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Determinants — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDeterminants3 Marks
Question
Let $A=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$ and $B=\left[\begin{array}{ll}6 & 8 \\ 7 & 9\end{array}\right]$ verify that $(A B)^{-1}=B^{-1} A^{-1}$
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Answer

Given: Matrix $A = \left[ {\begin{array}{*{20}{c}} 3&7 \\ 2&5 \end{array}} \right]$
$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 3&7 \\ 2&5 \end{array}} \right| = 15 - 14 = 1 \ne 0$
$\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A $ $= \frac{1}{1}\left[ {\begin{array}{*{20}{c}} 5&{ - 7} \\ { - 2}&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5&{ - 7} \\ { - 2}&3 \end{array}} \right]$
Matrix $B = \left[ {\begin{array}{*{20}{c}} 6&8 \\ 7&9 \end{array}} \right]$
$\therefore \left| B \right| = \left| {\begin{array}{*{20}{c}} 6&8 \\ 7&9 \end{array}} \right| = 54 - 56 = - 2 \ne 0$
$\therefore {B^{ - 1}} = \frac{1}{{\left| B \right|}}adj.B = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} 9&{ - 8} \\ { - 7}&6 \end{array}} \right]$
Now $AB = \left[ {\begin{array}{*{20}{c}} 3&7 \\ 2&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 6&8 \\ 7&9 \end{array}} \right] $ $= \left[ {\begin{array}{*{20}{c}} {18 + 49}&{24 + 63} \\ {12 + 35}&{16 + 45} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {67}&{87} \\ {47}&{61} \end{array}} \right]$
$\therefore \left| {AB} \right| = \left| {\begin{array}{*{20}{c}} {67}&{87} \\ {47}&{61} \end{array}} \right|  = 67(61) - 87(47) = 4087 - 4089 = - 2 \ne 0$
Now $\text{L.H.S.} = {\left( {AB} \right)^{ - 1}} = \frac{1}{{\left| {AB} \right|}}adj.\left( {AB} \right) $ $= \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} {61}&{ - 87} \\ { - 47}&{67} \end{array}} \right]….(i)$
$\text{R.H.S.} = B^{-1}A^{-1}​​​​​​​ = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} 9&{ - 8} \\ { - 7}&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5&{ - 7} \\ { - 2}&3 \end{array}} \right]$
$= \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} {45 + 16}&{ - 63 - 24} \\ { - 35 - 12}&{49 + 18} \end{array}} \right]$
$= \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} {61}&{ - 87} \\ { - 47}&{67} \end{array}} \right]….(ii)$
$\therefore$ From eq. $(i)$ and $(ii),$ we get
$\text{L.H.S. = R.H.S.}$
$ \Rightarrow {\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$

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