How to solve this part of deduction of determinant of transformation matrix? In part of the deduction for the determinant left or right handedness of a transformation matrix the author comes to this:
$$a_{21}a_{31}+a_{22}a_{32}+a_{23}a_{33}=0,\qquad i.e. \quad a_{2k}a_{3k}=0\qquad (4.1)\\
a_{31}a_{11}+a_{32}a_{12}+a_{33}a_{13}=0,\qquad i.e. \quad a_{3k}a_{1k}=0\qquad (4.2)\\
a_{11}a_{21}+a_{12}a_{22}+a_{13}a_{23}=0,\qquad i.e.\quad a_{1k}a_{2k}=0\qquad (4.3)$$
$a_{ij}$ are direction cosines for the transformation of one set of axes $Ox_i$ to another $Ox_i'$ given by the transformation matrix $$\begin{bmatrix}x_1'&x_2'&x_3'\end{bmatrix}=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$$so that $\langle x_1\,x_2\rangle=\langle x_2,x_3\rangle =\langle x_1,x_3\rangle =0$. This is shown in (4.2) and (4.3).
And then the author says "by solving (4.2) and (4.3) for the ratios $a_{11}$, $a_{12}$, $a_{13}$, we obtain:"
$$\frac{a_{11}}{a_{32}a_{23}-a_{33}a_{22}}=\frac{a_{12}}{a_{33}a_{21}-a_{31}a_{23}}=\frac{a_{13}}{a_{31}a_{22}-a_{32}a_{21}}=-\frac{1}{k}\qquad(5)$$
I can't figure out how the author got from (4.2) and (4.3) to (5)? Anyone help, please? 
 A: Your first set of equations is equivalent to the following matrix relationship:
$$\tag{1}\underbrace{\begin{pmatrix}a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}\end{pmatrix}}_M \underbrace{\begin{pmatrix}a_{11}&a_{21}&a_{31}\\
a_{12}&a_{22}&a_{32}\\
a_{13}&a_{23}&a_{33}\end{pmatrix}}_{M^T} = \underbrace{\begin{pmatrix}p&0&0\\
0&q&0\\0&0&r\end{pmatrix}}_{diag(p,q,r)} $$
for certain real numbers $p,q,r$.
Let us denote by $M_1,M_2,M_3$ the line vectors of $M$. 
The denominators of your second set of equations are the coordinates of the opposite of $M_2 \times M_3$ (their cross product), then : 
$$\tag{2}k M_1 = M_2 \times M_3 \ \ \text{(cross product)}$$
The logical connection between (1) and (2) is that the $M_k$s 
constitute an orthogonal basis (not necessarily orthonormal) ; thus one of the vectors is naturaly proportional to the cross product of the 2 others !
Edit : a different (computational) approach.
Multiply (4.2) by $a_{22}$,  multiply (4.3) by $a_{32}$ ; then subtract the resulting equations. After cancellation of the common term $a_{12}a_{22}a_{32}$ and convenient factorizations, we get:
$$-a_{11}(a_{31}a_{22} - a_{32}a_{21})+a_{13}(a_{32}a_{23}-a_{33}a_{22})=0$$
which is equivalent to :
$$\tag{3}\frac{a_{11}}{a_{32}a_{23}-a_{33}a_{22}}=\frac{a_{13}}{a_{31}a_{22}-a_{32}a_{21}}$$
Thus, we have obtained the equality of the first and the third ratios in (5).
A similar manipulation on the two first equations will give
$$\tag{4}\frac{a_{11}}{a_{32}a_{23}-a_{33}a_{22}}=\frac{a_{12}}{a_{33}a_{21}-a_{31}a_{23}}$$
Regarding the last ratio, the common value of (3) and (4) can be taken as being $-\dfrac{1}{k}$ for a certain $k$.
