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Determinants — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDeterminants2 Marks
Question
If $A=\left[\begin{array}{cc}2 & 3 \\ 1 & -4\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right]$ then verify that $(A B)^{-1}=B^{-1} A^{-1}$
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Answer

$AB = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 1&{ - 4} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ { - 1}&3 \end{array}} \right]= \begin{bmatrix}-1&5\\5&-14\end{bmatrix}$
$|AB| = 14 -25 = -11 $
$\Rightarrow (AB)$ is non singular and hence, $(AB)^{-1}$ exists
$\therefore (AB)^{-1}= \frac{1}{{|AB|}}adj(AB) = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} { - 14}&{ - 5} \\ { - 5}&{ - 1} \end{array}} \right]$
Now, $|A| = - 8 - 3 = - 11$
$ \Rightarrow A$ is non singular and hence $A^{-1}$ exists
Also, $|B| = 3 - 2 = 1$
$\Rightarrow B$ is non- singular and hence $B^{-1}$ exists
$A^{-1} = \frac{{ - 1}}{{11}}\left[ {\begin{array}{*{20}{c}} { - 4}&{ - 3} \\ { - 1}&2 \end{array}} \right]$
$B^{-1}= \frac{1}{1}\left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right]$
$B^{-1}A^{-1} = \frac{{ - 1}}{{11}}\left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 4}&{ - 3} \\ { - 1}&2 \end{array}} \right]$
$= \frac{{ - 1}}{{11}}\left[ {\begin{array}{*{20}{c}} { - 14}&{ - 5} \\ { - 5}&{ - 1} \end{array}} \right]$
Hence, $(AB)^{-1 }= B^{-1}A^{-1}$​​​​​​​

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