Determinants — Maths STD 12 Science — Question

$\Rightarrow \left[ {\begin{array}{*{20}{c}} 2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9 \end{array}} \right]$

Here $A = \left[ {\begin{array}{*{20}{c}} 2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]$and $B = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9 \end{array}} \right]$  

Here,$A_{11}=13,A_{12}=5,A_{13}=3\\ A_{21}=8,A_{22}=-10,A_{23}=-6\\ A_{31}=1,A_{32}=3,A_{33}=-5$

adj A=$\left[ {\begin{array}{*{20}{c}} 13&8&1 \\ 5&{ - 10}&{ 3} \\ 3&-6&{ - 5} \end{array}} \right]$

$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right| $ = 2(10 + 3) - 1(3 - 0)  = 26 + 5 + 3 $ = 34 \ne 0$

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

$\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}} {13}&8&1 \\ 5&{ - 10}&3 \\ 3&{ - 6}&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9 \end{array}} \right]$

$= \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}} {13 + 12 + 9} \\ {5 - 15 + 27} \\ {3 - 9 - 45} \end{array}} \right]$

$ = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}} {34} \\ {17} \\ { - 51} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{1}{2}} \\ {\frac{{ - 3}}{2}} \end{array}} \right]$

Therefore, $x = 1,y = \frac{1}{2}$ and $z = \frac{3}{2}$