Question

If $a^2+b^2+c^2=1,$ then the value of
$\left|\begin{array}{ccc}{a}^2+(b^2+c^2)\cos \varphi & ab(1-\cos \varphi )& ac(1-\cos \varphi )\\ ba(1-\cos \varphi )& b^2+(c^2+a^2)\cos \varphi & bc(1-\cos \varphi )\\ ca(1-\cos \varphi )& cb(1-\cos \varphi )& c^2+(a^2+b^2)\cos \varphi \end{array}\right|$ is:

• A. $\sin \varphi$
• B. ${\sin }^2\varphi$
• C. $\cos \varphi$
• D. ${\cos }^2\varphi$

Answer 1

The correct option is D ${\cos }^2\varphi$

Multiplying $C_1$ by $a$, $C_2$ by $b$ and $C_3$ by $c$, we have
$\Delta =\frac{1}{abc}\left|\begin{array}{ccc}{a}^3+a(b^2+c^2)\cos \varphi & a{b}^2(1-\cos \varphi )& a{c}^2(1-\cos \varphi )\\ b{a}^2(1-\cos \varphi )& b^3+b(c^2+a^2)\cos \varphi & b{c}^2(1-\cos \varphi )\\ c{a}^2(1-\cos \varphi )& c{b}^2(1-\cos \varphi )& c^3+c(a^2+b^2)\cos \varphi \end{array}\right|$

Now take $a,b$ and $c$ common from$R_1$, $R_2$ and $R_3$ respectively, we have:

$\Delta =\frac{abc}{abc}\left|\begin{array}{ccc}{a}^2+(b^2+c^2)\cos \varphi & b^2(1-\cos \varphi )& c^2(1-\cos \varphi )\\ a^2(1-\cos \varphi )& b^2+(c^2+a^2)\cos \varphi & c^2(1-\cos \varphi )\\ a^2(1-\cos \varphi )& b^2(1-\cos \varphi )& c^2+(a^2+b^2)\cos \varphi \end{array}\right|$

Applying $C_1\to C_1+C_2+C_3,$ we get

$\Delta =\frac{abc}{abc}\left|\begin{array}{ccc}1& b^2(1-\cos \varphi )& c^2(1-\cos \varphi )\\ 1& b^2+(c^2+a^2)\cos \varphi & c^2(1-\cos \varphi )\\ 1& b^2(1-\cos \varphi )& c^2+(a^2+b^2)\cos \varphi \end{array}\right|$

[$\therefore a^2+b^2+c^2=1$]
Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$

$\Delta =\left|\begin{array}{ccc}1& b^2(1-\cos \varphi )& c^2(1-\cos \varphi )\\ 0& (a^2+b^2+c^2)\cos \varphi & 0\\ 0& 0& (a^2+b^2+c^2)\cos \varphi \end{array}\right|$

Expanding along $C_1$ we get $\Delta =\left|\begin{array}{cc}\cos \varphi & 0\\ 0& \cos \varphi \end{array}\right|={\cos }^2\varphi$