A = $\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\ 3&2&{ - 4} \\ 1&1&{ - 2} \end{array}} \right]$$\therefore$ $|A| = 2(-4 + 4) + 3(-6 +4)+ 5(3 - 2) = 0 -6 +5 = -1$
Now, $A_{11} = 0, A_{12}= 2, A_{13}= 1$
$A_{21}= -1, A_{22} = -9, A_{23}= -5$
$A_{31}= 2, A_{32}= 23,A_{33} = 13$
$\therefore {A^{ - 1}} = \frac{1}{{|A|}}(adjA) = - \left[ {\begin{array}{*{20}{c}} 0&{ - 1}&2 \\ 2&{ - 9}&{23} \\ 1&{ - 5}&{13} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&1&{ - 2} \\ { - 2}&9&{ - 23} \\ { - 1}&5&{ - 13} \end{array}} \right]....(1)$
Now, the given system of equations can be written in the form of $AX = B$, where
$A = \left[ {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\ 3&2&{ - 4} \\ 1&1&{ - 2} \end{array}} \right]\;,X = \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]\;and\;B = \left[ {\begin{array}{*{20}{c}} {11} \\ { - 5} \\ { - 3} \end{array}} \right]$
The solution of the system of equations is given by $X = A^{-1}B$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&1&{ - 2} \\ { - 2}&9&{ - 23} \\ { - 1}&5&{ - 13} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {11} \\ { - 5} \\ { - 3} \end{array}} \right]$ [Using (1) ]
$ = \left[ {\begin{array}{*{20}{c}} {0 - 5 + 6} \\ { - 22 - 45 + 69} \\ { - 11 - 25 + 39} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \end{array}} \right]$
Hence,$ x = 1, y = 2$ and $z = 3.$

Determinants — Maths STD 12 Science — Question

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