Question 15 Marks
मान लीजिए A = $\left[\begin{array}{ccc} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]$हो तो सत्यापित कीजिए कि $(A^{-1})^{-1}= A$
Answerदिया है, A = $ \left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$ $\Rightarrow$ |A| = $ \left|\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right|$
= 1(15 - 1) - (- 2)(- 10 - 1) + 1(- 2 - 3) = 14 - 22 - 5 = - 13 $\neq$ 0
A के सहखण्ड निम्न हैं,
$A_{11} = 15 - 1 = 14, A_{12} = -(-10 - 1) = 11, A_{13} = -2 - 3 = -5,$
$A_{21} = -(-10 - 1) = 11, A_{22} = 5 - 1 = 4, A_{23} = -(1 + 2) = -3$
$A_{31}= (- 2 - 3) = - 5, A_{32}= - (1 + 2) = - 3, A_{33}= 3 - 4 = - 1$
adj (A) = $\left[\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right]^{T}$ = $\left[\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right]$
$\therefore A^{-1} = \frac{1}{|A|}$ (adj A) = $\frac{1}{-13}$ $\left[\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right]$= $\left[\begin{array}{rrr} -\frac{14}{13} & -\frac{11}{13} & \frac{5}{13} \\ -\frac{11}{13} & -\frac{4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13} \end{array}\right]$
$|A^{-1}| = - \frac{14}{13}$ (-$\frac{4}{169}$ - $\frac{9}{169}$) + $ \frac{11}{13}$(- $\frac{11}{169}$ - $\frac{15}{169}$) + $\frac{5}{13}$ (- $ \frac{33}{169}$ + $ \frac{20}{169}$)
= - $\frac{14}{13}$ (- $ \frac{13}{169}$) + $\frac{11}{13}$ (- $\frac{26}{169}$) + $\frac{5}{13}$ (- $\frac{13}{169}$)
= $\frac{14}{169}$ - $\frac{22}{169}$ - $\frac{5}{169}$ = - $\frac{13}{169}$ = - $\frac{1}{13}$
तथा adj $(A^{-1}) = \frac{1}{13}$$ \left[\begin{array}{ccc} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right]$
$\therefore (A^{-1})^{-1} = \frac{1}{\left|A^{-1}\right|}$ (adj $A^{-1}$)
= $\frac{1}{-\frac{1}{13}}$ $\times$ $\frac{1}{13}$ $\left[\begin{array}{ccc} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right]$= $ \left[\begin{array}{ccc} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]$ = A
View full question & answer→Question 25 Marks
मान लीजिए A = $\left[\begin{array}{ccc} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]$हो तो सत्यापित कीजिए कि $[adj A]^{-1}= adj (A^{-1})$
Answerदिया है, A = $ \left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$ $\Rightarrow$ |A| = $ \left|\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right|$
$= 1(15 - 1) - (- 2)(- 10 - 1) + 1(- 2 - 3) = 14 - 22 - 5 = - 13$ $\neq$ $0$
A के सहखण्ड निम्न हैं,
$A_{11} = 15 - 1 = 14, A_{12} = -(-10 - 1) = 11, A_{13} = -2 - 3 = -5,$
$A_{21} = -(-10 - 1) = 11, A_{22} = 5 - 1 = 4, A_{23} = -(1 + 2) = -3$
$A_{31}= (- 2 - 3) = - 5, A_{32}= - (1 + 2) = - 3, A_{33}= 3 - 4 = - 1$
adj (A) = $\left[\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right]^{T}$
= $\left[\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right]$
$\therefore$ $A^{-1}$ = $ \frac{1}{|A|}$ (adj A) = $\frac{1}{-13}$ $\left[\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right]$= $\left[\begin{array}{rrr} -\frac{14}{13} & -\frac{11}{13} & \frac{5}{13} \\ -\frac{11}{13} & -\frac{4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13} \end{array}\right]$
यहाँ, adj(A) = $\left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right]$= B (माना)
$\therefore$ |B| = |adj A| =$ \left|\begin{array}{ccc} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right|$= 14(- 4 - 9) - 11(- 11 - 15) - 5(- 33 + 20)
= - 182 + 286 + 65 = 169 $\neq$ 0
B के सहखण्ड निम्न हैं
$B_{11}= (- 4 - 9) = - 13, B_{12}= - (- 11 - 15) = 26,$
$B_{13}= (- 33 + 20) = - 13, B_{21}= - (- 11 - 15) = 26,$
$B_{22}= - 14 - 25 = - 39, B_{23}= - (- 42 + 55) = - 13,$
$B_{31}= (- 33 + 20) = - 13, B_{32}= - (- 42 + 55) = - 13,$
$B_{33}= 56 - 121 = -65$
adj(B) = adj (adj A)=$\left[\begin{array}{ccc} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{array}\right]^{\top}$
= $\left[\begin{array}{ccc} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{array}\right] $
$ \therefore$ $B^{-1}= [adj A]^{-1}$ = $\frac{1}{|\operatorname{adj} A|}$ {adj(adj A)}
$\Rightarrow$ $[adj A]^{-1}$ = $ \frac{1}{169}$ $\left[\begin{array}{ccc}-13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65\end{array}\right]$= $\frac{1}{13}$ $\left[\begin{array}{ccc}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right]$ ...(i)
$A^{-1}$ के सहखण्ड निम्न है,
$A_{11}$ = $\frac{-13}{169}$, $A_{12}$= $\frac{26}{169}$, $A_{13}$ = $\frac{-13}{169}$
$A_{21}$ = $\frac{26}{169}$, $A_{22}$ = $\frac{-39}{169}$, $A_{23}$= $\frac{-13}{169}$
$A_{31}$ = $\frac{-13}{169}$, $A_{32}$ = - $\frac{13}{169}$, $A_{33}$ = - $\frac{65}{169}$
अब, $adj (A^{-1})$ = $\left[\begin{array}{ccc} -\frac{13}{169} & \frac{26}{169} & -\frac{13}{169} \\ \frac{26}{169} & -\frac{39}{169} & -\frac{13}{169} \\ -\frac{13}{169} & -\frac{13}{169} & -\frac{65}{169} \end{array}\right]^{T}$= $\left[\begin{array}{ccc} -\frac{13}{169} & \frac{26}{169} & -\frac{13}{169} \\ \frac{26}{169} & -\frac{39}{169} & -\frac{13}{169} \\ -\frac{13}{169} & -\frac{13}{169} & -\frac{65}{169} \end{array}\right]$
