Answer to Question #198230 in Linear Algebra for FELICIA NDANGUMUNI

Question

Let A= 1 0 -2 - 1 2 -1 1 3 , and B = 8 3 0 1 4 -7 -5 2 6 Compute A--,(BT)- and B-1A-1. What do you observe about (7.1) (A-1)- in relation to A. (2) (7.2) ((BT)-1)" in relation to B-1. (2) (7.3) (AB)-' in relation to B-1A-1. (3)

Answer 1

$A=\begin{bmatrix} 1&-1&1\\0&2&-1\\-2&1&3\end{bmatrix} , B=\begin{bmatrix} 8&-3&-5\\0&+1&2\\4&-7&6\end{bmatrix}$

$|A|=\begin{vmatrix} 1&-1&1\\0&2&-1\\-2&1&3\end{vmatrix}$

  

  $=1(6+1)+1(0-2)+1(0+4) \\ =7-2+4=9$

$adj.(A)=\begin{bmatrix} 7&2&4\\4&5&1\\-1&1&2\end{bmatrix}^T =\begin{bmatrix} 7 &4&-1\\2&5&1\\4&1&2\end{bmatrix}$

    

$A^{-1}=\dfrac{adj.A}{|A|}=\dfrac{1}{9}\begin{bmatrix} 7 &4&-1\\2&5&1\\4&1&2\end{bmatrix}$

$=\begin{bmatrix} \dfrac{7}{9} &\dfrac{4}{9}&\dfrac{-1}{9}\\\\\dfrac{2}{9}&\dfrac{5}{9}&\dfrac{1}{9}\\\\\dfrac{4}{9}&\dfrac{1}{9}&\dfrac{2}{9}\end{bmatrix}$

     

Similiarly, $B=\begin{bmatrix} 8&-3&-5\\0&+1&2\\4&-7&6\end{bmatrix}$

   $\Rightarrow B^T=\begin{bmatrix} 8&0&4\\-3&1&-7\\-5&2&6\end{bmatrix}$

$|B^T|=8(6+14)-0+4(-6+5)=160-4=156$

$adjB^T=\begin{bmatrix} 20&53&-1\\8&68&-16\\-4&44&8\end{bmatrix}^T=\begin{bmatrix}20&8&-4\\53&68&44\\-1&-16&8\end{bmatrix}$

$(B^T)^{-1}=\dfrac{adj.B^T}{|B^T|}=\dfrac{1}{156}\begin{bmatrix}20&8&-4\\53&68&44\\-1&-16&8\end{bmatrix}$

            $=\begin{bmatrix}\dfrac{20}{156}&\dfrac{8}{156}&\dfrac{-4}{156}\\\\\dfrac{53}{156}&\dfrac{68}{156}&\dfrac{44}{156}\\\\\dfrac{-1}{156}&\dfrac{-16}{156}&\dfrac{8}{156}\end{bmatrix}$

Also, $|B|=|B^T|=156$

$adjB=\begin{bmatrix} 20&8&-4\\53&68&44\\-1&-16&8\end{bmatrix}^T=\begin{bmatrix}20&53&-1\\8&68&-16\\-4&44&8\end{bmatrix}$

So, $B^{-1}=\dfrac{adj. B}{|B|}=\dfrac{1}{156}\begin{bmatrix}20&53&-1\\8&68&-16\\-4&44&8\end{bmatrix}$

        

So, $B^{-1}A^{-1}=\dfrac{1}{156}\begin{bmatrix}20&53&-1\\8&68&-16\\-4&44&8\end{bmatrix}\dfrac{1}{9}\begin{bmatrix} 7 &4&-1\\2&5&1\\4&1&2\end{bmatrix}$

        $=\dfrac{1}{1404}\begin{bmatrix} 242&344&71\\128&356&28\\92&212&64\end{bmatrix}$

Yes All the relations are correct-

$(A^{-1})^{-1}=A \\[9pt] [(B^T)^{-1}]^T=B^{-1} \\[9pt] (AB)^{-1}=B^{-1}A^{-1}$