Based on the solution of Exercise 956, let's write down some second-order determinants whose value is equal to the reciprocal of the determinant

$$
D=\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|=a d-b c \neq 0
$$

Using second-order determinants, write down the ratio $x: y: z$ of the numbers $x, y, z$ that satisfy the system of equations

$$
\begin{aligned}
& a x+b y+c z=0 \\
& d x+e y+f z=0
\end{aligned}
$$

[^0]:    ${ }^{1}$ See K. M. L. 31 (1965) 119. o.
I. The requirement can be met in various ways. Any number $t$ can be written in the form of a second-order determinant:

$$
t=\left|\begin{array}{ll}
t & 0 \\
u & 1
\end{array}\right|, \quad \text { therefore the } \quad\left|\begin{array}{cc}
1 / D & 0 \\
u & 1
\end{array}\right|
$$

determinant - where $u$ is an arbitrary number - satisfies the requirement. - Below we provide an appropriate determinant through a different method.

Let the determinant written with elements $x, y, z, u$ be

$$
X=\left|\begin{array}{ll}
x & y \\
z & u
\end{array}\right|=\frac{1}{D}
$$

that is, the product of the two determinants, performed by row-row composition,

$$
D X=\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right| \cdot\left|\begin{array}{ll}
x & y \\
z & u
\end{array}\right|=\left|\begin{array}{ll}
a x+b y & a z+b u \\
c x+d y & c z+d u
\end{array}\right|=1=\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|
$$

The last two determinants are clearly equal if their elements are respectively equal, hence $x, y, z, u$ satisfy if

$$
\begin{aligned}
& a x+b y=1 \\
& c x+d y=0 \\
& a z+b u=0 \\
& c z+d u=1
\end{aligned}
$$

Subtract $d$ times (3) from $b$ times (4), and $c$ times (3) from $a$ times (4):

$$
\begin{gathered}
(a d-b c) x=D x=d, \quad(b c-a d) y=-D y=c \\
\text { thus } x=d / D, \quad y=-c / D
\end{gathered}
$$

and similarly from equations (5) and (6)

$$
z=-b / D, \quad u=a / D
$$

Thus, the condition is indeed satisfied by the transformations seen in the cited practice (8) and (6) equations:

$$
\begin{aligned}
& X=\left|\begin{array}{cc}
d / D & -c / D \\
-b / D & a / D
\end{array}\right|=\frac{1}{D} \cdot \frac{1}{D} \cdot\left|\begin{array}{cc}
d & -c \\
-b & a
\end{array}\right|= \\
& =\frac{1}{D^{2}} \cdot\left|\begin{array}{cc}
a & -b \\
-c & d
\end{array}\right|=\frac{1}{D^{2}} \cdot\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|=\frac{D}{D^{2}}=\frac{1}{D}
\end{aligned}
$$

In the last step, we wrote the negatives of the elements on the secondary diagonal, so the value of their product did not change.

From (2), we can obtain an arbitrary number of appropriate determinants by the same method, by writing a different appropriate one in the last transformation, for example

$$
1=\left|\begin{array}{ll}
2 & 1 \\
1 & 1
\end{array}\right|=\left|\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right|=\left|\begin{array}{cc}
n & n+1 \\
n-1 & n
\end{array}\right|
$$

II. Let us write a definite value $z_{0}(\neq 0)$ in place of $z$. Multiply the equations in (1) first by $e$ and $-b$, then by $-d$ and $a$. Adding them together,

$$
\begin{gathered}
(a e-b d) x-(b f-c e) z_{0}=0, \quad(a e-b d) y-(c d-a f) z_{0}=0 \\
x=\frac{b f-c e}{a e-b d} \cdot z_{0}, \quad y=\frac{c d-a f}{a e-b d} \cdot z_{0}
\end{gathered}
$$

provided the common denominator is not 0, otherwise for any $z_{0}$, either infinitely many triples of $x, y, z_{0}$ satisfy (1), or no triple does. The two numerators and the denominator can be written as second-order determinants, so the sought ratio cannot be other than

$$
x: y: z=\left|\begin{array}{ll}
b & c \\
e & f
\end{array}\right|:\left|\begin{array}{ll}
c & a \\
f & d
\end{array}\right|:\left|\begin{array}{ll}
a & b \\
d & e
\end{array}\right|
$$

At most one of the first two determinants in (7) can be 0. If both are 0, then the third one is also, because if $b f-c e=c d-a f=0$, then $e / b=f / c=d / a=k$, and thus $a e-b d=0$. In this case, the second equation in (1) can be obtained from the first by multiplying by $k$.

Further discussion of the question is not our task here.