$\Rightarrow $ $adj (A^{-1})$ = $\frac{1}{13}$$\left[\begin{array}{ccc} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right]$ ...(ii)
समी (i) तथा (ii) से, $adj (A^{-1}) = (adj A)^{-1}$
View full question & answer→Question 35 Marks
यदि $A^{-1} = \left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]$और B = $ \left[\begin{array}{ccc} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{array}\right] $, हो तो $AB)^{-1}$ का मान ज्ञात कीजिए।
Answerहम जानते हैं कि $(AB)^{-1} = B^{-1}A^{-1}$ यहाँ $A^{-1}$ दिया गया है अतः हम $B^{-1}$ ज्ञात करते हैं।
यहाँ, |B| = $\left|\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right|$ = 1(3 - 0) - 2(- 1 - 0) - 2(2 - 0) = 3 + 2 - 4 = 1 $\neq$ 0
$\therefore$ B व्युत्क्रमणीय आव्यूह है, अतः $B^{-1}$ विद्यमान है।
B के सहखण्ड निम्न हैं,
$B_{11}= (3 - 0) = 3, B_{12}= - (- 1 - 0) = 1, B_{13}= (2 - 0) = 2,$
$B_{21}= - (2 - 4) = 2, B_{22}= 1 - 0 = 1, B_{23}= - (- 2 - 0) = 2,$
$B_{31}= (0 + 6) = 6, B_{32}= - (0 - 2) = 2, B_{33}= 3 + 2 = 5,$
adj(B) = $ \left[\begin{array}{lll} 3 & 1 & 2 \\ 2 & 1 & 2 \\ 6 & 2 & 5 \end{array}\right]^{T} $= $\left[\begin{array}{lll} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right]$
$\therefore B^{-1} = \frac{1}{|B|}$ adj (B) = $\frac{1}{1}$$\left[\begin{array}{lll} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right]$ =$\left[\begin{array}{lll} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right] $
अब $(AB)^{-1}= B^{-1} A^{-1} = \left[\begin{array}{lll} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right]$$ \left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] $
= $\left[\begin{array}{ccc} 9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10 \end{array}\right]$= $\left[\begin{array}{ccc} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{array}\right]$
View full question & answer→Question 45 Marks
यदि a, b और c वास्तविक संख्याएँ हो और सारणिक
$ \Delta$ = $ \left|\begin{array}{lll} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right|$= 0
हो तो दर्शाइए कि या तो $a + b + c = 0$ या $a = b = c$ है।
Answer$\Delta$ $= 2(a + b + c) [(b - c) (c - b) - (b - a) (c - a)]$
$= 2(a + b + c)[bc - c^2- b^2+ bc - bc + ac + ab - a^2]$
$= 2(a + b + c) (ab + bc + ca - a^2- b^2- c^2)$
दिया है, $\Delta$ = 0
$\therefore$ (a + b + c) $ \times$ $2(ab + bc + ca - a^2- b^2- c^2) = 0$
$\Rightarrow$ या तो $( a + b + c) = 0$ या $2(ab + bc + ca - a^2 - b^2 - c^2) = 0$
यदि $2(ab + bc + ca - a^2 - b^2- c^2) = 0$
$ \Rightarrow$ -$(2 a^2+ 2b^2+ 2c^2- 2ab - 2bc - 2ca) = 0$ [(-) चिन्ह उभयनिष्ठ लेने पर]
$ \Rightarrow$ $2 a^2+ 2b^2+ 2c^2- 2ab - 2bc - 2ca = 0$
$ \Rightarrow$ $(a^2+ b^2- 2ab) + (b^2+ c^2- 2bc) + (c^2+ a^2- 2ca) = 0$
$ \Rightarrow$ $(a - b)^2+ (b - c)^2+ (c - a)^2= 0$ [$\because$ $(x - y)^2= x^2+ y^2- 2xy]$
$\Rightarrow $ $(a-b)^2= (b - c)^2= (c - a)^2 = 0$ [$\because$$(a - b)^2, (b - c)^2, (c - a)^2$ धनात्मक हैं]
$\Rightarrow $$(a - b) = (b - c) = (c - a) = 0$
$\Rightarrow $ $a = b = c$
अतः यदि $\Delta$ $= 0$ तो $a + b + c = 0$ या $a = b = c$
View full question & answer→Question 55 Marks
निम्नलिखित समीकरण निकाय को हल कीजिए
$ \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4$
$ \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1$
$ \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$
Answerमाना $\frac{1}{x} = p, \frac{1}{y} = q$ तथा $\frac{1}{z} = r$ तब दिया गया समीकरण निकाय निम्न हैं
$2p + 3q + 10r = 4, 4p - 6q + 5r = 1, 6p + 9q - 20r = 2$
इस समीकरण निकाय को निम्न रूप में लिख सकते हैं $AX = B$, जहाँ
$A = \left[\begin{array}{ccc} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{array}\right] , X = \left[\begin{array}{l} p \\ q \\ r \end{array}\right]$
तथा $B = \left[\begin{array}{l} 4 \\ 1 \\ 2 \end{array}\right]$
यहाँ,$ |A| = \left|\begin{array}{ccc} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{array}\right| $
$= 2(120 - 45) - 3(- 80 - 30) + 10(36 + 36)$
$= 150 + 330 + 720 = 1200$
$\therefore A$ व्युत्क्रमणीय आव्यूह है।
$\Rightarrow A^{-1}$ विद्यमान है।
अतः दिए गए समीकरण निकाय का अद्वितीय हल निम्न है $X = A^{-1}B $
$A$ के सहखण्ड निम्न है,
$A_{11}= 120 - 45 = 75, A_{12}= - (- 80 - 30) = 110, A_{13}= (36 + 36) = 72, $
$ A_{21}= - (- 60 - 90) = 150, A_{22}= (- 40 - 60) = - 100, A_{23}= - (18 - 18) = 0, $
$ A_{21}= 15 + 60 = 75, A_{22}= - (10 - 40) = 30, A_{23}= - 12 - 12 = - 24$
$\therefore adj(A) = \left[\begin{array}{ccc} 75 & 110 & 72 \\ 150 & -100 & 0 \\ 75 & 30 & -24 \end{array}\right]^{\top}$
$= \left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right]$
अब, $A^{-1} = \frac{1}{|A|} (adj A) = \frac{1}{1200} \left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right]$
$\because X = A^{-1} B \Rightarrow \left[\begin{array}{l} p \\ q \\ r \end{array}\right] = \frac{1}{1200} \left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right] \left[\begin{array}{l} 4 \\ 1 \\ 2 \end{array}\right] $
$= \frac{1}{1200} \left[\begin{array}{c} 300+150+150 \\ 440-100+60 \\ 288+0-48 \end{array}\right] $
