Operations upon the determinant of a matrix

what is the value of the determinant in the last question

I am having trouble finding the determinant in the last question.

I tried det = 15 because the first row is multiplied by 5, which multiples the original det value of 3 by 5. Then, I changed 15 into -15 because the rows are swapped.

I know that adding a scalar multiple of one row or column to another does not change the determinant. I assume if I factor out the negative sign from the last row so that it fits this rule. I end up with det = 15 again.

Where in my logic am I going awry?

Thanks,

Interchange the first and third row, this gives a $$-1$$. Now remove the $$5$$ from the third row, so you have a $$5$$ coming out, and the last two rows are $$[d,e,f]$$ and $$[g,h,i]$$, as with the usual matrix. Now, you must split the first row using linearity of the determinant (in its first row in our case): $$\det\begin{bmatrix} -3a-2d & -3b-2e & -3c - 2f \\ d&e&f \\ g&h&i \end{bmatrix} \\= \det\begin{bmatrix} -3a & -3b & -3c \\ d&e&f \\ g&h&i \end{bmatrix} + \det\begin{bmatrix} -2d & -2e & - 2f \\ d&e&f \\ g&h&i \end{bmatrix}$$
The second matrix has determinant zero, and the first has determinant $$-3 \times -3 = 9$$. Multiplying this with the $$-1$$ and $$5$$ we got earlier gives the result $$-45$$.
I would start from the third question. Multiplying the top row by $$5$$ produces a determinant of $$15$$. Then, multiplying the bottom row by $$-3$$ (the step you missed) makes a determinant of $$-45$$. Finally, subtract two lots of the second row from the third, which leaves the determinant at $$-45$$. As this produces the matrix in question, $$-45$$ is your answer.