MADEEASY TESTSERIES

Question

To justify the OPTION B they gave an example of 2*2 matrix. However we can see that row 2 is linearly dependent on row1. Even though the 2nd row looks non-zero it can be made into zero. SO am I wrong or the explanation is wrong?

Comments

this is asking about original matrix .so explanation is right.

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In row echelon form row will be zero, may or may not be in original matrix

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Can you explain, I’m not able to understand.

My logic was, since if one row of matrix is zero the rank of matrix is reduced by 1 which is what the option is stating.

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@DAWID15 It means:-

 Suppose the original matrix may or may not have any row with all zeros.

Now you perform some row operations, to convert the original matrix to echelon form.

Now you found that one row became all zeros. Thus, you can conclude rank = n-1.

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The second row of the matrix is indeed linearly dependent on the first row. Linear dependence means that one row (or column) of a matrix can be expressed as a linear combination of the other rows (or columns). In other words, if one row of a matrix can be expressed as a multiple of another row, then the rows are linearly dependent.

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@shishir__roy bro for option c answer would be min(n-k,k). right?

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@*nikhil* can you prove your claim?

Moreover, $rank(I_4 + 0) ≠ min(4,0)$

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okay i got it.so we can't comment anything on rank of A+B just from given information.and suppose instead of A+B we have AB then only we can write min(n-k, k)?

Answer 1

If determinant if A is equals to zero then only we can say rank less than order(i.e, rows are linearly dependent) but there is no relation between rows and rank.

Yes if after the row Matrix operations you get 1 row entirely becoming zero then we guarantee rank equals n-1