Let A = [ ccc 1 & 2 & 1 2 & 3 & 1 1 & 1 & 5 ]. verify that (A–1)–1 = A

Let A= [ ccc 1 & -2 & 1 -2 & 3 & 1 1 & 1 & 5 ] |A|=1(15-1) + 2(-10-1) + 1(-2-3) = 14-22-5 = -13 Now, &A_ 11 =14, A_ 12 =11, A_ 13 =-5 &A_ 21 =11, A_ 22 =4, A_ 23 =-3 &A_ 31 =-5, A_ 12 =-3, A_ 13 =-1 a d j A= [ ccc 14 & 11 & -5 11 & 4 & -3 -5 & -3 & -1 ] A^ -1 =1 |A| (a d j A) = -1 13 [ ccc 14 & 11 & -5 11 & 4 & -3 -5 & -3 & -1 ]=1 13 [ ccc -14 & -11 & 5 -11 & -4 & 3 5 & 3 & 1 ] We have shown that A^ -1 =1 13 [ ccc -14 & -11 & 5 -11 & -4 & 3 5 & 3 & 1 ] and, Adj A-1 = 1 13 [ ccc -1 & 2 & -1 2 & -3 & -1 -1 & -1 & -5 ] Now |A^ -1 |= (1 13 )^ 3 [-14 (-13)+11 (-26)+5 (-13)]= (1 13 )^ 3 (-169)=-1 13 (A^ -1 )^ -1 = a d j A^ -1 |A^ -1 | =1 (-1 13 ) 1 13 [ ccc -1 & 2 & -1 2 & -3 & -1 -1 & -1 & -5 ]= [ ccc 1 & -2 & 1 -2 & 3 & 1 1 & 1 & 5 ]=A (A-1)-1 = A

Let A = [ ccc 1 & 2 & 1 2 & 3 & 1 1 & 1 & 5 ]. verify that (A–1)–…

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Question

Let A = $\left[\begin{array}{ccc} {1} & {2} & {1} \\ {2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]$. verify that (A^–1)^–1 = A

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Answer

Let $A=\left[\begin{array}{ccc} {1} & {-2} & {1} \\ {-2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]$
$\therefore $ |A|=1(15-1) + 2(-10-1) + 1(-2-3) = 14-22-5 = -13
Now, $\begin{aligned} &A_{11}=14, A_{12}=11, A_{13}=-5\\ &A_{21}=11, A_{22}=4, A_{23}=-3\\ &A_{31}=-5, A_{12}=-3, A_{13}=-1 \end{aligned}$
$\therefore $ $a d j A=\left[\begin{array}{ccc} {14} & {11} & {-5} \\ {11} & {4} & {-3} \\ {-5} & {-3} & {-1} \end{array}\right]$
$\therefore $ $A^{-1}=\frac{1}{|A|}(a d j A)$
= $-\frac{1}{13}\left[\begin{array}{ccc} {14} & {11} & {-5} \\ {11} & {4} & {-3} \\ {-5} & {-3} & {-1} \end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc} {-14} & {-11} & {5} \\ {-11} & {-4} & {3} \\ {5} & {3} & {1} \end{array}\right]$
We have shown that
$A^{-1}=\frac{1}{13}\left[\begin{array}{ccc} {-14} & {-11} & {5} \\ {-11} & {-4} & {3} \\ {5} & {3} & {1} \end{array}\right]$
and, Adj A^-1 = $\frac{1}{13}\left[\begin{array}{ccc} {-1} & {2} & {-1} \\ {2} & {-3} & {-1} \\ {-1} & {-1} & {-5} \end{array}\right]$
Now
$\left|A^{-1}\right|=\left(\frac{1}{13}\right)^{3}[-14 \times(-13)+11 \times(-26)+5 \times(-13)]=\left(\frac{1}{13}\right)^{3} \times(-169)=-\frac{1}{13}$
$\therefore $ $\left(A^{-1}\right)^{-1}=\frac{a d j A^{-1}}{\left|A^{-1}\right|}=\frac{1}{\left(-\frac{1}{13}\right)} \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$
$\Rightarrow$ (A^-1)^-1 = A

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