Question

1.9.given A $\begin{bmatrix} 1 & 1& -1 \\ 2 & 2 & 3 \\ 4 & 0 &0 \\ \end{bmatrix}$ B=$\begin{bmatrix} 1 & 1 & -1 \\ 2 & 2 & 3 \\ 4 & 0 & 0 \\ \end{bmatrix}$

1.9.1. find -A ^-1 +3B^T

1.9.2. Find B^-1+(A^T+A^-1)

Answer 1

Remind that for $3\times3$ matrix $T=\left(\begin{array}{lll}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right)$ the inverse can be computed using the formula: $T^{-1}=\frac{1}{|T|}Adj\,\,T$, where $|T|$ is the determinant of matrix $T$. It has the form: $|T|=a_{11}a_{22}a_{33}+a_{21}a_{13}a_{32}+a_{31}a_{12}a_{23}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}-a_{31}a_{13}a_{22}$ ; $Adj\,\,T=\left(\begin{array}{lll}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{array}\right)$ and $A_{11}=\left|\begin{array}{lll}a_{22}&a_{23}\\a_{32}&a_{33}\end{array}\right|$, $A_{12}=-\left|\begin{array}{lll}a_{12}&a_{13}\\a_{32}&a_{33}\end{array}\right|$, $A_{13}=\left|\begin{array}{lll}a_{12}&a_{13}\\a_{22}&a_{23}\end{array}\right|$, $A_{21}=-\left|\begin{array}{lll}a_{21}&a_{23}\\a_{31}&a_{33}\end{array}\right|$, $A_{22}=\left|\begin{array}{lll}a_{11}&a_{13}\\a_{31}&a_{33}\end{array}\right|$, $A_{23}=-\left|\begin{array}{lll}a_{11}&a_{13}\\a_{21}&a_{23}\end{array}\right|$ , $A_{31}=\left|\begin{array}{lll}a_{21}&a_{22}\\a_{31}&a_{32}\end{array}\right|$, $A_{32}=-\left|\begin{array}{lll}a_{11}&a_{12}\\a_{31}&a_{32}\end{array}\right|$, $A_{33}=\left|\begin{array}{lll}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right|.$

For the given matrices we have:

a). $A^{-1}=\left(\begin{array}{lll}0&0&\frac{1}{4}\\\frac{3}{5}&\frac{1}{5}&-\frac{1}{4}\\-\frac{2}{5}&\frac{1}{5}&0\end{array}\right)$; $B^{\top}=\left(\begin{array}{lll}1&2&4\\1&2&0\\-1&3&0\end{array}\right)$;

$-A^{-1}+3B^{\top}=\left(\begin{array}{lll}3&6&\frac{47}{4}\\\frac{12}{5}&\frac{29}{5}&\frac{1}{4}\\-\frac{13}{5}&\frac{44}{5}&0\end{array}\right)$;

b). $B^{-1}=A^{-1}$ . We receive:

$B^{-1}+(A^{\top}+A^{-1})=2A^{-1}+A^{\top}=\left(\begin{array}{lll}1&2&\frac{9}{2}\\\frac{11}{5}&\frac{12}{5}&-\frac{1}{2}\\-\frac{9}{5}&\frac{17}{5}&0\end{array}\right)$.