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।
Exercise-4.4-15
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दिया 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) $
$\Rightarrow A^{-1}=\frac{1}{11}\left(-A^{2}+6 A-5 l\right)$
$= \frac{1}{11} \left\{-\left[\begin{array}{rrr} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]+6\left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]-5\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\right\}$
$= \frac{1}{11} \left\{\left[\begin{array}{rrr} -4 & -2 & -1 \\ 3 & -8 & 14 \\ -7 & 3 & -14 \end{array}\right]+\left[\begin{array}{rrr} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{array}\right]-\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]\right\}$
$= \frac{1}{11} \left[\begin{array}{rrr} -4+6-5 & -2+6-0 & -1+6-0 \\ 3+6-0 & -8+12-5 & 14-18-0 \\ -7+12-0 & 3-6+0 & -14+18-5 \end{array}\right] = \frac{1}{11} \left[\begin{array}{rrr} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{array}\right]$
$\therefore A^{2 }= A A = \left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]\left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right]$
$= \left[\begin{array}{rrr} 1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9 \end{array}\right] = \left[\begin{array}{rrr} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right]$
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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