$= \frac{1}{1200} \left[\begin{array}{c} 600 \\ 400 \\ 240 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5} \end{array}\right] $
$ \Rightarrow p = \frac{1}{2} , q = \frac{1}{3} , r = \frac{1}{5}$
$\Rightarrow \frac{1}{x} = \frac{1}{2} , \frac{1}{y} = \frac{1}{3} , \frac{1}{z} = \frac{1}{5}$
$\Rightarrow x = 2, y = 3 $ तथा $z = 5$
View full question & answer→Question 65 Marks
दी गई समीकरण निकायों का संगत अथवा असंगत के रूप में वर्गीकरण कीजिए: $3x - y - 2z = 2 , 2y - z = - 1 , 3x - 5y = 3$
Answerदिया गया समीकरण निकाय निम्न है $3x - y - 2z = 2,$
$0x + 2y - z = - 1$ तथा $3x - 5y + 0z = 3,$
जिसे निम्न रूप से लिख सकते हैं $AX = B,$
जहाँ $A =\left[\begin{array}{rrr} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right], X = \left[\begin{array}{l} x \\ y \\ z \end{array}\right]$ तथा $B =\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right] $
यहाँ, $|A| = \left|\begin{array}{rrr} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right|$
$= 3(0 - 5) + 1(0 + 3) - 2(0 - 6) $
$= - 15 + 3 + 12 $
$= 0$
$\therefore A$ अव्युत्क्रमणीय आव्यूह है।
अब, $A$ के सहखण्ड निम्न हैं
$A_{11}= - 5, A_{12}= -3, A_{13}= - 6, A_{21}= 10, A_{22}= 6, A_{23}= 12$
$A_{31}= 5, A_{32}= 3, A_{33}= 6$
$\therefore adj \ (A) = \left[\begin{array}{rrr} -5 & -3 & -6 \\ 10 & 6 & 12 \\ 5 & 3 & 6 \end{array}\right]^{T} = \left[\begin{array}{rrr} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{array}\right]$
$\therefore (adj \ A)B = \left[\begin{array}{rrr} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{array}\right] \left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]$
$= \left[\begin{array}{c} -10-10+15 \\ -6-6+9 \\ -12-12+18 \end{array}\right]$
$=\left[\begin{array}{c} -5 \\ -3 \\ -6 \end{array}\right] \neq 0$
View full question & answer→Question 75 Marks
दी गई समीकरण निकायों का संगत अथवा असंगत के रूप में वर्गीकरण कीजिए: $x + y + z = 1 , 2x + 3y + 2z = 2 , ax + ay + 2az = 4$
Answerदिए गए समीकरण निकाय को निम्न रूप में लिखा जा सकता है
$A =\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a \end{array}\right] , X = \left[\begin{array}{l} x \\ y \\ z \end{array}\right]$ तथा $B = \left[\begin{array}{l} 1 \\ 2 \\ 4 \end{array}\right]$
यहाँ, $|A| =\left|\begin{array}{lll} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a \end{array}\right| $
$= 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a)$
$= 4a - 2a - a $
$= 4a - 3a$
$= a \neq 0$
$\therefore A$ व्युत्क्रमणीय आव्यूह है।
$\therefore A^{-1}$ विद्यमान है। अतः समीकरण निकाय संगत है।
View full question & answer→Question 85 Marks
$4 \ kg$ प्याज, $3 \ kg$ गेहूँ और $2 \ kg$ चावल का मूल्य $₹. 60$ है। $2 \ kg$ प्याज, $4 \ kg$ गेहूँ और $6\ kg$ चावल का मूल्य $₹. 90$ है। $6 \ kg$ प्याज, $2 \ kg$ और $3 \ kg$ चावल का मूल्य $₹. 70$ है। आव्यूह विधि द्वारा प्रत्येक का मूल्य प्रति $kg$ ज्ञात कीजिए।
Answerमाना प्याज गेहूँ तथा चावल का मूल्य क्रमश: $₹. x, ₹. y$ तथा $₹. z$ है, तब
$4x + 3y + 2z = 60, 2x + 4y + 6z = 90, 6x + 2y + 3z = 70$
इस समीकरण निकाय को निम्न रूप में लिख सकते हैं $AX = B,$ जहाँ
$A = \left[\begin{array}{lll}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right], X = \left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ तथा $B = \left[\begin{array}{l}60 \\ 90 \\ 70\end{array}\right]$
यहाँ, $|A| =\left|\begin{array}{lll} 2 & 4 & 6 \\ 6 & 2 & 3 \end{array}\right|$
$= 4(12 - 12) - 3(6 - 36) + 2(4 - 24)$
$= 0 + 90 - 40 $
$= 50 \neq 0$
$\therefore A$ व्युत्क्रमणीय आव्यूह है।
$\Rightarrow A^{-1}$ विद्यमान है।
अतः दिया गया समीकरण निकाय संगत है जिसका अद्वितीय हल निम्न है
$X = A^{-1} B$
$A$ के सहखण्ड निम्न हैं
$A_{11}= 12 - 12 = 0, A_{12}= - (6 - 36) = 30, A_{13}= 4 - 24 = - 20,$
$A_{21}= - (9 - 4) = - 5, A_{22}= 12 - 12 = 0, A_{23}= - (8 - 18) = 10,$
$A_{31}= (18 - 8) = 10, A_{32} = - (24 - 4) = - 20, A_{33}= 16 - 6 = 10$
$adj \ (A) =\left[\begin{array}{rrr} 0 & 30 & -20 \\ -5 & 0 & 10 \\ 10 & -20 & 10 \end{array}\right]^{T}= \left[\begin{array}{rrr} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array}\right] $
$\therefore A^{-1} = \frac{1}{|A|} (adj \ A) = \frac{1}{50} \left[\begin{array}{rrr} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array}\right] $
अब, $X = A^{-1} B \Rightarrow \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \frac{1}{50} \left[\begin{array}{rrr} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array}\right] \left[\begin{array}{l} 60 \\ 90 \\ 70 \end{array}\right] $
$= \frac{1}{50} \left[\begin{array}{c} 1800+0-1400 \\ -1200+900+700 \end{array}\right] $
$\Rightarrow \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \frac{1}{50} \left[\begin{array}{l} 250 \\ 400 \\ 400 \end{array}\right] = \left[\begin{array}{l} 5 \\ 8 \\ 8 \end{array}\right]$
$\Rightarrow \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \left[\begin{array}{l} 5 \\ 8 \\ 8 \end{array}\right]$
$\therefore x = 5, y = 8$ तथा $z = 8$
अतः $1$ किग्रा प्याज का मूल $₹. 5,1$ किग्रा गेहूँ का मूल्य $₹. 8$ तथा $1$ किग्रा चावल का मूल्य $₹. 8$ है।
View full question & answer→Question 95 Marks
