Find the inverse using elementary row transformations: `[1 6-3 5]` (ii) `[2 5 1 3]`

Explanation

(i)`[(1 ,6),(-3, 5)]`


we know that,

`A``A^(-1)=I`

so,

`[(1 ,6),(-3, 5)]``A^(-1)`=`[(1 ,0),(0, 1)]`

Applying `R1->R1-6/(23)R2 ` and `R2->R2+3R1`

we get,

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