Question

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Let A be an r × r matrix and suppose there are r-1 rows (columns) such that all row's (columns) are linear combinations of these r- 1 rows (columns). Show det (A) 0

Please explain, and show the details of steps!

Answer 1

We know that the determinant of a matrix is unchanged when we add a multiple of one row of a matrix to another row. Let A be a rxr matrix such that a row of A is a linear combination of r-1 rows i.e. let R_i = a₁ R₁+ a₂ R₂ +…+ a_i-1R_i-1+a_i+1R_i+1 +…+a_r R_r, where R_i denotes the ith row of A. Now, if we add –a₁times 1^st row,…- a_i-1 times row R_i-1, -a_i+1 times row Ri₊₁,…,-a_r times row R_r to the ith row, we get a row of zeros. Further, if we carry out a cofactor expandion along this zero row, we get det(A) = 0.