यदि $ A =\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]$ है तो $A^{-1}$ ज्ञात कीजिए। $A^{-1}$ का प्रयोग करके निम्नलिखित समीकरण निकाय को हल कीजिए। $2x - 3y + 5z = 11 , 3x + 2y - 4z = - 5 , x + y - 2z = - 3$
Answerदिए गए समीकरण निकाय को निम्न रूप में लिख सकते हैं $AX = B$ जहाँ
$A = \left[\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right], X = \left[\begin{array}{l} x \\ y \\ z \end{array}\right] $ तथा $B =\left[\begin{array}{c} 11 \\ -5 \\ -3 \end{array}\right]$
यहाँ, $|A| =\left|\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right|$
$= 2(- 4 + 4) - (- 3)(-6 + 4) + 5(3 - 2)$
$= 0 - 6 + 5$
$= - 1\neq 0$
$\therefore A$ व्युत्क्रमणीय आव्यूह है।
$\Rightarrow A^{-1}$ विद्यमान है।
अतः दिया गया समीकरण निकाय संगत है जिसका अद्वितीय हल निम्न है
$X = A^{-1} B$
$A$ के सहखण्ड निम्न हैं
$A_{11}= - 4 + 4 = 0, A_{12}= - (- 6 + 4) = 2, A_{13}= 3 - 2 = 1,$
$A_{21}= - (6 - 5) = - 1, A_{22}= - 4 - 5 = - 9, A_{23}= -(2 + 3) = - 5,$
$A_{31}= (12 - 10) = 2, A_{32}= - (- 8 - 15) = 23, A_{33}= 4 + 9 = 13$
$adj\ (A) =\left[\begin{array}{rrr} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{array}\right]^{T} = \left[\begin{array}{rrr} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] $
$\therefore A^{-1} = \frac{1}{|A|} (adj) \ A) = \frac{1}{-1} \left[\begin{array}{rrr} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] = \left[\begin{array}{rrr} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right] $
$X = A^{-1}B \Rightarrow \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \left[\begin{array}{rrr} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right] \left[\begin{array}{l} 11 \\ -5 \\ -3 \end{array}\right] $
$\Rightarrow \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \left[\begin{array}{c} 0-5+6 \\ -22-45+69 \\ -11-25+39 \end{array}\right] = \left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$
अतः $x = 1, y = 2$ तथा $z = 3$
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समीकरण निकाय को आव्यूह विधि से हल कीजिए। $x - y + 2z = 7 , 3x + 4y - 5z = - 5 , 2x - y + 3z = 12$
Answerदिए गए समीकरण निकाय को निम्न रूप में लिख सकते हैं $AX = B$ जहाँ
$A = \left[\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right], X = \left[\begin{array}{l} x \\ y \\ z \end{array}\right] $ तथा $B =\left[\begin{array}{c} 11 \\ -5 \\ -3 \end{array}\right]$
यहाँ, $|A| =\left|\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right|$
$= 2(- 4 + 4) - (- 3)(-6 + 4) + 5(3 - 2)$
$= 0 - 6 + 5$
$= - 1\neq 0$
$\therefore A$ व्युत्क्रमणीय आव्यूह है।
$\Rightarrow A^{-1}$ विद्यमान है।
अतः दिया गया समीकरण निकाय संगत है जिसका अद्वितीय हल निम्न है
$X = A^{-1} B$
$A$ के सहखण्ड निम्न हैं
$A_{11}= - 4 + 4 = 0, A_{12}= - (- 6 + 4) = 2, A_{13}= 3 - 2 = 1,$
$A_{21}= - (6 - 5) = - 1, A_{22}= - 4 - 5 = - 9, A_{23}= -(2 + 3) = - 5,$
$A_{31}= (12 - 10) = 2, A_{32}= - (- 8 - 15) = 23, A_{33}= 4 + 9 = 13$
$adj\ (A) =\left[\begin{array}{rrr} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{array}\right]^{T} = \left[\begin{array}{rrr} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] $
$\therefore A^{-1} = \frac{1}{|A|} (adj) \ A) = \frac{1}{-1} \left[\begin{array}{rrr} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] = \left[\begin{array}{rrr} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right] $
$X = A^{-1}B \Rightarrow \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \left[\begin{array}{rrr} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right] \left[\begin{array}{l} 11 \\ -5 \\ -3 \end{array}\right] $
$\Rightarrow \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \left[\begin{array}{c} 0-5+6 \\ -22-45+69 \\ -11-25+39 \end{array}\right] = \left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$
अतः $x = 1, y = 2$ तथा $z = 3$
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समीकरण निकाय को आव्यूह विधि से हल कीजिए: $2x + 3y + 3z = 5 ,x - 2y + z = - 4 , 3x - y - 2z = 3$
Answerदिए गए समीकरण निकाय को निम्न रूप में लिख सकते हैं $AX = B,$ जहाँ
$A =\left[\begin{array}{rrr} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{array}\right] , X = \left[\begin{array}{l} x \\ y \\ z \end{array}\right]$ तथा $B = \left[\begin{array}{c} 5 \\ -4 \\ 3 \end{array}\right]$
यहाँ, $|A| =\left|\begin{array}{rrr} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{array}\right| $
$= 2(4 + 1) - 3(- 2 - 3) + 3(- 1 + 6)$
$= 10 + 15 + 15 $
$= 40 \neq 0$
$\therefore A$ व्युत्क्रमणीय अव्यूह है।
$\therefore A^{-1}$ विद्यमान है।
अतः दिया गया समीकरण निकाय संगत है जिसका अद्वितीय हल निम्न है $X = A^{-1} B$
$A$ के सहखण्ड निम्न हैं $A_{11}= 4 + 1 = 5, A_{12}= - (- 2 - 3) = 5, A_{13}= (- 1 + 6) = 5,$
$A_{21}= - (- 6 + 3) = 3, A_{22}= (- 4 - 9) = - 13, A_{23}= - (- 2 - 9) = 11,$
$A_{31}= 3 + 6 = 9, A_{32}= - (2 - 3) = 1, A_{33}= - 4 - 3 = - 7$
$ adj \ (A) =\left[\begin{array}{rrr}5 & 5 & 5 \\ 3 & -13 & 11 \\ 9 & 1 & -7\end{array}\right]^{T} = \left[\begin{array}{rrr}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right]$
