Answer to Question #216070 in Algebra for Tunde

Question

Let a, b, c, d be a set of real numbers, show that a²+b²+c²+d²=1

If

a²-1 ab ac ad

ba b²-1 bc bc

ca cb c²-1 cd

da db dc d²-1

equal to zero.

Answer 1

$\begin{vmatrix} (a^2-1) & ab & ac & ad\\ ba & (b^2-1) & bc & bd\\ ca & cb & (c^2-1) & cd\\ ad & db & dc & (d^2-1) \end{vmatrix}=0$ using Laplace method to find the determinant

$\begin{vmatrix} a^2-1 & ab \\ ba & b^2-1 \end{vmatrix}\begin{vmatrix} c^2-1 & cd \\ dc & d^2-1 \end{vmatrix}-\\ \\ \begin{vmatrix} a^2-1 & ac \\ ba & bc \end{vmatrix}\begin{vmatrix} cb & cd \\ bd & d^2-1 \end{vmatrix}+\\ \begin{vmatrix} a^2-1 & ad \\ ba & bd \end{vmatrix}\begin{vmatrix} cb & c^2-1 \\ db & dc \end{vmatrix}+\\ \begin{vmatrix} ab & ac \\ b^2-1 & bc \end{vmatrix}\begin{vmatrix} ca & cd \\ da & d^2-1 \end{vmatrix}-\\ \begin{vmatrix} ab & ad \\ b^2-1 & bd \end{vmatrix}\begin{vmatrix} ca & c^2-1\\ da & dc \end{vmatrix}+\\ \begin{vmatrix} ac & ad \\ bc & bd \end{vmatrix}\begin{vmatrix} ca & cb \\ da & db \end{vmatrix}=0 \\ (1-(a^2+b^2))(1-(c^2+d^2))-(bc)^2-(bd)^2-\\(ac)^2-(ad)^2+0=0\\ 1-c^2-d^2-a^2-b^2+(cb)^2+(ac)^2+(ad)^2+\\(bd)^2-(bc)^2-(bd)^2-(ac)^2-(ad)^2=0\\ \therefore\\ 1=a^2+b^2+c^2+d^2\\ a^2+b^2+c^2+d^2=1$