Question #192262

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 , (BT )-1 and B-1A-1. What do you observe about

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

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

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



1
Expert's answer
2021-05-13T13:13:31-0400

A= [111021213]\begin{bmatrix} 1 & -1 & 1\\ 0 & 2 & -1 \\ 2 & 1 & 3\end{bmatrix} , B= [835012476]\begin{bmatrix} 8 & -3 & -5\\ 0 & 1 & 2 \\ 4 & -7 & 6\end{bmatrix}


Now

A1=A^{-1}= [754515251515453525]=[1.40.80.20.40.20.20.80.60.4]\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]

BT=B^T= [147258369]\left[\begin{array}{ccc}1 & 4 & 7\\2 & 5 & 8\\3 & 6 & 9\end{array}\right] and (BT)1=(B^T)^{-1}= then the determinant of the matrix equals 0 ,Thus, the matrix is not invertible.



B1=B^{-1}=

[53953156115623917394391391139239][0.120.330.0060.050.430.100.020.280.05]\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]


B1×A1=B^{-1}\times A^{-1}= [754515251515453525][325331003500120043100110150725120]=[22125341231250511000110019250010710004110053625]=[0.1760.750.09840.0510.010.00760.1070.410.0848]\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 (A1)1=A(A^{-1})^{-1}= A


(7.2)

we got

((BT)1)T=B1((B^T)^{-1})^T=B^{-1}


(7.8)

we got

(AB) -1 = B-1 A-1



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