Question

If $a^2+b^2+c^2=1$ then is the value of the determinant $\left|\begin{array}{ccc}{a}^2+(b^2+c^2)cos\theta & ba(1-cos\theta )& ca(1-cos\theta )\\ ab(1-cos\theta )& b^2(c^2+a^2)cos\theta & cb(1-cos\theta )\\ ac(1-cos\theta )& bc(1-cos\theta )& c^2+(a^2+b^2)cos\theta \end{array}\right|$ independent of a,b,c?
If yes enter 1 else enter 0.

Answer 1

Let $\Delta =\left|\begin{array}{ccc}{a}^2+(b^2+c^2)\cos \varphi & ab(1-\cos \varphi )& ac(1-\cos \varphi )\\ ba(1-\cos \varphi )& b^2+(c^2+a^2)\cos \varphi & bc(1-\cos \varphi )\\ ca(1-\cos \varphi )& cb(1-\cos \varphi )& c^2+(a^2+b^2)\cos \varphi \end{array}\right|$
Multiplying $C_1$ by $a$, $C_2$ by $b$ and $C_3$ by $c$ we get
$\Delta =\frac{1}{abc}\left|\begin{array}{ccc}{a}^3+a(b^2+c^2)\cos \varphi & a{b}^2(1-\cos \varphi )& a{c}^2(1-\cos \varphi )\\ b{a}^2(1-\cos \varphi )& b^3+b(c^2+a^2)\cos \varphi & b{c}^2(1-\cos \varphi )\\ c{a}^2(1-\cos \varphi )& c{b}^2(1-\cos \varphi )& c^3+c(a^2+b^2)\cos \varphi \end{array}\right|$
Taking $a$ common from $R_1$, $b$ from $R_2$ and $c$ from $R_3$, we get
$\Delta =\frac{abc}{abc}\left|\begin{array}{ccc}{a}^2+(b^2+c^2)\cos \varphi & b^2(1-\cos \varphi )& c^2(1-\cos \varphi )\\ a^2(1-\cos \varphi )& b^2+(c^2+a^2)\cos \varphi & c^2(1-\cos \varphi )\\ a^2(1-\cos \varphi )& b^2(1-\cos \varphi )& c^2+(a^2+b^2)\cos \varphi \end{array}\right|$
Applying $C_1\to C_1+C_2+C_3$
$\Delta =\left|\begin{array}{ccc}1& b^2(1-\cos \varphi )& c^2(1-\cos \varphi )\\ 1& b^2+(c^2+a^2)\cos \varphi & c^2(1-\cos \varphi )\\ 1& b^2(1-\cos \varphi )& c^2+(a^2+b^2)\cos \varphi \end{array}\right|$
As $a^2+b^2+c^2=1$
Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$
$\Delta =\left|\begin{array}{ccc}1& b^2(1-\cos \varphi )& c^2(1-\cos \varphi )\\ 0& (b^2+c^2+a^2)\cos \varphi & 0\\ 1& 0& (c^2+a^2+b^2)\cos \varphi \end{array}\right|$
Expanding along $C_1$
We get $\Delta ={\cos }^2\varphi$