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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants3 Marks
Question
Let $A = \left[ {\begin{array}{*{20}{c}} 3&7 \\ 2&5 \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 6&8 \\ 7&9 \end{array}} \right]$verify that $(AB)^{-1} = B^{-1}A^{-1}$
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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 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)
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
L.H.S. = R.H.S.
$ \Rightarrow {\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$

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