Answer :
Let A =
To apply elementary row transformations we write:
A = IA where I is the identity matrix
We proceed with operations in such a way that LHS becomes I and the transformations in I give us a new matrix such that
I = XA
And this X is called inverse of A = A-1
Note: Never apply row and column transformations simultaneously over a matrix.
So we have:
Applying R2→ R2 – (5/2)R1
⇒
Applying R3→ R3 - R2
⇒
Applying R1→ R1 + R2
⇒ =
Applying R2→ R2 - 5R3
=
Applying R1→ R1 + 2R3
⇒ =
Applying R1→ (1/2)R1 and R3→ 2R3
⇒ =
As we got Identity matrix in LHS.
∴ A-1 =
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