Computing determinant of matrix - confused with solution

Question

Given that $$\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=9$$

Compute \begin{vmatrix}a&b&c\\4g&4h&4i\\d+2a&e+2b&f+2c\end{vmatrix}

I have the solution my teacher made below but I'm having trouble understanding how he got the values in front of each matrix (which I highlighted). Like why does $4$ eventually turn into $-4$? I don't understand where he's pulling these values from and I'd appreciate it if someone explained.

\begin{align} \begin{vmatrix}a&b&c\\4g&4h&4i\\d+2a&e+2b&f+2c\end{vmatrix}\hspace{4mm}&\vcenter{\underrightarrow{R_2\leftarrow{\textstyle{\frac{1}{4}}}R_2}}\hspace{4mm}\bbox[yellow,3px]4\begin{vmatrix}a&b&c\\g&h&i\\d+2a&e+2b&f+2c\end{vmatrix} \\[2ex] &\vcenter{\underrightarrow{R_3\leftarrow R_3-2R_1}}\hspace{4mm}\bbox[yellow,3px]4\begin{vmatrix}a&b&c\\g&h&i\\d&e&f\end{vmatrix} \\[2ex] &\vcenter{\underrightarrow{R_2\leftrightarrow R_3}}\hspace{3mm}\bbox[yellow,3px]{-4}\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=-4\cdot9=-36 \end{align}

Answer 1

The key is to know how elementary row operations affect the determinant:

• If two rows are swapped, the determinant changes sign i.e. is multiplied by $-1$
• If a row is multiplied by a constant $c$, then so is the determinant
• If a multiple of one row is added to another, the determinant is unaffected

We are given that your matrix has determinant $9$. Notice that the matrix in question is obtained from your original matrix as follows:

• Multiply the third row by $4$ (so multiply determinant by $4$ to get $36$)
• Add twice row $1$ to row $2$ (no effect on determinant)
• Swap the second and third rows (multiply determinant by $-1$ to get $-36$)