$\therefore A^{-1} = \frac{1}{|A|} (adj \ A) = \frac{1}{40}\left[\begin{array}{rrr}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right]$
अब, $X = A^{-1}B$
$\Rightarrow \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \frac{1}{40} \left[\begin{array}{rrr} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{array}\right] \left[\begin{array}{c} 5 \\ -4 \\ 3 \end{array}\right] $
$= \frac{1}{40} \left[\begin{array}{c} 25-12+27 \\ 25+52+3 \\ 25-44-21 \end{array}\right]$
$=\frac{1}{40}\left[\begin{array}{c} 40 \\ 80 \\ -40 \end{array}\right]$
$=\left[\begin{array}{c} 1 \\ 2 \\ -1 \end{array}\right]$
अतः $x = 1, y = 2$ तथा $z = - 1$
View full question & answer→Question 125 Marks
समीकरण निकाय को आव्यूह विधि से हल कीजिए। $x - y + z = 4 , 2x + y - 3z = 0 , x + y + z = 2$
Answerदिए गए समीकरण निकाय को निम्न रूप में लिख सकते हैं $AX = B,$
जहाँ $A =\left[\begin{array}{rrr}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right] , X = \left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ तथा $B = \left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$
यहाँ $|A| =\left|\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right|$
$= 1(1 + 3) - (- 1)(2 + 3) + 1(2 - 1) $
$= 4 + 5 + 1$
$= 10 \neq 0$
$\therefore A$ व्युत्क्रमणीय आव्यूह है।
$\Rightarrow A^{-1}$ विद्यमान है।
अतः दिया गया समीकरण निकाय संगत है। जिसका अद्वितीय हल निम्न है
$X = A^{-1} B$
$A$ के सहखण्ड निम्न हैं $A_{11}= 1 + 3 = 4, A_{12}= - (2 + 3) = - 5, A_{13}= 2 - 1 = 1,$
$A_{21}= - (- 1 - 1) = 2, A_{22}= 1 - 1 = 0, A_{23}= - (1 + 1) = - 2,$
$A_{31}= 3 - 1 = 2, A_{32}= - (- 3 - 2) = 5, A_{33}= 1 + 2 = 3$
$adj\ (A) =\left[\begin{array}{rrr} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{array}\right]^{T} = \left[\begin{array}{rrr} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array}\right]$
$ \therefore A^{-1} = \frac{1}{|A|} (adj \ A) = \frac{1}{10} \left[\begin{array}{rrr} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array}\right] $
अब, $X = A^{-1} B \Rightarrow \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \frac{1}{10} \left[\begin{array}{rrr} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array}\right] \left[\begin{array}{l} 4 \\ 0 \\ 2 \end{array}\right] $
$\Rightarrow \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \frac{1}{10} \left[\begin{array}{c} 16+0+4 \\ -20+0+10 \\ 4+0+6 \end{array}\right] $
$\Rightarrow \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \frac{1}{10} \left[\begin{array}{c} 20 \\ -10 \\ 10 \end{array}\right] $
$= \left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right] $
अतः $x = 2, y = - 1$ तथा $z = 1$
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समीकरण निकाय को आव्यूह विधि से हल कीजिए: $2x + y + z = 1 , x - 2y - z = \frac{3}{2} , 3y - 5z = 9$
Answerदिए गए समीकरण निकाय को निम्न रूप में लिख सकते हैं $AX = B$, जहाँ
$A = \left[\begin{array}{rrr} 2 & 1 & 1 \\ 2 & -4 & -2 \\ 0 & 3 & -5 \end{array}\right] , X = \left[\begin{array}{l} x \\ y \\ z \end{array}\right]$ तथा $ B =\left[\begin{array}{l} 1 \\ 3 \\ 9 \end{array}\right] $
यहाँ, $|A| = \left[\begin{array}{rrr}2 & 1 & 1 \\ 2 & -4 & -2 \\ 0 & 3 & -5\end{array}\right]$
$= 2(20 + 6) - 1(- 10 - 0) + 1(6 - 0) $
$= 52 + 10 + 6$
$= 68 \neq 0$
$\therefore A$ व्युत्क्रमणीय आव्यूह है।
$\Rightarrow A^{-1}$ विद्यमान है।
अतः दिया गया समीकरण निकाय संगत है जिसका अद्वितीय हल निम्न है $X = A^{-1}B$
$A$ के सहखण्ड निम्न हैं
$A_{11}= 20 + 6 = 26, A_{12}= -(- 10 + 0) = 10, A_{13}= 6 + 0 = 6$
$A_{21}= - (- 5 - 3) = 8, A_{22}= - 10 - 0 = - 10, A_{23}= - (6 - 0) = - 6$
$A_{31}=(- 2 + 4) = 2, A_{32}= - (- 4 - 2) = 6, A_{33}= - 8 - 2 = - 10$
$ adj \ (A) =\left[\begin{array}{rrr} 26 & 10 & 6 \\ 8 & -10 & -6 \\ 2 & 6 & -10 \end{array}\right]^{T} = \left[\begin{array}{rrr} 26 & 8 & 2 \\ 10 & -10 & 6 \\ 6 & -6 & -10 \end{array}\right] $
$\therefore A^{-1} = \frac{1}{|A|} (adj \ A) = \frac{1}{68} \left[\begin{array}{rrr} 26 & 8 & 2 \\ 10 & -10 & 6 \\ 6 & -6 & 10 \end{array}\right] $
अब, $X = A^{-1} B \Rightarrow \left[\begin{array}{l}x \\ y \\ z\end{array}\right] = \frac{1}{68}\left[\begin{array}{rrr}26 & 8 & 2 \\ 10 & -10 & 6 \\ 6 & -6 & -10\end{array}\right] \left[\begin{array}{l}1 \\ 3 \\ 9\end{array}\right]$
$\Rightarrow \left[\begin{array}{l}x \\ y \\ z\end{array}\right] = \frac{1}{68} \left[\begin{array}{c}26+24+18 \\ 10-30+54 \\ 6-18-90\end{array}\right] = \frac{1}{68} \left[\begin{array}{c}68 \\ 34 \\ -102\end{array}\right]$
$\Rightarrow \left[\begin{array}{l}x \\ y \\ z\end{array}\right] = \left[\begin{array}{c}1 \\ \frac{1}{2} \\ \frac{-3}{2}\end{array}\right]$
अतः $x = 1, y = \frac{1}{2}$ तथा $ z =\frac{-3}{2}$
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आव्यूह के व्युत्क्रम (जिनका अस्तित्व हो) ज्ञात कीजिए: $\left[\begin{array}{ccc} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{array}\right]$
Answerमाना A = $ \left[\begin{array}{rrr} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{array}\right]$ $\therefore$ |A| = $ \left|\begin{array}{rrr} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{array}\right|$
