मान लीजिए $A =\left[\begin{array}{ccc} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]$हो तो सत्यापित कीजिए कि $[adj A]^{-1 }= adj (A^{-1})$
Miscellaneous Exercise-8(1)
Download our app for free and get started
दिया 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,
$A11 = \frac{-13}{169}, A12 = \frac{26}{169}, A13 = \frac{-13}{169}$
$A21 = \frac{26}{169}, A22 = \frac{-39}{169}, A23 = \frac{-13}{169}$
$A31 = \frac{-13}{169}, A32 = - \frac{13}{169}, A33 = - \frac{65}{169}$
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, $adj (A-1) = (adj A)-1$
$= 1(15 - 1) - (- 2)(- 10 - 1) + 1(- 2 - 3) = 14 - 22 - 5 = - 13 \neq 0$
$A$ 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,
$A11 = \frac{-13}{169}, A12 = \frac{26}{169}, A13 = \frac{-13}{169}$
$A21 = \frac{26}{169}, A22 = \frac{-39}{169}, A23 = \frac{-13}{169}$
$A31 = \frac{-13}{169}, A32 = - \frac{13}{169}, A33 = - \frac{65}{169}$
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, $adj (A-1) = (adj A)-1$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1समीकरण निकाय को आव्यूह विधि से हल कीजिए।View Solution
$x - y + 2z = 7$
$3x + 4y - 5z = - 5$
$2x - y + 3z = 12$ - 2आव्यूह के व्युत्क्रम $($जिनका अस्तित्व हो$)$ ज्ञात कीजिए: $\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]$View Solution
- 3आव्यूह के व्युत्क्रम $($जिनका अस्तित्व हो$)$ ज्ञात कीजिए: $\left[\begin{array}{ccc} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{array}\right]$View Solution
- 4यदि $A = \left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]$ है तो $A^{-1}$ ज्ञात कीजिए। $A^{-1}$ का प्रयोग करके निम्नलिखित समीकरण निकाय को हल कीजिए। $2x - 3y + 5z = 11 , 3x + 2y - 4z = - 5 , x + y - 2z = - 3$View Solution
- 5दी गई समीकरण निकायों का संगत अथवा असंगत के रूप में वर्गीकरण कीजिए:View Solution
$x + y + z = 1$
$2x + 3y + 2z = 2$
$ax + ay + 2az = 4$ - 6समीकरण निकाय को आव्यूह विधि से हल कीजिए:View Solution
$2x + y + z = 1$
$x - 2y - z = \frac{3}{2} $
$3y - 5z = 9$ - 7यदि $A = \left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$, तो सत्यापित कीजिए कि $\ce{A^{3 }- 6A^{2 }+ 9A - 4I = O}$ है तथा इसकी सहायता से $A^{-1}$ ज्ञात कीजिए।View Solution
- 8आव्यूह के व्युत्क्रम $($जिनका अस्तित्व हो$)$ ज्ञात कीजिए: $\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}\right]$View Solution
- 9आव्यूहों 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:View Solution
$x - y + 2z = 1 , 2y - 3z = 1 , 3x - 2y + 4z = 2$ - 10समीकरण निकाय को आव्यूह विधि से हल कीजिए।View Solution
$x - y + z = 4$
$2x + y - 3z = 0$
$x + y + z = 2$