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    i cant do second part

    i tried adding a new row and collumn to C with 0 entries for ith entry where i>n+1 and tried to found the determinant of that

    any solutions?

    HINT: Consider the Laplace expansions, and notice that since all other elements of the matrix are zero apart from the parts of A and B there is a certain invariance (Subbing detB out).


    Suppose det_{a_j}(a'_1) is the determinant of the n \times n matrix A, where the a_jth column of the original matrix A is replaced by the new column a'_1 of A', except for a_{11}. So the original matrix A has a principal diagonal of a_{22},a_{33},...,a_{nn}.

    Then consider the Laplace expansion of your new matrix with an (n + 1) \times (n + 1) matrix A' up top.

    Then det(A') = a_{11}M_{11} + a_{12}M_{12} + ... + a_{1n}M_{1n}. Where M_{1k} are the cofactors of a_{1k}.

    Now by the inductive hypothesis M_{1k} = det_{a_k}(a'_1)detB, as M_{1k} is always an (n + m) \times (n + m) matrix.

    Now take C' as your new matrix, then you have that det(C') = a_{11}detA detB + a_{12}det_{a_2}(a'_1)detB + ... + a_{1n}det_{a_n}(a'_1)detB = (a_{11}detA + a_{12}det_{a_2}(a'_1) + ... + a_{1n}det_{a_n}(a'_1))detB, by the Laplace expansion.

    Now considering the Laplace expansion of detA', it is clear that detA' = a_{11}detA + a_{12}det_{a_2}(a'_1) + ... + a_{1n}det_{a_n}(a'_1)), thus det(C') = det(A')det(B).

    In fact there's a really easy way to prove the fact that the determinant of C is indeed det(A)det(B) if you know the idea behind Leibniz's formula for determinants, but I guess it doesn't really work for an induction problem like this one.
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