Question

Compute $\left(AB\right)^{-1}$, if $A = \left[\begin{matrix}1&1&2\\ 0&2&-3\\ 3&-2&4\end{matrix}\right]$ and $B^{-1}=\left[\begin{matrix}1&2&0\\ 0&3&-1\\ 1&0&2\end{matrix}\right]$

• A. $\frac{1}{19}\left[\begin{matrix}16&12&1\\ 21&11&-7\\ 10&-2&3\end{matrix}\right]$
• B. $\frac{1}{19}\left[\begin{matrix}16&12&10\\ 21&11&-2\\ 1&-7&3\end{matrix}\right]$
• C. $\frac{1}{19}\left[\begin{matrix}16&12&1\\ -21&-11&7\\ 10&-2&3\end{matrix}\right]$
• D. $\frac{1}{19}\left[\begin{matrix}16&-21&1\\ 21&11&7\\ 10&-2&3\end{matrix}\right]$

Answer 1

Answer: (A)

$\left|A\right|=\left|\begin{matrix}1&1&2\\ 0&2&-3\\ 3&-2&4\end{matrix}\right|=1\left(8-6\right)-0+3\left(-3-4\right)=19\ne0$(Expanding along $C_{1})$
$\therefore \quad A^{-1}$ exists.
Now, adj(A) $=\left[\begin{matrix}2&-9&-6\\ -8&-2&5\\ -7&3&2\end{matrix}\right]^{T}=\left[\begin{matrix}2&-8&-7\\ -9&-2&3\\ -6&5&2\end{matrix}\right]$

$\therefore \quad A^{-1}=\frac{1}{\left|A\right|}=adj\,\left(A\right)=\frac{-1}{19} \left[\begin{matrix}2&-8&-7\\ -9&-2&3\\ -6&5&2\end{matrix}\right]$

So, $\left(AB\right)^{-1}=B^{-1}A^{-1}=\left[\begin{matrix}1&2&0\\ 0&3&-1\\ 1&0&2\end{matrix}\right]\cdot\frac{-1}{19}\left[\begin{matrix}2&-8&-7\\ -9&-2&3\\ -6&5&2\end{matrix}\right]$

$=\frac{-1}{19} \left[\begin{matrix}2-18-0&-8-4+0&-7+6+0\\ 0-27+6&0-6-5&0+9-2\\ 2-0-12&-8-0+10&-7+10+4\end{matrix}\right]$

$=\frac{-1}{19}\left[\begin{matrix}-16&-12&-1\\ -21&-11&7\\ -10&2&-3\end{matrix}\right]=\frac{1}{19} \left[\begin{matrix}16&12&1\\ 21&11&-7\\ 10&-2&3\end{matrix}\right]$