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Find inverse, by elementary row operations (if possible), of the matrices.

\(\begin{bmatrix} 1 & -3 \\[0.3em] -2 & 6 \end{bmatrix}\)

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To apply elementary row transformations we write:

B = IB 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 = XB

And this X is called inverse of B = B-1

So we have:

As we got all zeroes in one of the row of matrix in LHS.

So by any means we can make identity matrix in LHS.

∴ inverse of B does not exist.

B-1 does not exist. …ans

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