Question

Find inverse, by elementary row operations (if possible), of the following matrix:

[1−3−26]

Solution

Let A=[1−3−26]

In order to use elementary row operations, we write A = IA

⇒[1−3−26]=[1001]A

⇒[1−300]=[1021]A [∵R2→R2+2R1]

Since, we obtain all zeroes in a row of the matrix A on LHS, A−1 does not exist.

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