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If `{:A=[(a,b),(c,d)]:}` such that `ad - bc ne 0`, then `A^(-1)`, isA. `1/(ad-bc){:[(a,-b),(-c,a)]:}`B. `1/(ad-bc){:[(a,-b),(-c,a)]:}`C. `{:[(d,b),(-c,a)]:}`D. none of these |
Answer» Correct Answer - A We have, `absA=ad-bcne0` Cofactor of `a_11=d," Cofactor of "a_12=-c` Cofactor of `a_21=-b," Cofactor of "a_22 =a` `:. A^(-1) =1/absAadjA =1/(ad-bc){:[(d,-b),(-c,a)]:}` |
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