Determinants — Class 12 Maths Solution

The given system of equations can be written in the form $AX = B$ ,

where
$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&2\\7&3\end{array}} \right],X = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\end{array}} \right]$

and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4\\5\end{array}} \right]$

Now, $|A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&2\\7&3\end{array}} \right| = 15 - 14 = 1 \ne 0$

$\Rightarrow$ A is non-singular and so given system has a unique solution.

Cofactors of elements of A are given by

${A_{11}} = {( - 1)^{1 + 1}}(3) = 3,{A_{12}} = {( - 1)^{1 + 2}}(7) = - 7,$

${A_{21}} = {( - 1)^{2 + 1}}(2) = - 2,{A_{22}} = {( - 1)^{2 + 2}}(5) = 5$

% $\therefore$ $adj\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 7}\\{ - 2}&5\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 2}\\{ - 7}&5\end{array}} \right]$

and ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\;A) = \cfrac{1}{2}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 2}\\{ - 7}&5\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 2}\\{ - 7}&5\end{array}} \right]$

Solution of given system is given by $X = {A^{ - 1}}B.$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 2}\\{ - 7}&5\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4\\5\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{12}&{ - 10}\\{ - 28}&{ + 25}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2\\{ - 3}\end{array}} \right]$

Hence; $x = 2,y = - 3$