$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2\\3&4\end{array}} \right]$
$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2\\3&4\end{array}} \right]$
Official Solution
Let $P = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2\\3&4\end{array}} \right]$ . Let ${A_{ij}}$ be cofactors of ${a_{ij}}$ in P.
Then, the cofactors of elements of P are given by
${A_{11}} = {( - 1)^{1 + 1}}(4) = 4$,
${A_{12}} = {( - 1)^{1 + 2}}(3) = - 3,$
${A_{21}} = {( - 1)^{2 + 1}}(2) = - 2$
${A_{22}} = {( - 1)^{2 + 2}}(1) = 1$
% $\therefore$ $adj$ $P = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&{ - 3}\\{ - 2}&1\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&{ - 2}\\{ - 3}&1\end{array}} \right]$
No comments yet — start the discussion.