Question

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

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

Answer 1

Answer: (A)

$\left|A\right|=\left|\begin{array}{ccc}1& 1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right|=1\left(8-6\right)-0+3\left(-3-4\right)=19\ne 0$(Expanding along $C_1)$
$\therefore A^{-1}$ exists.
Now, adj(A) $={\left[\begin{array}{ccc}2& -9& -6\\ -8& -2& 5\\ -7& 3& 2\end{array}\right]}^T=\left[\begin{array}{ccc}2& -8& -7\\ -9& -2& 3\\ -6& 5& 2\end{array}\right]$

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

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

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

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