrix}\ +\ 5\begin{vmatrix} 3 & 1\\ -1 & 1 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta =1(1\ -\ 4)\ – 2 (3\ -\ (-4)) + 5(3\ -\ (-1))\ \hspace{9cm}\]
\[\Delta =1(-3)\ – 2 (7) + 5(4)\ \hspace{13cm}\]
\[\Delta =-3\ -14 + 20\ \hspace{14cm}\]
\[\Delta =3\ \hspace{17cm}\]
\[\Delta_x = \begin{vmatrix} 4 & 2 & 5 \\ 6 & 1 & 4 \\ -1 & 1 & 1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =4\begin{vmatrix} 1 & 4 \\ 1 & 1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 6 & 4 \\ -1 & 1 \\ \end{vmatrix}\ +\ 5\begin{vmatrix} 6 & 1\\ -1 & 1 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x =4(1\ -\ 4)\ – 2 (6\ -\ (-4)) + 5(6\ -\ (-1))\ \hspace{9cm}\]
\[\Delta_x =4(-3)\ – 2 (10) + 5(7)\ \hspace{13cm}\]
\[\Delta_x =-12\ -20 + 35\ \hspace{14cm}\]
\[\Delta_x =3\ \hspace{17cm}\]
\[\Delta_y = \begin{vmatrix} 1 & 4 & 5 \\ 3 & 6 & 4 \\ -1 & -1 & 1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y =1\begin{vmatrix} 6 & 4 \\ -1 & 1 \\ \end{vmatrix}\ -\ 4\begin{vmatrix} 3 & 4 \\ -1 & 1 \\ \end{vmatrix}\ +\ 5\begin{vmatrix} 3 & 6\\ -1 & -1 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y =1(6\ -\ (-4))\ – 4 (3\ -\ (-4)) + 5(-3\ -\ (-6))\ \hspace{9cm}\]
\[\Delta_y =1(10)\ – 4 (7) + 5(3)\ \hspace{13cm}\]
\[\Delta_y = 10\ -28\ + 15\ \hspace{14cm}\]
\[\Delta_y =-3\ \hspace{17cm}\]
\[\Delta_z = \begin{vmatrix} 1 & 2 & 4 \\ 3 & 1 & 6 \\ -1 & 1 & -1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z =1\begin{vmatrix} 1 & 6 \\ 1 & -1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 3 & 6 \\ -1 & -1 \\ \end{vmatrix}\ +\ 4\begin{vmatrix} 3 & 1\\ -1 & 1 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z =1(-1\ -\ 6)\ – 2 (-3\ -\ (-6)) + 4(3\ -\ (-1))\ \hspace{9cm}\]
\[\Delta_z =1(-7)\ – 2 (3) + 4(4)\ \hspace{13cm}\]
\[\Delta_z =-7\ -6 + 16\ \hspace{14cm}\]
\[\Delta_z =3\ \hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{3}{3} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{-3}{3} =\ -1\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{3}{3} =\ 1\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x =1\ y = -1\ z = 1\ in\ equation (1)\ \hspace{18cm}\]
\[LHS = 1 + 2(-1) + 5(1)\]\[ = 1 – 2 + 5 = 4\]\[ = RHS\]
\[Example\ 2:\ Solve\ the\ following\ equations\ using\ Cramers\ Rule\ \hspace{20cm}\]
\[x + 2y – z=-1,\ 3x + 8y + 2z = 28\ and\ 4x + 9y – z = 14\ \hspace{15cm}\]
\[\color {black}{Solution:}\ \hspace{20cm}\]
\[x + 2y – z = – 1\ ————— (1) \hspace{10cm}\]
\[3x + 8y + 2z = 28\ \hspace{15cm}\]
\[4x + 9y – z = 14\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix} 1 & 2 & -1 \\ 3 & 8 & 2 \\ 4 & 9 & -1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta =1\begin{vmatrix} 8 & 2 \\ 9 & – 1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 3 & 2 \\ 4 & -1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 3 & 8\\ 4 & 9 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta =1(-8\ -\ 18)\ – 2 (-3\ -\ 8) – 1(27\ -\32)\ \hspace{9cm}\]
\[\Delta =1(-26)\ – 2 (-11) – 1(-5)\ \hspace{13cm}\]
\[\Delta =-26\ +22 + 5\ \hspace{14cm}\]
\[\Delta =1\ \hspace{17cm}\]
\[\Delta_x = \begin{vmatrix} -1 & 2 & -1 \\ 28 & 8 & 2 \\ 14 & 9 & -1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =-1\begin{vmatrix} 8 & 2 \\ 9 & -1 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 28 & 2 \\ 14 & -1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 28 & 8\\ 14 & 9 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x =-1(-8\ -\ 18)\ – 2 (-28\ -\ 28) – 1(252\ -\ 112)\ \hspace{9cm}\]
\[\Delta_x =-1(-26)\ – 2 (-56) -1 (140)\ \hspace{13cm}\]
\[\Delta_x =26\ + 112 – 140\ \hspace{14cm}\]
\[\Delta_x = -2\ \hspace{17cm}\]
\[\Delta_y = \begin{vmatrix} 1 & -1 & -1 \\ 3 & 28 & 2 \\ 4 & 14 & -1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y =1\begin{vmatrix} 28 & 2 \\ 14 & -1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 3 & 2 \\ 4 & -1 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 3 & 28\\ 4 & 14 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y =1(-28\ -\ 28)\ + 1 (-3\ -\ 8) – 1(42\ -\ 112)\ \hspace{9cm}\]
\[\Delta_y =1(-56)\ + 1 (-11) – 1(-70)\ \hspace{13cm}\]
\[\Delta_y = -56\ -11\ + 70\ \hspace{14cm}\]
\[\Delta_y =3\ \hspace{17cm}\]
\[\Delta_z = \begin{vmatrix} 1 & 2 & – 1 \\ 3 & 8 & 28 \\ 4 & 9 & 14 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z =1\begin{vmatrix} 8 & 28 \\ 9 & 14 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 3 & 28 \\ 4 & 14 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 3 & 8\\ 4 & 9 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z =1(112\ -\ 252)\ – 2 (42\ -\ 112) – 1(27\ -\ 32)\ \hspace{9cm}\]
\[\Delta_z =1(-140)\ – 2 (-70) – 1(-5)\ \hspace{13cm}\]
\[\Delta_z =-140\ + 140 + 5\ \hspace{14cm}\]
\[\Delta_z =5\ \hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{-2}{1} =\ -2\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{3}{1} =\ 3\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{5}{1} =\ 5\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x =-2\ y = 3\ z = 5\ in\ equation (1)\ \hspace{18cm}\]
\[LHS = -2 + 2(3) – 5\]\[ = -2 + 6 – 5 = -1\]\[ = RHS\]

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