If A = [ * 20 c 2&3 1& - 4 ] and B = [ * 20 c 1& - 2 - 1 &3 ] then verify that (AB)^ -1 = B^ -1 A^ -1

AB = [ * 20 c 2&3 1& - 4 ] [ * 20 c 1& - 2 - 1 &3 ]= -1&5 5&-14 |AB| = 14 -25 = -11 (AB) is non singular and hence, (AB)^ -1 exists (AB)^ -1 = 1 |AB| adj(AB) = 1 11 [ * 20 c - 14 & - 5 - 5 & - 1 ] Now, |A| = - 8 - 3 = - 11 A is non singular and hence A^ -1 exists Also, |B| = 3 - 2 = 1 B is non- singular and hence B^ -1 exists A^ -1 = - 1 11 [ * 20 c - 4 & - 3 - 1 &2 ] B^ -1 = 1 1 [ * 20 c 3&2 1&1 ] B^ -1 A^ -1 = - 1 11 [ * 20 c 3&2 1&1 ] [ * 20 c - 4 & - 3 - 1 &2 ] = - 1 11 [ * 20 c - 14 & - 5 - 5 & - 1 ] Hence, (AB)^ -1 = B^ -1 A^ -1

If A = [ * 20 c 2&3 1& - 4 ] and B = [ * 20 c 1& - 2 - 1 &3 ] the…

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Question

If $A = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 1&{ - 4} \end{array}} \right]$ and $ B = \left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ { - 1}&3 \end{array}} \right]$ then verify that $(AB)^{-1} = B^{-1} A^{-1}$

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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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