0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

$$\det(A)=\det(A^T)$$ 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).

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.