Replacing $R_3$ with $2R_1 - 3R_3$ is changing the matrix's determinant. Why isn't the determinant the same from summing multiples of rows?

Question

I am trying to solve the determinant

$$ \begin{vmatrix} 3 & 10 & 1 \\ 0 & 3 & -3 \\ 2 & 2 & 5 \\ \end{vmatrix} $$

My plan was to add another 0 below the existing 0 in the matrix, so I can get a nice co-factor expansion down the first column.

So, I replaced the $R_3$ with $2R_1 - 3R_3$:

$$ \begin{vmatrix} 3 & 10 & 1 \\ 0 & 3 & -3 \\ 0 & 14 & -13 \\ \end{vmatrix} $$

However, the determinant has been changed. The determinant above is -3 while the one below is 9.

May I ask if my understanding of the rule of replacing a row with the sum of scalar multiples of other rows flawed?

Answer 1

The determinant is unchanged by adding a multiple of a row to a given row, so for example, $$ R_3 + 2R_1 \to R_3 $$ (or $R_i + t R_j \to R_i$ for any row indices $i, j$ , any scalar $t$) doesn’t change the determinant.

However, the determinant scales by scaling a given row, so for example $$ R_3 \to -3R_3 $$ changes the determinant by a factor of $-3$ (or $R_i \to sR_i$ for any row index $i$, any scalar $s$ changes the determinant by a factor of $s$).

Your modification of the matrix is the composition of both, so the determinant is scaled by $1 \cdot (-3) = -3$, agreeing with your two determinant calculations.

Answer 2

If we add a scalar multiple of one row to another row, the determinant does not change.

Note that this is less general than replacing a row by a sum of multiples of other rows.