= 2(- 1 - 0) - 1(4 - 0) + 3(8 - 7) = - 2 - 4 + 3 = - 3
A के सहखण्ड निम्न हैं
$A_{11}= - 1 - 0 = - 1, A_{12}= - (4 - 0) = - 4, A_{13}= 8 - 7 = 1$
$A_{21}= - (1 - 6) = 5, A_{22} = 2 + 21 = 23, A_{23}= - (4 + 7) = - 11$
$A_{31}= 0 + 3 = 3, A_{32}= - (0 - 12) = 12, A_{33}= - 2 - 4 = - 6$
$\therefore $ adj (A) =$ \left[\begin{array}{rrr} -1 & -4 & 1 \\ 5 & 23 & -11 \\ 3 & 12 & -6 \end{array}\right]^{T}$= $\left[\begin{array}{rrr} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{array}\right]$
अब, $A^{-1} =\frac{1}{|A|}$ (adj A) = $\frac{1}{-3}$$ \left[\begin{array}{rrr} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{array}\right]$= $ \left[\begin{array}{rrr} \frac{1}{3} & -\frac{5}{3} & -1 \\ \frac{4}{3} & -\frac{23}{3} & -4 \\ -\frac{1}{3} & \frac{11}{3} & 2 \end{array}\right]$
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आव्यूह के व्युत्क्रम (जिनका अस्तित्व हो) ज्ञात कीजिए: $\left[\begin{array}{ccc} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{array}\right]$
Answerमाना A = $\left[\begin{array}{rrr} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{array}\right]$ $\therefore$ |A| = $\left|\begin{array}{rrr} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{array}\right|$ = 1(- 3 - 0) - 0 + 0 = - 3
A के सहखण्ड निम्न हैं
$A_{11}= - 3 - 0 = - 3, A_{12}= - (- 3 - 0) = 3, A_{13} = 6 - 15 = - 9$
$A_{21}= - (0 - 0) = 0, A_{22}= - 1 - 0 = - 1, A_{23}= - (2 - 0) = - 2$
$A_{31}= 0 - 0 = 0, A_{32}= - (0 - 0) = 0, A_{33}= 3 - 0 = 3$
$\therefore $ adj (A) = $\left[\begin{array}{rrr} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \end{array}\right]$ = $\left[\begin{array}{rrr} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{array}\right]$
अब, $A^{-1} = \frac{1}{|A|}$ (adj A) = $\frac{1}{-3}$ $\left[\begin{array}{rrr} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{array}\right]$= $ \left[\begin{array}{rrr} 1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1 \end{array}\right] $
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आव्यूह के व्युत्क्रम (जिनका अस्तित्व हो) ज्ञात कीजिए: $\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}\right]$
Answerमाना A = $ \left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}\right]$
$ \therefore |A| = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) = 10$
$A $ के सहखण्ड निम्न हैं
$A_{11}= 10 - 0 = 10, A_{12}= - (0 - 0) = 0, A_{13}= 0 - 0 = 0$
$A_{21}= - (10 - 0) = - 10$
$A_{22}= 5 - 0 = 5, A_{23}= -(0 - 0) = 0$
$A_{31}= 8 - 6 = 2, A_{32}= - (4 - 0) = - 4, A_{33}= 2 - 0 = 2$
$ \therefore$ adj (A) = $\left[\begin{array}{rrr} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2 \end{array}\right]^{T}$ = $\left[\begin{array}{rrr} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right]$
अब $A^{-1}= \frac{1}{|A|}$ adj (A) =$ \frac{1}{10}$ $ \left[\begin{array}{rrr} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right] $ $= \left[\begin{array}{ccr} 1 & -1 & \frac{1}{5} \\ 0 & \frac{1}{2} & -\frac{2}{5} \\ 0 & 0 & \frac{1}{5} \end{array}\right]$
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यदि A = $\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$, तो सत्यापित कीजिए कि $A^3- 6A^2+ 9A - 4I = O$ है तथा इसकी सहायता से $A^{-1}$ ज्ञात कीजिए।
Answerदिया है, A = $\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$
$\therefore$$ A^2= A A =$ $\left[\begin{array}{rrr} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right]$$\left[\begin{array}{rrr} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right]$
= $\left[\begin{array}{rr} 4+1+1 & -2-2-1 & 2+1+2 \\ -2-2-1 & 1+4+1 & -1-2-2 \\ 2+1+2 & -1-2-2 & 1+1+4 \end{array}\right]$= $\left[\begin{array}{rrr} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right] $
तथा $A^3= A^2 A$ = $\left[\begin{array}{rrr} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right]$ $\left[\begin{array}{rrr} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right]$
= $\left[\begin{array}{rrr} 12+5+5 & -6-10-5 & 6+5+10 \\ -10-6-5 & 5+12+5 & -5-6-10 \\ 10+5+6 & -5-10-6 & 5+5+12 \end{array}\right]$= $\left[\begin{array}{rrr} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{array}\right]$
$ \therefore $$ A^3 - 6A^2+ 9A - 4I$
= $ \left[\begin{array}{rrr} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{array}\right]$- $6\left[\begin{array}{rrr} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right]$ + $9\left[\begin{array}{rrr} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right]$ - $ 4\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] $
= $ \left[\begin{array}{rrr} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{array}\right]$ - $ \left[\begin{array}{rrr} 36 & -30 & 30 \\ -30 & 36 & -30 \\ 30 & -30 & 36 \end{array}\right]$+ $\left[\begin{array}{rrr} 18 & -9 & 9 \\ -9 & 18 & -9 \\ 9 & -9 & 18 \end{array}\right]$ - $\left[\begin{array}{lll} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right] $
