If A= \begin{bmatrix} 5 &-3 \\ -3 &2 \end{bmatrix}  and B^{-1}= \begin{bmatrix} 3 &2 \\ 0& -1 \end{bmatrix}, find A-1 and hence find (AB)-1.

 

 

 

 
 
 
 
 

Answers (1)

A= \begin{bmatrix} 5 &-3 \\ -3 & 2 \end{bmatrix}\: \: \: \: B^{-1}= \begin{bmatrix} 3 &2 \\ 0& -1 \end{bmatrix}
A^{-1}\Rightarrow \frac{1}{\left | A \right |}\left ( adj\, A \right )
\left ( adj\, A \right )= \begin{bmatrix} 2 &3 \\ 3 & 5 \end{bmatrix}   & \left | A \right |= 10-9\Rightarrow 1\neq 0\therefore A^{-1}\: exist\cdot
adj\, \left ( A \right )=\begin{bmatrix} 2 & 3\\ 3 &5 \end{bmatrix}\: \: \: \: \left | A \right |= 1
\therefore A^{-1}= \frac{1}{\left | A \right |}\left ( adj\, A \right )\Rightarrow \frac{1}{1}\begin{bmatrix} 2 & 3\\ 3& 5 \end{bmatrix}
                                  A^{-1}\Rightarrow \begin{bmatrix} 2 & 3\\ 3& 5 \end{bmatrix}
\left ( AB \right )^{-1}= B^{-1}A^{-1}             B^{-1}= \begin{bmatrix} 3& 2\\ 0& -1 \end{bmatrix}
\Rightarrow \begin{bmatrix} 3 & 2\\ 0& -1 \end{bmatrix}\begin{bmatrix} 2 & 3\\ 3& 5 \end{bmatrix}
\Rightarrow \begin{bmatrix} 6+6 & 9+10\\ -3 &0- 5 \end{bmatrix} = \begin{bmatrix} 12 & 19\\ -3& -5 \end{bmatrix}

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