310. Volume of a parallelepiped. Calculate the determinant

$$
\left|\begin{array}{ccc}
a^{2}+b^{2}-c^{2}-d^{2} & 2 b c-2 a d & 2 b d+2 a c \\
2 b c+2 a d & a^{2}-b^{2}+c^{2}-d^{2} & 2 c d-2 a b \\
2 b d-2 a c & 2 c d+2 a b & a^{2}-b^{2}-c^{2}+d^{2}
\end{array}\right|
$$
310. I. The determinant is equal to the oriented volume of the parallelepiped constructed on vectors, the projections of which in the Cartesian coordinate system coincide with the rows of this determinant. The square of the length of the first vector (and, by symmetry, the square of the length of each of the other two vectors) is equal to $\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}$. The scalar product of the first pair of vectors is 0, so by symmetry all our vectors are pairwise orthogonal. Therefore, our parallelepiped is a cube with volume $\pm\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{3}$. Since it is directly visible that the coefficient of $a^{6}$ in this determinant is +1, the determinant is equal to $\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{3}$.

Using the same method, more general results can be obtained. For example, let $s_{n}=x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}$, and let $I_{n}$ be the identity matrix of order $n$, and $A_{n}$ be the matrix of order $n$ with element $a_{i j}=(-1)^{i+j+1} x_{i} x_{j}$. Then for $n=1,2, \ldots$ we get

$$
\left|s_{n} I_{n}+2 A_{n}\right|=-s_{n}^{n}
$$

[N. Nettheim, M. M., 43, 229 (May 1970).]

II. Let $D$ be the given determinant and let

$$
A=\left|\begin{array}{rrr}
a & d & -c \\
-d & a & b \\
c & -b & a
\end{array}\right|
$$

It is easy to see that the determinant $A^{\prime}$ of the adjugate matrix ${ }^{*}$ is

$$
A^{\prime}=\left|\begin{array}{lll}
a^{2}+b^{2} & b c+a d & b d-a c \\
b c-a d & a^{2}+c^{2} & a b+c d \\
b d+a c & c d-a b & a^{2}+d^{2}
\end{array}\right|
$$

and that

$A \cdot A^{\prime}=$

$$
=\left|\begin{array}{ccc}
a\left(a^{2}+b^{2}-c^{2}-d^{2}\right) & 2 a(b c+a d) & 2 a(b d-a c) \\
2 a(b c-a d) & a\left(a^{2}+c^{2}-b^{2}-d^{2}\right) & 2 a(a b+c d) \\
2 a(a c+b d) & 2 a(c d-a b) & a\left(a^{2}+d^{2}-b^{2}-c^{2}\right)
\end{array}\right|
$$

that is, $A \cdot A^{\prime}=a^{3} D$. By elementary properties of determinants, we get that $A^{\prime}=A^{2}$, so $a^{3} D=A^{3}$.

Since $A=a\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$, we immediately find $D=\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{3}$

[E. Morrison, M. M., 43, 230 (May 1970).]

III. Denoting the original determinant by $\Delta$, and the determinant of the transposed matrix by $\Delta_{T}$, we get that

$$
\begin{aligned}
& \Delta^{2}=\Delta \cdot \Delta_{T}= \\
& =\left|\begin{array}{ccc}
\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2} & 0 & 0 \\
0 & \left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2} & 0 \\
0 & 0 & \left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}
\end{array}\right| .
\end{aligned}
$$

Therefore,

$$
\Delta=\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{3}
$$

[B. Iansson, M. M., 43, 230 (May 1970).]