(i) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 4}\\0&3\end{array}} \right|$
(ii) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&c\\b&d\end{array}} \right|$
(i) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 4}\\0&3\end{array}} \right|$
(ii) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&c\\b&d\end{array}} \right|$
Official Solution
(i) Let $P = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 4}\\0&3\end{array}} \right|$
Minor of the element ${a_{ij}}$ is ${M_{ij}},$.
Here,
${M_{11}} = 3,{M_{12}} = 0,{M_{21}} = - 4,{M_{22}} = 2$
For cofactors, we know that ${P_{ij}} = {( - 1)^{i + j}}{M_{ij}}$
$\therefore$ ${P_{11}} = 3,{P_{12}} = - 0 = 0,{P_{21}} = 4,{P_{22}} = 3$
(ii) Let $P = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&c\\b&d\end{array}} \right|$ ,
Minor of the element ${a_{ij}}$ is ${M_{ij}},$
Here ${M_{11}} = d,{M_{12}} = b,{M_{21}} = c,{M_{22}} = a$
For cofactors,
we know that ${P_{ij}} = {( - 1)^{i + j}}{M_{ij}}$
$\therefore {P_{11}} = d,{P_{12}} = - b,{P_{21}} = - c,{P_{22}} = a$
No comments yet — start the discussion.