Question

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

If the matrix

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

$\begin{vmatrix} a²-1 &ab &ac &ad\\ ba &b²-1 &bc &bd\\ ca &cb &c²-1 &cd\\ da &db &dc &d²-1\\ \end{vmatrix}= (a^2-1)\begin{vmatrix} b^2-1&bc&bd\\cb&c^2-1&cd\\db&dc&d^2-1 \end{vmatrix} -ab \begin{vmatrix} ba&bc&bd\\ca&c^2-1&cd\\da& dc&d^2-1 \end{vmatrix}+ac \begin{vmatrix} ba&b^2-1&bd\\ca&cb &cd\\da& db&d^2-1 \end{vmatrix}-ad \begin{vmatrix} ba&b^2-1&bc\\ca&cb&c^2-1\\da& db&dc \end{vmatrix}=$

$=(a^2-1)\bigg[(b^2-1)\big((c^2-1)(d^2-1)-c^2d^2\big)-bc\big(cb(d^2-1)-cbd^2\big)+bd\big(c^2bd-bd(c^2-1)\big)\bigg]-ab\bigg[ba\big((c^2-1)(d^2-1)-c^2d^2\big)-bc\big(ca(d^2-1)-acd^2\big)+bd\big(ac^2d-ad(c^2-1)\big)\bigg]+ac\bigg[ba\big(cb(d^2-1)-bcd^2\big)-(b^2-1)\big(ac(d^2-1)-acd^2\big)+bd(abcd-abcd)\bigg]-ad\bigg[ba\big(bc^2d-bd(c^2-1)\big)-ac\big(dc(b^2-1)-b^2cd\big)+ad\big((b^2-1)(c^2-1)-b^2c^2\big)\bigg]=$

$=(a^2-1)\bigg[-1+b^2+c^2+d^2\bigg]-a^2b^2-a^2c^2-a^2d^2=$

$=1-a^2-b^2-c^2-d^2=0$

So, $a^2+b^2+c^2+d^2=1$ .