Question

let

Matrix

A = 1 -1 1 B = 8 -3 -5

0 2 -1 0 1 2

2 1 3 4 -7 6

Compute A ^-1 , (B^T )^-1 and B^-1A^-1. What do you observe about

(7.1) (A ^-1 )^-1 in relation to A.

(7.2) ((B^T )^-1 ) T in relation to B^-1 .

(7.3) (AB) ^-1 in relation to B^-1 A^-1 .

Answer 1

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

Now

$A^{-1}=$ $\left[\begin{array}{ccc}\frac{7}{5} & \frac{4}{5} & - \frac{1}{5}\\- \frac{2}{5} & \frac{1}{5} & \frac{1}{5}\\- \frac{4}{5} & - \frac{3}{5} & \frac{2}{5}\end{array}\right] = \left[\begin{array}{ccc}1.4 & 0.8 & -0.2\\-0.4 & 0.2 & 0.2\\-0.8 & -0.6 & 0.4\end{array}\right]$

$B^T=$ $\left[\begin{array}{ccc}1 & 4 & 7\\2 & 5 & 8\\3 & 6 & 9\end{array}\right]$ and $(B^T)^{-1}=$ then the determinant of the matrix equals 0 ,Thus, the matrix is not invertible.

$B^{-1}=$

$\left[\begin{array}{ccc}\frac{5}{39} & \frac{53}{156} & - \frac{1}{156}\\\frac{2}{39} & \frac{17}{39} & - \frac{4}{39}\\- \frac{1}{39} & \frac{11}{39} & \frac{2}{39}\end{array}\right] \\ \\\approx \left[\begin{array}{ccc}0.12& 0.33& -0.006\\0.05 & 0.43 & -0.10\\-0.02 & 0.28 & 0.05\end{array}\right]$

$B^{-1}\times A^{-1}=$ $\left[\begin{array}{ccc}\frac{7}{5} & \frac{4}{5} & - \frac{1}{5}\\- \frac{2}{5} & \frac{1}{5} & \frac{1}{5}\\- \frac{4}{5} & - \frac{3}{5} & \frac{2}{5}\end{array}\right]\cdot \left[\begin{array}{ccc}\frac{3}{25} & \frac{33}{100} & - \frac{3}{500}\\\frac{1}{200} & \frac{43}{100} & - \frac{1}{10}\\- \frac{1}{50} & \frac{7}{25} & \frac{1}{20}\end{array}\right] = \left[\begin{array}{ccc}\frac{22}{125} & \frac{3}{4} & - \frac{123}{1250}\\- \frac{51}{1000} & \frac{1}{100} & - \frac{19}{2500}\\- \frac{107}{1000} & - \frac{41}{100} & \frac{53}{625}\end{array}\right] = \left[\begin{array}{ccc}0.176 & 0.75 & -0.0984\\-0.051 & 0.01 & -0.0076\\-0.107 & -0.41 & 0.0848\end{array}\right]$

(7.1)

we got $(A^{-1})^{-1}= A$

(7.2)

we got

$((B^T)^{-1})^T=B^{-1}$

(7.8)

we got

(AB) ^-1 = B^-1 A^-1