$\left[\begin{array}{cc} 2 & 3 \\ -4 & -6 \end{array}\right]$ में सत्यापित कीजिए कि $A(adj A) = (adj A)\cdot A = |A|\cdot I$ है।
Exercise-4.4-3
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माना $A = \left[\begin{array}{cc} 2 & 3 \\ -4 & -6 \end{array}\right], |A| = \left|\begin{array}{cc} 2 & 3 \\ -4 & -6 \end{array}\right| = - 12 - (- 12) = -12 + 12 = 0$
$\therefore |A| I_2 = 0 \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] = \left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]= 0 $
$A$ के सहखण्ड $A_{11} = - 6, A_{12 }= 4, A_{21 }= - 3, A_{22 }= 2$
$\therefore adj A = \left[\begin{array}{cc} -6 & 4 \\ -3 & 2 \end{array}\right]^{T} = \left[\begin{array}{cc} -6 & -3 \\ 4 & 2 \end{array}\right]$
अब, $(adj A) A = \left[\begin{array}{cc}-6 & -3 \\ 4 & 2\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right] = \left[\begin{array}{cc}-12+12 & -18+18 \\ 8-8 & 12-12\end{array}\right] = \left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] = 0$
तथा $A(adj A) = \left[\begin{array}{cc} 2 & 3 \\ -4 & -6 \end{array}\right] \left[\begin{array}{cc} -6 & -3 \\ 4 & 2 \end{array}\right] = \left[\begin{array}{cc} -12+12 & -6+6 \\ 24-24 & 12-12 \end{array}\right] = \left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] = 0$
अतः $A(adj A) = (adj A) A = |A| I_2$
$\therefore |A| I_2 = 0 \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] = \left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]= 0 $
$A$ के सहखण्ड $A_{11} = - 6, A_{12 }= 4, A_{21 }= - 3, A_{22 }= 2$
$\therefore adj A = \left[\begin{array}{cc} -6 & 4 \\ -3 & 2 \end{array}\right]^{T} = \left[\begin{array}{cc} -6 & -3 \\ 4 & 2 \end{array}\right]$
अब, $(adj A) A = \left[\begin{array}{cc}-6 & -3 \\ 4 & 2\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right] = \left[\begin{array}{cc}-12+12 & -18+18 \\ 8-8 & 12-12\end{array}\right] = \left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] = 0$
तथा $A(adj A) = \left[\begin{array}{cc} 2 & 3 \\ -4 & -6 \end{array}\right] \left[\begin{array}{cc} -6 & -3 \\ 4 & 2 \end{array}\right] = \left[\begin{array}{cc} -12+12 & -6+6 \\ 24-24 & 12-12 \end{array}\right] = \left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] = 0$
अतः $A(adj A) = (adj A) A = |A| I_2$
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