A Question About A Calculation Of A Determinant. 
Calculate $\det(A)$.
$$A =\begin{bmatrix}
a&b&c&d\\
-b&a&-d&c\\
-c&d&a&-b\\
-d&-c&b&a\\
\end{bmatrix}.$$

This is an answer on a book:
$$A A^T = (a^2+b^2+c^2+d^2) I.$$
$$\det(A) = \det(A^T).$$
$$\det(A)^2 = (a^2+b^2+c^2+d^2)^4.$$
The coefficient of $a^4$ in $\det(A)$ is $1$.
Therefore,
$$\det(A) = (a^2+b^2+c^2+d^2)^2.$$

I cannot understand the above answer.
It is obvious that $\det(A)$ is a multivariable polynomial $p(a,b,c,d)$.
Maybe, the following is true:
$p(a,b,c,d) = (a^2+b^2+c^2+d^2)^2$ for $(a,b,c,d) \in A$,
$p(a,b,c,d) = -(a^2+b^2+c^2+d^2)^2$ for $(a,b,c,d) \in B$,
where, $A \cup B = \mathbb{R}^4, A \cap B = \emptyset$.

Please prove this is impossible.

Thank you very much, Mr. Kavi Rama Murthy.

Proof:
$p(a,b,c,d)$ is a continuous function.
$(a,b,c,d)=(0,0,0,0)$ is the only solution for $p(a,b,c,d)^2 = (a^2+b^2+c^2+d^2)^4 = 0$.
so,
$(a,b,c,d)=(0,0,0,0)$ is the only solution for $p(a,b,c,d) = 0$.
$p(1,1,0,0) = \begin{vmatrix}
1&1&0&0\\
-1&1&0&0\\
0&0&1&-1\\
0&0&1&1\\
\end{vmatrix} = \begin{vmatrix}
1&1&0&0\\
0&2&0&0\\
0&0&1&-1\\
0&0&0&2\\
\end{vmatrix}
= 4 > 0$.
Let $(w_0, x_0, y_0, z_0) \in \mathbb{R}^4 - (0,0,0,0)$.
We can assume $w_0 \ne 0$ without loss of generality.
$f(x) := p(x, 1, 0, 0)$ is a one variable continuous function.
By the intermediate-value theorem, $f(w_0) = p(w_0, 1, 0, 0) > 0$.
$g(x, y, z) := p(w_0, x,y,z)$ is a three variable continuous funtion.
Again, by the intermediate-value theorem, $g(x_0, y_0, z_0) = p(w_0,x_0,y_0,z_0) > 0$.
Hence, $p(w, x, y, z) \geq 0$ for all $(w,x,y,z) \in \mathbb{R}^4$.
So, $p(w, x, y, z) = (w^2+x^2+y^2+z^2)^2$ for all $(w,x,y,z) \in \mathbb{R}^4$.
 A: The book tells you that $A A^\top = x I$ with $x = (a^2+b^2+c^2+ d^2)$. It follows that
$$\det(A A^\top)= \det(x I) = x^4$$
But one has
$$\det(A A^\top) = \det(A)\det(A^\top) = (\det(A))^2 $$
Hence $\det(A)=\pm x^2$. The coefficient of $a^4$ shows that it is $+x^2$.
A: why did I do like this? because above answer was formed by multiplying with transpose. if someone is interested in the direct solution, one can proceed as follows. It is too broad. I have learned about this technique today. I don't recommend this. Till $3\times 3$ this method is fine for this type of problems. For fun, I have solved like this.
Let $F$ be the field and let$D$ be an alternating $4-$linear function on $4\times 4$ matrices over the polynomial ring $F[x]$.
Let
$$A =\begin{bmatrix}
a&b&c&d\\
-b&a&-d&c\\
-c&d&a&-b\\
-d&-c&b&a\\
\end{bmatrix}.$$ If we denonte the rows of the identity matrix by $\epsilon_1$,$\epsilon_2$,$\epsilon_3$ and $\epsilon_4$. Then $$D(A)=D(a\epsilon_1-b\epsilon_2-c\epsilon_3-d\epsilon_4,b\epsilon_1+a\epsilon_2+d\epsilon_3-c\epsilon_4,c\epsilon_1-d\epsilon_2+a\epsilon_3+b\epsilon_4,d\epsilon_1+c\epsilon_2-b\epsilon_3+a\epsilon_4)$$
$$D(A)=aD(\epsilon_1,b\epsilon_1+a\epsilon_2+d\epsilon_3-c\epsilon_4,c\epsilon_1-d\epsilon_2+a\epsilon_3+b\epsilon_4,d\epsilon_1+c\epsilon_2-b\epsilon_3+a\epsilon_4)-bD(\epsilon_2,b\epsilon_1+a\epsilon_2+d\epsilon_3-c\epsilon_4,c\epsilon_1-d\epsilon_2+a\epsilon_3+b\epsilon_4,d\epsilon_1+c\epsilon_2-b\epsilon_3+a\epsilon_4)-cD(\epsilon_3,b\epsilon_1+a\epsilon_2+d\epsilon_3-c\epsilon_4,c\epsilon_1-d\epsilon_2+a\epsilon_3+b\epsilon_4,d\epsilon_1+c\epsilon_2-b\epsilon_3+a\epsilon_4)-dD(\epsilon_4,b\epsilon_1+a\epsilon_2+d\epsilon_3-c\epsilon_4,c\epsilon_1-d\epsilon_2+a\epsilon_3+b\epsilon_4,d\epsilon_1+c\epsilon_2-b\epsilon_3+a\epsilon_4)$$
By simplifying using the properties, we get $D(A)=a^4+b^4+c^4+d^4+2a^2b^2+2a^2c^2+2a^2d^2+2b^2c^2+2b^2d^2+2c^2d^2=(a^2+b^2+c^2+d^2)^2.$
