00²(1+ v) ] [¯xy]Invert the above to get the below form:w w&128x3bMbE0₂1-UEy 1-v²Lo-D101 v0v 1 0002Mb bсон&2855xyGive a step by step procedure.

Answered: &$ धनु 28x9 0 bbb 1 Ox E 1 -U -U 1 0 oy…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Related questions

Question

0
0
²(1+ v) ] [¯xy]
Invert the above to get the below form:
w w
&
128x3
b
M
b
E
0₂
1
-U
E
y 1-v²
Lo
-D
1
0
1 v
0
v 1 0
00
2
M
b b
сон
&
2855
xy
Give a step by step procedure.

Expert Solution

Step 1: We write the given system and give Gauss - Jordan method to find out the inverse of a matrix.

(.) Given,

$open square brackets table row cell epsilon subscript x end cell row cell epsilon subscript y end cell row cell 2 epsilon subscript x y end subscript end cell end table close square brackets space equals space 1 over E open square brackets table row 1 cell negative nu end cell 0 row cell negative nu end cell 1 0 row 0 0 cell 2 open parentheses 1 plus nu close parentheses end cell end table close square brackets open square brackets table row cell sigma subscript x end cell row cell sigma subscript y end cell row cell sigma subscript x y end subscript end cell end table close square brackets$

(.) Gauss - Jordan method: In gauss - Jordan method we write the $n cross times n$ square matrix $A$ as $A equals open square brackets A space space colon space space I close square brackets$ . Then reduce the augmented matrix $open square brackets A space space colon space space space I close square brackets$ into the form $open square brackets I space colon space space A to the power of negative 1 end exponent close square brackets$ by using row operations.

Here $I$ is an $apostrophe n cross times n apostrophe$ identity matrix.