Question
Solve the system of linear equation, using matrix method $x - y + 2z = 7; 3x + 4y - 5z = -5; 2x - y + 3z = 12$

Answer

Matrix form of given equations is $AX = B \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&4&{ - 5} \\ 2&{ - 1}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 7 \\ { - 5} \\ {12} \end{array}} \right]$
Here $A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&4&{ - 5} \\ 2&{ - 1}&3 \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 7 \\ { - 5} \\ {12} \end{array}} \right]$
$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&4&{ - 5} \\ 2&{ - 1}&3 \end{array}} \right|$
$ =1(12-5)-(-1)(9+10)+2(-3-8) $
$ =7+19-22=4 \neq 0 $
$ A_{11}=7, A_{12}=-19, A_{13}=-11 $
$ A_{21}=1, A_{22}=-1, A_{23}=-1 $
$ A_{31}=-3, A_{32}=11, A_{33}=7$
adj A=$\left[ {\begin{array}{*{20}{c}} 7&1&{ - 3} \\ { - 19}&{ - 1}&{11} \\ { - 11}&{ - 1}&7 \end{array}} \right]$
Therefore, solution is unique and $X = {A^{ - 1}}B = \frac{1}{{\left| A \right|}}\left( {adj.A} \right)B$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{c}} 7&1&{ - 3} \\ { - 19}&{ - 1}&{11} \\ { - 11}&{ - 1}&7 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 7 \\ { - 5} \\ {12} \end{array}} \right]$
$= \frac{1}{4}\left[ {\begin{array}{*{20}{c}} {49 - 5 - 36} \\ { - 133 + 5 + 132} \\ { - 77 + 5 + 84} \end{array}} \right]$
$= \frac{1}{4}\left[ {\begin{array}{*{20}{c}} 8 \\ 4 \\ {12} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \\ 1 \\ 3 \end{array}} \right]$
Therefore, $x = 2, y = 1$ and $z = 3$

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