= $ \left[\begin{array}{rr} 22-36+18-4 & -21+30-9-0 & 21-30+9-0 \\ -21+30-9-0 & 22-36+18-4 & -21+30-9-0 \\ 21-30+9-0 & -21+30-9-0 & 22-36+18-4 \end{array}\right]$
= $\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$= 0
$ \therefore $ $A^3- 6A^2+ 9A - 4I = 0$ $\Rightarrow$ $(A A A) A^{-1}- 6(A A) A^{-1}+ 9 A A^{-1}- 4I A^{-1}= 0 (A^{-1}$ से पूर्व गुणा करने पर क्योंकि |A| $\neq$ 0)
[$\because$ | A | = $\left|\begin{array}{rrr}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right|$ $= 2(4 - 1) + 1(- 2 + 1) + 1(1 - 2) = 6 - 1 - 1 = 4$ $\neq$ 0]
$\Rightarrow$ $A A (A A^{-1}) - 6A(A A^{-1}) + 9(A A^{-1}) - 4(IA^{-1}) = 0$
$\Rightarrow$ $A AI - 6AI + 9I - 4 A^{-1}= 0$ ($\because $ $A A^{-1}= I$ तथा $IA^{-1}= A^{-1}$ से)
$\Rightarrow$ $A^2- 6A + 9I = 4A^{-1}$ ($\because $ $A^2I = A^2$ तथा $AI = A$ से)
$\Rightarrow$ $A^{-1}=$ $ \frac{1}{4}$ $(A^2- 6A + 9)$
= $\frac{1}{4}$ $\left\{\left[\begin{array}{rrr} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right]-\left[\begin{array}{rrr} 12 & -6 & 6 \\ -6 & 12 & -6 \\ 6 & -6 & 12 \end{array}\right]+\left[\begin{array}{lll} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right]\right\}$
= $\frac{1}{4}$ $\left[\begin{array}{rrr} 6-12+9 & -5+6+0 & 5-6+0 \\ -5+6+0 & 6-12+9 & -5+6+0 \\ 5-6+0 & -5+6+0 & 6-12+9 \end{array}\right]$ = $\frac{1}{4}$$\left[\begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right] $
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आव्यूह A = $\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]$ के लिए दर्शाइए कि $A^3- 6A^2+ 5A + 11I = O$ है। इसकी सहायता से $A^{-1}$ ज्ञात कीजिए।
Answerदिया है, A = $ \left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]$
$\therefore A^2= A A = \left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]$$\left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]$
= $\left[\begin{array}{rrr} 1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9 \end{array}\right]$ = $\left[\begin{array}{rrr} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]$
तथा $A^3= A^2 A$ =$ \left[\begin{array}{rrr} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]$ $\left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]$
= $\left[\begin{array}{rrr} 4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+16+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42 \end{array}\right]$ = $\left[\begin{array}{rrr} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{array}\right]$
$\therefore$ $e\ A^3- 6A^2+ 5A + 11I$
= $ \left[\begin{array}{rrr} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{array}\right]$- $6\left[\begin{array}{rrr} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]$ + $5\left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]$ + $11\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] $
= $\left[\begin{array}{rrr} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{array}\right]$ - $\left[\begin{array}{rrr} 24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84 \end{array}\right]$ + $\left[\begin{array}{rrr} 5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15 \end{array}\right]$+ $\left[\begin{array}{rrr} 11& 0& 0\\ 0 & 11 & 0 \\ 0 & 0& 11 \end{array}\right]$
= $\left[\begin{array}{rrr} 8-24+5+11 & 7-12+5+0 & 1-6+5+0 \\ -23+18+5+0 & 27-48+10+11 & -69+84-15+0 \\ 32-42+10+0 & -13+18-5+0 & 58-84+15+11 \end{array}\right]$ = $\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$ = 0
अब, $A^3- 6A^2+ 5A + 11I = 0$ $\Rightarrow$ $(A A A) A^{-1}- 6(A A) A^{-1}+ 5 A A^{-1}+ 11IA^{-1}= 0$
$(A^{-1}$ से पूर्व गुणा करने पर क्योंकि |A| $\neq$ 0)
[$\because$| A | = $ \left|\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right|$ = 1(6 - 3) - 1(3 + 6) +1(- 1 - 4) = 3 - 9 - 5 = - 11 $\neq$ 0]
$\Rightarrow$ $A A (A A^{-1}) - 6A (A A^{-1}) + 5(A A^{-1}) + 11(I A^{-1}) = 0$
$\Rightarrow$ $AAI -6A I + 5I + 11 A^{-1}= 0$ ($\because$ $AAI = A^2$ तथा $AI = A$ से)
$\Rightarrow$ $A^2 - 6A + 5I = -11A^{-1}$ ($\because$ $AAI = A^2$ तथा $AI = A$ से)
$\Rightarrow$ $A^{-1} =$ $-\frac{1}{11}\left(A^{2}-6 A+5 l\right)$
$\Rightarrow$ $A^{-1}=\frac{1}{11}\left(-A^{2}+6 A-5 l\right)$
= $\frac{1}{11}$ $\left\{-\left[\begin{array}{rrr} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]+6\left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]-5\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\right\}$
= $\frac{1}{11}$ $\left\{\left[\begin{array}{rrr} -4 & -2 & -1 \\ 3 & -8 & 14 \\ -7 & 3 & -14 \end{array}\right]+\left[\begin{array}{rrr} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{array}\right]-\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]\right\}$
= $\frac{1}{11}$ $\left[\begin{array}{rrr} -4+6-5 & -2+6-0 & -1+6-0 \\ 3+6-0 & -8+12-5 & 14-18-0 \\ -7+12-0 & 3-6+0 & -14+18-5 \end{array}\right]$ = $\frac{1}{11}$ $\left[\begin{array}{rrr} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{array}\right]$
