Question

find the inverse of the matrix using elementary operations?

Answer 1

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HOPE THIS HELPS YOU ...

Answer:

This can be solved by either row operation or by coloumn peration ,but now I am going to use row operation which showa that . A=IA

Step-by-step explanation:

We have to find the inverse of the matrix ,

$\left[\begin{array}{ccc}1&5&2\\1&1&7\\0&-3&4\end{array}\right]$

Applying the row operation,

A = IA

=> $\left[\begin{array}{ccc}1&5&2\\1&1&7\\0&-3&4\end{array}\right]$ = $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ .A

R2 ---> R2 - R1

=> $\left[\begin{array}{ccc}1&5&2\\0&-4&5\\0&-3&4\end{array}\right]$ = $\left[\begin{array}{ccc}1&0&0\\-1&1&0\\0&0&1\end{array}\right]$ .A

R1 ---> R1 - R2

=> $\left[\begin{array}{ccc}1&1&7\\0&-4&5\\0&-3&4\end{array}\right]$ = $\left[\begin{array}{ccc}0&1&0\\-1&1&0\\0&0&1\end{array}\right]$ .A

R2 ---> R2 - R3

=> $\left[\begin{array}{ccc}1&1&7\\0&-1&1\\0&-3&4\end{array}\right]$ = $\left[\begin{array}{ccc}0&1&0\\-1&1&-1\\0&0&1\end{array}\right]$ .A

R1 ---> R1 + R2

=> $\left[\begin{array}{ccc}1&0&8\\0&-1&1\\0&-3&4\end{array}\right]$ = $\left[\begin{array}{ccc}-1&2&-1\\-1&1&-1\\0&0&1\end{array}\right]$ .A

R1 ---> R1 - 2R3

=> $\left[\begin{array}{ccc}1&6&0\\0&-1&1\\0&-3&4\end{array}\right]$ = $\left[\begin{array}{ccc}1&0&1\\-1&1&-1\\0&0&1\end{array}\right]$ .A

R3 ---> R3 - R1/2

=> $\left[\begin{array}{ccc}1&6&0\\0&-1&1\\0&0&4\end{array}\right]$ = $\left[\begin{array}{ccc}1&0&1\\-1&1&-1\\1/2&0&1/2\end{array}\right]$ .A

R2 ---> R2 - R3/4

=> $\left[\begin{array}{ccc}1&6&0\\0&-1&1\\0&0&4\end{array}\right]$ = $\left[\begin{array}{ccc}1&0&1\\-1&1&3/4\\1/2&0&1/2\end{array}\right]$ .A

R1 ---> R1 + R2/6

=> $\left[\begin{array}{ccc}1&0&0\\0&-1&1\\0&0&4\end{array}\right]$ = $\left[\begin{array}{ccc}5/6&1/6&9/8\\-1&1&3/4\\1/2&0&1/2\end{array}\right]$ .A

R2 ---> R2/(-1) and R3 ---> R3/4

=> $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ = $\left[\begin{array}{ccc}5/6&1/6&9/8\\1&-1&-3/4\\1/8&0&1/8\end{array}\right]$ .A

Hence, inverse of the given matrix A i.e, A inverse = $\left[\begin{array}{ccc}5/6&1/6&9/8\\1&-1&-3/4\\1/8&0&1/8\end{array}\right]$

THANK YOU FOR THE WONDERFUL QUESTION ...

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