If B, C, D, are square n x n invertible matrices and BC = BD, then C = D.
To prove that C = D, we need to show that all corresponding entries in C and D are equal
To prove that C = D, we need to show that all corresponding entries in C and D are equal.
Let’s assume C and D are not equal, so there exists at least one entry (i, j) in C which is not equal to the corresponding entry in D.
Let’s denote the entries of C and D as cij and dij respectively. Since BC = BD, we can write the entry (i, j) of BC as Σcik * dkj, where the summation is taken over the index k from 1 to n. Similarly, the entry (i, j) of BD can be written as Σdik * dkj.
Since BC = BD, we have Σcik * dkj = Σdik * dkj.
Now let’s consider two cases:
Case 1: cij ≠ dij. In this case, we can say that cik ≠ dik for at least one value of k.
The equation Σcik * dkj = Σdik * dkj implies that Σcik * dkj = Σdik * dkj – (dik – cik) * dkj.
If cik ≠ dik, then (dik – cik) ≠ 0. Therefore, the expression on the right-hand side is non-zero.
However, this contradicts the assumption that BC = BD, because the entry (i, j) of BC and BD should be equal. Thus, cij cannot be unequal to dij, and we can conclude that C = D.
Case 2: cij = dij. In this case, we have shown that the entry (i, j) of both C and D are equal, and we can conclude that C = D.
In both cases, we have shown that if BC = BD, then C = D. Therefore, the statement is proven.
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