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