# Determinant properties doubt

Knowing that $$\begin{vmatrix} 2 & 2 & 3\\ x & y & z\\ a & 2b & 3c \end{vmatrix}=10$$ and $x,y,z,a,b,c \in \mathbb{R}$, calculate $$\begin{vmatrix} 0 & 3x & y & z\\ 0 & 3a & 2b & 3c\\ 0 & 6 & 2 & 3\\ 5 & 0 & 0 & 0 \end{vmatrix}$$

The solution is

$$\begin{vmatrix} 0 & 3x & y & z\\ 0 & 3a & 2b & 3c\\ 0 & 6 & 2 & 3\\ 5 & 0 & 0 & 0 \end{vmatrix}=(-5)\begin{vmatrix} 3x & y & z\\ 3a & 2b & 3c\\ 6 & 2 & 3 \end{vmatrix}=(-5)\cdot 3\begin{vmatrix} x & y & z\\ a & 2b & 3c\\ 2 & 2 & 3 \end{vmatrix}=(-5)\cdot 3\cdot (-1)\begin{vmatrix} x & y & z\\ 2 & 2 & 3\\ a & 2b & 3c \end{vmatrix}$$ $$=(-5)\cdot 3\cdot (-1)(-1)\begin{vmatrix} 2 & 2 & 3\\ x & y & z\\ a & 2b & 3c \end{vmatrix}=(-5)\cdot 3\cdot (-1)(-1)\cdot 10=\boxed{-150}$$

I have a doubt with this determinant in the first step. It is resolved, but I do not understand why, only the first step (because it is part of a major determinant), the other properties are clear to me.

$$\det(A)=\det(A^T)$$ Taking transpose of the $4\times4$ matrix and you will see that the first row has only a $5$, I think it should be rather easy to understand the calculation then(-5 times the minor of the original matrix).

They expand the order $4$ determinant by the first column, that's all. The rules for the signs are the same as for an expansion by a row.

$A = \begin{vmatrix} 0 & 3x & y & z\\ 0 & 3a & 2b & 3c\\ 0 & 6 & 2 & 3\\ 5 & 0 & 0 & 0 \end{vmatrix}$

For a $4\times 4$ matrix the signs of the minors are;

$A = \begin{vmatrix}+&&-&&+&&-\\-&&+&&-&&+\\+&&-&&+&&-\\-&&+&&-&&+\end{vmatrix}$

Expand along $R_4$ gives ;

$A = (-5)\begin{vmatrix}3x&&y&&z\\3a&&2b&&3c\\6&&2&&3\end{vmatrix}$

interchanging the rows $R_1\leftrightarrow R_3$ and $R_2\leftrightarrow R_3$

so, $A = (-5)\begin{vmatrix}6&&2&&3\\3x&&y&&z\\3a&&2b&&3c\end{vmatrix}$

$A = -5\cdot3\cdot\begin{vmatrix}2&&2&&3\\x&&y&&z\\a&&2b&&3c\end{vmatrix}$

$A = -15\cdot 10= -150$