Prove that det(A)=0 when ak , the k-th row vector of A, is equal to λ ai + µ aj , where ai , aj ∈ R^n

Question

Hi can anybody help me with this problem?

Let A be an n × n matrix, let i, j, k be pairwise distinct indices, 1 ≤ i, j, k ≤ n, and let λ, µ ∈ R be arbitrary real numbers. Suppose that ak , the k-th row vector of A, is equal to λ ai + µ aj , where ai , aj ∈ R^n denote the i-th and the j-th row vectors of A respectively. Prove that det(A) = 0.

Answer 1

You may know that:

• the determinant doesn't change when you add a multiple of a row (or column) to another row (or column);
• the determinant of a matrix with a row (or column) full of zeros, is $0$ (you can expand along this row/column).

Hint: with the information given in your question: subtract $\lambda$ times the $i$th row and $\mu$ times the $j$th row from the $k$th row.

Answer 2

Since the $k^{th}$ row of $A$ is a linear combination of the $i^{th}$ row and $j^{th}$ row so $\text{Rank A}<n$ and hence $\det A=0$