Determinants — Maths STD 12 Science — Question

$\Rightarrow \left[ {\begin{array}{*{20}{c}} 2&3&3 \\ 1&{ - 2}&1 \\ 3&{ - 1}&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5 \\ { - 4} \\ 3 \end{array}} \right]$

Here $\left| A \right| = \left[ {\begin{array}{*{20}{c}} 2&3&3 \\ 1&{ - 2}&1 \\ 3&{ - 1}&{ - 2} \end{array}} \right],X\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 5 \\ { - 4} \\ 3 \end{array}} \right]$

$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&3&3 \\ 1&{ - 2}&1 \\ 3&{ - 1}&{ - 2} \end{array}} \right|$

= 2(4 + 1) - 3( - 2 - 3) + 3( - 1 + 6)

$ = 10 + 15 + 15 = 40 \ne 0$

Therefore, solution is unique and $X = {A^{ - 1}}B = \frac{1}{{\left| A \right|}}\left( {adj.A} \right)B$

$\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}} 5&3&9 \\ 5&{ - 13}&1 \\ 5&{11}&{ - 7} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5 \\ { - 4} \\ 3 \end{array}} \right]$

$= \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}} {25 - 12 + 27} \\ {25 + 52 + 3} \\ {25 - 44 - 21} \end{array}} \right]$

$= \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}} {40} \\ {80} \\ { - 40} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right]$

Therefore, x = 1, y = 2 and z = - 1 .