Question

Let A = ${\displaystyle \left[\begin{array}{ccc}1& -2& 1\\ -2& 3& 1\\ 1& 1& 5\end{array}\right]}$ verify that

1. [adj A]^–1 = adj (A^–1)
2. (A^–1)^–1 = A

Answer 1

A = ${\displaystyle \left[\begin{array}{ccc}1& -2& 1\\ -2& 3& 1\\ 1& 1& 5\end{array}\right]}$ 

∴ |A| = 1(15 - 1) + 2(-10 - 1) + 1(-2 - 3)

= 14 - 22 - 5

= -13

Now, ${\displaystyle A_{11}=14,A_{12}=11,A_{13}=-5}$

${\displaystyle A_{21}=11,A_{22}=4,A_{23}=-3}$

${\displaystyle A_{31}=-5,A_{32}=-3,A_{33}=-1}$

∴ adj A = ${\displaystyle \left[\begin{array}{ccc}14& 11& -5\\ 11& 4& -3\\ -5& -3& -1\end{array}\right]}$

∴ ${\displaystyle A^{-1}=\frac{1}{\left|A\right|}}$(adj A)

= ${\displaystyle -\frac{1}{13}\left[\begin{array}{ccc}14& 11& -5\\ 11& 4& -3\\ -5& -3& -1\end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc}-14& -11& 5\\ -11& -4& 3\\ 5& 3& 1\end{array}\right]}$

(i) |adj A| = 14(-4 - 9) - 11(-11 - 15)-5(-33 + 20)

= 14(-13) - 11(-26) - 5(-13)

= -183 + 286 + 65 = 169

we have,

adj(adj A) = ${\displaystyle \left[\begin{array}{ccc}-13& 26& -13\\ 26& -39& -13\\ -13& -13& -65\end{array}\right]}$

∴ ${\displaystyle {\left[adjA\right]}^{-1}=\frac{1}{\left|adjA\right|}\left(adj\left(adjA\right)\right)}$

= ${\displaystyle \frac{1}{169}\left[\begin{array}{ccc}-13& 26& -13\\ 26& -39& -13\\ -13& -13& -65\end{array}\right]}$

= ${\displaystyle \frac{1}{13}\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]}$

Now, ${\displaystyle A^{-1}=\frac{1}{13}\left[\begin{array}{ccc}-14& -11& 5\\ -11& -4& 3\\ 5& 3& 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{14}{13}& -\frac{11}{13}& \frac{5}{13}\\ -\frac{11}{13}& -\frac{4}{13}& \frac{3}{13}\\ \frac{5}{13}& \frac{3}{13}& \frac{1}{13}\end{array}\right]}$

∴ ${\displaystyle adj\left(A^{-1}\right)=\left[\begin{array}{ccc}-\frac{4}{169}-\frac{9}{169}& -(-\frac{11}{169}-\frac{15}{169})& -\frac{33}{169}+\frac{20}{169}\\ -(-\frac{11}{169}-\frac{15}{169})& (-\frac{14}{169}-\frac{25}{169})& -(\frac{42}{169}+\frac{55}{169})\\ -\frac{33}{169}+\frac{20}{169}& -(\frac{42}{169}+\frac{55}{169})& \frac{56}{169}-\frac{121}{169}\end{array}\right]}$

= ${\displaystyle \frac{1}{169}\left[\begin{array}{ccc}-13& 26& -13\\ 26& -39& -13\\ -13& -13& -65\end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]}$

So, ${\displaystyle {\left[adjA\right]}^{-1}=adj\left(A^{-1}\right)}$. 

(ii) We have shown that:

${\displaystyle A^{-1}=\frac{1}{13}\left[\begin{array}{ccc}-14& -11& 5\\ -11& -4& 3\\ 5& 3& 1\end{array}\right]}$

or, ${\displaystyle adj{A}^{-1}=\frac{1}{13}\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]}$

Now,

${\displaystyle \left|A^{-1}\right|={\left(\frac{1}{13}\right)}^3[-14\times (-13)+11\times (-26)+5\times (-13)]}$

${\displaystyle ={\left(\frac{1}{13}\right)}^3\times (-169)}$

${\displaystyle =-\frac{1}{13}}$

∴ ${\displaystyle {\left(A^{-1}\right)}^{-1}=\frac{adj{A}^{-1}}{\left|A^{-1}\right|}}$

${\displaystyle =\frac{1}{(-\frac{1}{13})}\times \frac{1}{13}\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{ccc}1& -2& 1\\ -2& 3& 1\\ 1& 1& 5\end{array}\right]}$ = A

Hence, (A^-1)^-1 = A