We are given the determinant of the matrix:

$\det \begin{bmatrix} a & b & c \\ p & q & r \\ x & y & z \end{bmatrix} = -1,$

and we are asked to compute the determinant of the matrix:

$\det \begin{bmatrix} -2a & -2b & -2c \\ 2p + x & 2q + y & 2r + z \\ 3x & 3y & 3z \end{bmatrix}.$

Step 1: Factor out constants from rows

We can factor out constants from the rows of the matrix.

• In the first row, each element has a factor of $-2$, so we can factor out $-2$ from the first row.
• In the second row, we have $2p + x$, $2q + y$, and $2r + z$. This is a sum of two terms, so we can't factor out any constant from the whole row.
• In the third row, each element has a factor of $3$, so we can factor out $3$ from the third row.

So, we have:

= (-2) \cdot 3 \cdot \det \begin{bmatrix} a & b & c \\ 2p + x & 2q + y & 2r + z \\ x & y & z \end{bmatrix}.$$ Now, the expression simplifies to: $$-6 \cdot \det \begin{bmatrix} a & b & c \\ 2p + x & 2q + y & 2r + z \\ x & y & z \end{bmatrix}.$$ ### Step 2: Perform row operation to simplify the determinant Next, observe that we can simplify the second row by subtracting twice the third row from it. Specifically: - Replace the second row $$[2p + x, 2q + y, 2r + z]$$ by $$[2p + x - 2x, 2q + y - 2y, 2r + z - 2z] = [2p - x, 2q - y, 2r - z]$$. This operation does not change the determinant because it is a row operation that corresponds to adding a multiple of one row to another. The matrix becomes: $$\begin{bmatrix} a & b & c \\ 2p - x & 2q - y & 2r - z \\ x & y & z \end{bmatrix}.$$ ### Step 3: Factor out constants again Notice that the second row is now $$[2p - x, 2q - y, 2r - z]$$, which can be rewritten as \( 2 \cdot [p - \frac{x}{2}, q - \frac{y}{2}, r - \frac{z}{2]} \,.