0
$\begingroup$

How to prove that by adding to a row a multiple of another row, the determinant remains the same ?

  • 0
    You mean *adding* a row... – 2017-02-19
  • 0
    @DavidMitra ... of a _matrix_ – 2017-02-19
  • 0
    I mean adding a row of a determinant of a matrix with a multiple of a second row or a column with a multiple of a second column – 2017-02-19
  • 0
    @Fifi A determinant is a single number. It doesn't have rows. What you mean is "Adding a multiple of one row of a matrix to another row of that matrix". – 2017-02-19
  • 0
    ok you understood what I mean – 2017-02-19
  • 0
    @Arthur It is written in my book "we call row(resp.column)of det(A) every row (resp. Column) of A." – 2017-02-19
  • 0
    Well, ok, then. If it's a defined term, you may use it. I just haven't seen it used that way before. – 2017-02-19

1 Answers 1

2

We have that $$\Delta = \begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix}\tag{1}$$ $$\Rightarrow \text{ (Performing }R_1 \to R_1 + kR_2) \, \, \Delta_1 = \begin{vmatrix} a_1 + kb_1 & a_2+kb_2 & a_3 + kb_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix} \text{ (say) }$$ $$\Rightarrow \Delta_1 = \begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix} + \begin{vmatrix} kb_1 & kb_2 & kb_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix}$$ $$\Rightarrow \Delta_1 = \begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix} + k\begin{vmatrix} b_1 & b_2 & b_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix}$$ $$\Rightarrow \Delta_1 = \begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{vmatrix} + 0$$ $$\Rightarrow \Delta_1 = \Delta$$

Similarly, we can show the same for any column operation. Thus, for any such column and row operation, the determinant remains the same. Hope it helps.

  • 0
    Are you assuming the determinant is linear here? – 2017-02-19