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आव्यूह के व्युत्क्रम (जिनका अस्तित्व हो) ज्ञात कीजिए: $\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]$
Answerमाना A = $ \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]$
$\therefore$ |A| = $ \left|\begin{array}{rrr} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right|$ $= 1 (- \cos^2 \alpha - \sin^2 \alpha)$
$= - (\cos ^2 \alpha + \sin^2 \alpha) = - 1$ ($\because cos^2 \theta + \sin^2 \theta = 1$)
A के सहखण्ड निम्न हैं
$A_{11}= -\cos^2 \alpha - \sin^2 \alpha= -1, A_{12}= - (0 - 0) = 0, A_{13}= 0 - 0 = 0$
$A_{21}= - (0 - 0) = 0, A_{22}= - \cos \alpha - 0 = - \cos \alpha$,
$A_{23}= - (\sin \alpha - 0) = - \sin \alpha$
$A_{31}= 0 - 0 = 0, A_{32}= - (\sin \alpha - 0) = - \sin \alpha, A_{33}= \cos \alpha 0 = \cos \alpha$
$\therefore$ adj (A) = $ \left[\begin{array}{rrr} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right]^{T}$ = $ \left[\begin{array}{rrr} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right]$
अब $A^{-1} = \frac{1}{|A|}$ (adj A) = $ \frac{1}{-1}$ $ \left[\begin{array}{rrr} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right]$ = $\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right] $
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आव्यूह के व्युत्क्रम (जिनका अस्तित्व हो) ज्ञात कीजिए: $\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]$
Answerमाना A = $\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]$
$\therefore$ |A| = $\left|\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right| = 1(8 - 6) - (-1)(0 + 9) + 2(0 - 6) = 2 + 9 - 12 = - 1$
A के सहखण्ड निम्न हैं
$A_{11}= 8 - 6 = 2, A_{12}= - (0 + 9) = - 9, A_{13}= 0 - 6 = - 6$
$A_{21}= - (- 4 + 4) = 0, A_{22}= 4 - 6 = - 2, A_{23}= - (- 2 + 3) = - 1$
$A_{31}= 3 - 4 = 0, A_{32}= -(-3 - 0) = 3, A_{33}= 2 - 0 = 2$
$\therefore$ adj (A) = $\left[\begin{array}{rrr} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \end{array}\right]^{\top}$ = $\left[\begin{array}{rrr} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{array}\right]$
अब, $ A^{-1}= \frac{1}{|A|}$(adj A) = $ \frac{1}{-1}$$ \left[\begin{array}{rrr} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{array}\right]$ = $\left[\begin{array}{rrr} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right] $
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आव्यूहों के गुणनफल $\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$$\left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$ का प्रयोग करते हुए निम्नलिखित समीकरण निकाय को हल कीजिए: $x - y + 2z = 1 , 2y - 3z = 1 , 3x - 2y + 4z = 2$
Answerदिया गया गुणनफल $\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right] \left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$
$= \left[\begin{array}{ccc} -2-9+12 & 0-2+2 & 1+3-4 \\ 0+18-18 & 0+4-3 & 0-6+6 \\ -6-18+24 & 0-4+4 & 3+6-8 \end{array}\right]$ $=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] $
अतः $\left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]^{-1} = \left[\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right] $
अब दिए गए समीकरण निकाय को आव्यूह के रूप निम्नलिखित रूप में लिखा जा सकता है
$\left[\begin{array}{l} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right] \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right]$
या $ \left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]^{-1} \left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right] $
$= \left[\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right] \left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right] $
$= \left[\begin{array}{c} -2+0+2 \\ 9+2-6 \\ 6+1-4 \end{array}\right] $
$= \left[\begin{array}{l} 0 \\ 5 \\ 3 \end{array}\right] $
अतः $x = 0, y = 5$ और $z = 3$
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निम्नलिखित समीकरण निकाय $3x - 2y + 3z = 8 , 2x + y - z = 1 , 4x - 3y + 2z = 4$ को आव्यूह विधि से हल कीजिए।
Answerसमीकरण निकाय को $AX = B$ के रूप में व्यक्त किया जा सकता है जहाँ $A =\left[\begin{array}{ccc} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{array}\right]$,
$X =\left[\begin{array}{l} x \\ y \\ z \end{array}\right] $और $B =\left[\begin{array}{l} 8 \\ 1 \\ 4 \end{array}\right]$
हम देखते हैं कि
$|A| = 3(2 - 3) + 2(4 + 4) + 3(-6 - 4) = - 17 \neq 0$ है।
अतः $A$ व्युत्क्रमणीय है, और इसके व्युत्क्रम का अस्तित्व है।
$A_{11}= - 1, A_{12}= - 8, A_{13}= -10$
$A_{21}= - 5, A_{22}= - 6, A_{23}= 1$
$A_{31}= - 1, A_{32}= 9, A_{33}= 7$
इसलिए $A^{-1} = - \frac{1}{17} \left[\begin{array}{ccc}-1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7\end{array}\right]$
और $X = A^{-1}B = - \frac{1}{17} \left[\begin{array}{ccc} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{array}\right] \left[\begin{array}{l} 8 \\ 1 \\ 4 \end{array}\right]$
अतः $ \left[\begin{array}{l}x \\ y \\ z\end{array}\right] = - \frac{1}{17} \left[\begin{array}{c}-17 \\ -34 \\ -51\end{array}\right] $
$= \left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$
अतः $x = 1, y = 2$ व $z = 3$
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