Question

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

• A. ${\cos }^2\theta$
• B. 0
• C. 1
• D. ${\sin }^2\theta$

Answer 1

Answer: (A)

First multiply $C_1$ by $a,C_2$ by $b,C_3$ by $c$, followed by
multiplying $R_1$ by $1/a,R_2$ by $1/b$ and $R_3$ by $1/c$, we get $\left|\begin{array}{ccc}{a}^2+\left(b^2+c^2\right)\cos \theta & b^2\left(1-\cos \theta \right)& c^2\left(1-\cos \theta \right)\\ a^2\left(1-\cos \theta \right)& b^2+\left(c^2+a^2\right)\cos \theta & c^2\left(1-\cos \theta \right)\\ a^2\left(1-\cos \theta \right)& b^2\left(1-\cos \theta \right)& c^2+\left(a^2+b^2\right)\cos \theta \end{array}\right|$
Using $C_1\to C_1+C_2+C_3$ and $a^2+b^2+c^2=1$, we get $\Delta =\left|\begin{array}{ccc}1& b^2(1-\cos \theta )& c^2(1-\cos \theta )\\ 1& b^2+\left(1-b^2\right)\cos \theta & c^2(1-\cos \theta )\\ 1& b^2(1-\cos \theta )& c^2+\left(1-c^2\right)\cos \theta \end{array}\right|$
Using $R_2\to R_2-R_1,R_3\to R_3-R_1$, we get
$\Delta =\left|\begin{array}{ccc}1& b^2(1-\cos \theta )& c^2(1-\cos \theta )\\ 0& \cos \theta & 0\\ 0& 0& \cos \theta \end{array}\right|={\cos }^2\theta$