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

Answer

See the solution below.
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Solution

Let \(A=\left[\begin{array}{ccc} {1} & {2} & {1} \\ {2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]\)
\(\therefore |A|=1(15-1) - 2(10-1) + 1(2-3) = 14-18-1= -5\)
Now, \(\begin{aligned} &A_{11}=14, A_{12}=-9, A_{13}=-1\\ &A_{21}=-9, A_{22}=4, A_{23}=1\\ &A_{31}=-1, A_{12}=1, A_{13}=-1 \end{aligned}\)
\(\therefore a d j A=\left[\begin{array}{ccc} {14} & {-9} & {-1} \\ {-9} & {4} & {1} \\ {-1} & {1} & {-1} \end{array}\right]\)
So, we have \(A^{-1}=\frac{1}{|A|}(a d j A)\)
= \(-\frac{1}{5}\left[\begin{array}{ccc} {14} & {-9} & {-1} \\ {-9} & {4} & {1} \\ {-1} & {1} & {-1} \end{array}\right]\)
Now,
\( |adj A| = 14(-4-1)+9(9+1)-1(-9+4)\)
\(= 14(-5)+9(10)-1(-5)\)
\(=-70+90+5=25\)
we have,
\( a d j A=\left[\begin{array}{ccc} {14} & {-9} & {-1} \\ {-9} & {4} & {1} \\ {-1} & {1} & {-1} \end{array}\right]\)
Now,
\(\begin{aligned} &A_{11}=5, A_{12}=-10, A_{13}=-5\\ &A_{21}=10, A_{22}=-15, A_{23}=5\\ &A_{31}=-5, A_{32}=-5, A_{33}=-25 \end{aligned}\)
So,
\(adj (adj A) = \left[\begin{array}{ccc} {5} & {-10} & {-5} \\ {10} & {-15} & {5} \\ {-5} & {-5} & {-25} \end{array}\right]\)
\(\therefore [\operatorname{adj} A]^{-1}=\frac{1}{|a d j A|}(\operatorname{adj}(\operatorname{adj} A))\)
\(= \frac{1}{25} \left[\begin{array}{ccc} {5} & {-10} & {-5} \\ {10} & {-15} & {5} \\ {-5} & {-5} & {-25} \end{array}\right]\)
\(= \frac{1}{5}\left[\begin{array}{ccc} {-1} & {2} & {-1} \\ {2} & {-3} & {-1} \\ {-1} & {-1} & {-5} \end{array}\right]\)
\(=\frac{-1}{5}\left[\begin{array}{lll} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]\)\(\dots(i)\)
Now, \(A^{-1}=-\frac{1}{5}\left[\begin{array}{ccc} {14} & {-9} & {-1} \\ {-9} & {4} & {1} \\ {-1} & {1} & {-1} \end{array}\right]\)
\(\begin{array}{l} \text { Also, } \operatorname{adj}\left(A^{-1}\right)=\operatorname{adj}\left(\frac{-1}{5}\left[\begin{array}{rrr} 14 & -9 & -1 \\ -9 & 4 & 1 \\ -1 & 1 & -1 \end{array}\right]\right) \\ =\operatorname{adj}\left[\begin{array}{rrr} -\frac{14}{5} & \frac{9}{5} & \frac{1}{5} \\ \frac{9}{5} & -\frac{4}{5} & \frac{-1}{5} \\ \frac{1}{5} & \frac{-1}{5} & \frac{1}{5} \end{array}\right]=\left[\begin{array}{rrr} -\frac{5}{25} & \frac{-10}{25} & -\frac{5}{25} \\ \frac{-10}{25} & -\frac{15}{25} & -\frac{5}{25} \\ \frac{-5}{25} & -\frac{5}{25} & -\frac{25}{25} \end{array}\right]^{t} \\ \Rightarrow\left(\operatorname{adj} A^{-1}\right)=\frac{-1}{5}\left[\begin{array}{lll} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]\dots(ii) \end{array}\)
Using equation \((i)\)and, \((ii)\), we say that\( [adj A] ^{-1} = adj(A ^{-1})\).
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