Determinants — Class 12 Maths Solution

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

where
$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 1}&{ - 2}\\0&2&{ - 1}\\3&{ - 5}&0\end{array}} \right],X\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\\z\end{array}} \right]$ and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2\\{ - 1}\\3\end{array}} \right]$

Now, $|A| = 3\left( {0 - 5} \right) + 1\left( {0 + 3} \right) - 2\left( {0 - 6} \right) = - 15 + 3 + 12 = 0$

Here, A is a singular matrix, so we will compute (adj A )B.

For adj A, cofactors of elements of A are given by

${A_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 1}\\{ - 5}&0\end{array}} \right| = - 5,$

${A_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 1}\\3&0\end{array}} \right| = - 3,$

${A_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&2\\3&{ - 5}\end{array}} \right| = - 6,$

${A_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&{ - 2}\\{ - 5}&0\end{array}} \right| = - ( - 10) = 10,$

${A_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 2}\\3&0\end{array}} \right| = 6,1$

${A_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 1}\\3&{ - 5}\end{array}} \right| = - ( - 15 + 3) = 12$,

${A_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&{ - 2}\\2&{ - 1}\end{array}} \right| = 1 + 4 = 5,$

${A_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 2}\\0&{ - 1}\end{array}} \right| = 3,$

${A_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 1}\\0&2\end{array}} \right| = 6$

$\therefore$

Now, $(adj\;A) \cdot B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&{10}&5\\{ - 3}&6&3\\{ - 6}&{12}&6\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2\\{ - 1}\\3\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 10 - 10 + 15}\\{ - 6 - 6 + 9}\\{ - 12 - 12 + 18}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}\\{ - 3}\\{ - 6}\end{array}} \right] \ne 0$

Hence, system of equations is inconsistent with no solution.