यदि $A = \left[\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right]$ और B = $ \left[\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right]$ है तो सत्यापित कीजिए कि $(AB)^{-1 }= B^{-1}A^{-1 }$ है।
Exercise-4.4-12
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दिया है, $A = \left[\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right], \because |A| = \left|\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right| = 15 - 14 = 1$
$A$ के सहखण्ड निम्न हैं $A_{11 }= 5, A_{12 }= - 2, A_{21 }= - 7, A_{22 }= 3$
$\therefore adj(A) = \left[\begin{array}{rr} 5 & -2 \\ -7 & 3 \end{array}\right]^{T} = \left[\begin{array}{rr} 5 & -7 \\ -2 & 3 \end{array}\right]$
अब, $A^{-1 }= \frac{1}{|A|} adj(A) = \frac{1}{1} \left[\begin{array}{rr} 5 & -7 \\ -2 & 3 \end{array}\right]$
यहाँ, $B = \left[\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right] \therefore |B| = \left|\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right| = 54 - 56 = - 2$
$B$ के सहखण्ड निम्न हैं $B_{11 }= 9, B_{12 }= - 7, B_{21 }= - 8, B_{22 }= 6$
$adj(A) = \left[\begin{array}{rr}9 & -7 \\ -8 & 6\end{array}\right]^{T} = \left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right], अब B^{-1 }= \frac{1}{|B|} (adj B) = \frac{1}{-2} \left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right]\left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right]$
अतः $B^{-1} A^{-1 }= \frac{1}{-2} \left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right] \left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right] = \frac{1}{-2} \left[\begin{array}{rr}45+16 & -63-24 \\ -35-12 & 49+18\end{array}\right]$
$= \frac{1}{-2} \left[\begin{array}{rr} 61 & -87 \\ -47 & 67 \end{array}\right] = \left[\begin{array}{rr} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{array}\right] ...(i)$
तथा
$AB = \left[\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right] \left[\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right] = \left[\begin{array}{cc} 18+49 & 24+63 \\ 12+35 & 16+45 \end{array}\right] = \left[\begin{array}{ll} 67 & 87 \\ 47 & 61 \end{array}\right] $
$\therefore | AB| = \left|\begin{array}{ll} 67 & 87 \\ 47 & 61 \end{array}\right| = 67 \times 61 - 47 \times 87 = 4087 - 4089 = - 2$
$AB$ के सहखण्ड निम्न हैं
$adj(AB) = \left[\begin{array}{rr}61 & -47 \\ -87 & 67\end{array}\right]^{\top} = \left[\begin{array}{rr}61 & -87 \\ -47 & 67\end{array}\right]$
$\therefore (A B)^{-1 }= \frac{1}{|A B|} (adj AB) = \frac{1}{-2} \left[\begin{array}{rr} 61 & -87 \\ -47 & 67 \end{array}\right] = \left[\begin{array}{rr} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{array}\right] ..(ii)$
समी $(i)$ तथा $(ii)$ से, $(A B)^{-1 }= B^{-1 }A^{-1}$
$A$ के सहखण्ड निम्न हैं $A_{11 }= 5, A_{12 }= - 2, A_{21 }= - 7, A_{22 }= 3$
$\therefore adj(A) = \left[\begin{array}{rr} 5 & -2 \\ -7 & 3 \end{array}\right]^{T} = \left[\begin{array}{rr} 5 & -7 \\ -2 & 3 \end{array}\right]$
अब, $A^{-1 }= \frac{1}{|A|} adj(A) = \frac{1}{1} \left[\begin{array}{rr} 5 & -7 \\ -2 & 3 \end{array}\right]$
यहाँ, $B = \left[\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right] \therefore |B| = \left|\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right| = 54 - 56 = - 2$
$B$ के सहखण्ड निम्न हैं $B_{11 }= 9, B_{12 }= - 7, B_{21 }= - 8, B_{22 }= 6$
$adj(A) = \left[\begin{array}{rr}9 & -7 \\ -8 & 6\end{array}\right]^{T} = \left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right], अब B^{-1 }= \frac{1}{|B|} (adj B) = \frac{1}{-2} \left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right]\left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right]$
अतः $B^{-1} A^{-1 }= \frac{1}{-2} \left[\begin{array}{rr}9 & -8 \\ -7 & 6\end{array}\right] \left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right] = \frac{1}{-2} \left[\begin{array}{rr}45+16 & -63-24 \\ -35-12 & 49+18\end{array}\right]$
$= \frac{1}{-2} \left[\begin{array}{rr} 61 & -87 \\ -47 & 67 \end{array}\right] = \left[\begin{array}{rr} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{array}\right] ...(i)$
तथा
$AB = \left[\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right] \left[\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right] = \left[\begin{array}{cc} 18+49 & 24+63 \\ 12+35 & 16+45 \end{array}\right] = \left[\begin{array}{ll} 67 & 87 \\ 47 & 61 \end{array}\right] $
$\therefore | AB| = \left|\begin{array}{ll} 67 & 87 \\ 47 & 61 \end{array}\right| = 67 \times 61 - 47 \times 87 = 4087 - 4089 = - 2$
$AB$ के सहखण्ड निम्न हैं
$adj(AB) = \left[\begin{array}{rr}61 & -47 \\ -87 & 67\end{array}\right]^{\top} = \left[\begin{array}{rr}61 & -87 \\ -47 & 67\end{array}\right]$
$\therefore (A B)^{-1 }= \frac{1}{|A B|} (adj AB) = \frac{1}{-2} \left[\begin{array}{rr} 61 & -87 \\ -47 & 67 \end{array}\right] = \left[\begin{array}{rr} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{array}\right] ..(ii)$
समी $(i)$ तथा $(ii)$ से, $(A B)^{-1 }= B^{-1 }A^{-1}$
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