STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

$C_{21}=1, C_{22}=-2, C_{23}=3$

$C_{31}=1, C_{32}=-2, C_{33}=-3$

$\Rightarrow$ $det \,A=|A|$ = $\left[ {\begin{array}{*{20}{c}}
  3&{ - 3}&4 \\ 
  2&{ - 3}&4 \\ 
  0&{ - 1}&1 
\end{array}} \right]$

$\Rightarrow$ $A^{-1} = {1\over{|A|}} $ .$(Adj A) = {1\over1}$  $\left[ {\begin{array}{*{20}{c}}
  {{C_{11}}}&{{C_{21}}}&{{C_{31}}} \\ 
  {{C_{12}}}&{{C_{22}}}&{{C_{32}}} \\ 
  {{C_{13}}}&{{C_{23}}}&{{C_{33}}} 
\end{array}} \right]$

Now, ${A^2} = \left[ {\,\begin{array}{*{20}{c}}3&{ - 4}&4\\0&{ - 1}&0\\{ - 2}&2&{ - 3}\end{array}\,} \right]$

and ${A^3} = {A^2}.A = \left[ {\begin{array}{*{20}{c}}3&{ - 4}&4\\0&{ - 1}&0\\{ - 2}&2&{ - 3}\end{array}} \right]\, \times \,\left[ {\begin{array}{*{20}{c}}3&{ - 3}&4\\2&{ - 3}&4\\0&{ - 1}&1\end{array}} \right]$

$ = \left[ {\,\begin{array}{*{20}{c}}1&{ - 1}&0\\{ - 2}&3&{ - 4}\\{ - 2}&3&{ - 3}\end{array}\,} \right] = {A^{ - 1}}$.