Question

Using properties of determinants, prove that $\left|\begin{array}{ccc}\frac{(a+b{)}^2}{c}& c& c\\ a& \frac{(b+c{)}^2}{a}& a\\ b& b& \frac{(c+a{)}^2}{b}\end{array}\right|=2(a+b+c{)}^3$

OR

If $p\ne 0,q\ne 0$ and $\left|\begin{array}{ccc}p& q& p\alpha +q\\ q& r& p\alpha +r\\ p\alpha +q& q\alpha +r& 0\end{array}\right|=0$, then, using properties of determinants.
Prove that at least one of the following statements is true:

(a) p, q, r are in G.P.

(b) $\alpha$ is a root of the equation $p{x}^2+2qx+r=0$

Answer 1

LHS : Let $\Delta =\left|\begin{array}{ccc}\frac{(a+b{)}^2}{c}& c& c\\ a& \frac{(b+c{)}^2}{a}& a\\ b& b& \frac{(c+a{)}^2}{b}\end{array}\right|=2(a+b+c{)}^3$

By $R_1\to c{R}_1,R_2\to a{R}_2,R_3\to b{R}_3\Rightarrow \Delta =\frac{1}{abc}\left|\begin{array}{ccc}(a+b{)}^2& c^2& c^2\\ a^2& (b+c{)}^2& a^2\\ b^2& b^2& (c+a{)}^2\end{array}\right|$

By $C_1\to C_1-C_{3}and{C}_2\to C_2-C_3,\Rightarrow \Delta =\frac{1}{abc}\left|\begin{array}{ccc}(a+b+c)(a+b-c)& 0& c^2\\ 0& (b+c+a)(b+c-a)& a^2\\ (b+c+a)(b-c-a)& (b+c+a)(b-c-a)& (c+a{)}^2\end{array}\right|$

Taking (a+b+c) common form $C_1$ and $C_2$ both

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

By $R_3\to R_3-(R_1+R_2)$,
$\Rightarrow \Delta =\frac{(a+b+c{)}^2}{abc}\left|\begin{array}{ccc}(a+b-c)& 0& c^2\\ 0& (b+c-a)& a^2\\ -2a& -2c& 2ca\end{array}\right|$

By $C_1\to c{C}_1,C_2\to a{C}_2$
$\Rightarrow \Delta =\frac{(a+b+c{)}^2}{abcac}\left|\begin{array}{ccc}(ac+bc-c^2)& 0& c^2\\ 0& (ba+ca-a^2)& a^2\\ -2ca& -2ca& 2ca\end{array}\right|$

By $C_1\to C_1+C_3,C_2\to C_2+C_3\Rightarrow \Delta =\frac{(a+b+c{)}^2}{abcac}\left|\begin{array}{ccc}ac+bc& c^2& c^2\\ a^2& ba+ca& a^2\\ 0& 0& 2ca\end{array}\right|$

Take c,a, 2ca common from $R_1,R_2,and{R}_3$,

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

Expanding along $R_3$,

$\Rightarrow \Delta =\frac{2(a+b+c{)}^2}{b}[0-0+1(ab+ac+b^2+bc-ac\left)\right]$

$\therefore \Delta =\frac{2(a+b+c{)}^2}{b}(ab+b^2+bc)=2(a+b+c{)}^2=RHS$.

OR We have

$\left|\begin{array}{ccc}p& q& p\alpha +q\\ q& r& q\alpha +r\\ p\alpha +q& q\alpha +r& 0\end{array}\right|=0$

By $R_1\to \alpha R_1,\frac{1}{\alpha }\left|\begin{array}{ccc}p\alpha & q\alpha & p{\alpha }^2+q\alpha \\ q& r& q\alpha +r\\ p\alpha +q& q\alpha +r& 0\end{array}\right|=0$

By $R_3\to R_3-(R_1+R_2),\frac{1}{\alpha }\left|\begin{array}{ccc}p\alpha & q\alpha & p{\alpha }^2+q\alpha \\ q& r& q\alpha +r\\ 0& 0& -p{\alpha }^2-2q\alpha -r\end{array}\right|=0$

Expanding along $R_1,\frac{1}{\alpha }[0-0+(-p{\alpha }^2-2q\alpha -r\left)\right(p\alpha r-q^2\alpha \left)\right]=0$

$\Rightarrow -\frac{1}{\alpha }(-\alpha (q^2-pr\left)\right)(2q\alpha +r+p{\alpha }^2)=0$

Either $(q^2-pr)=0$ or $(2q\alpha +r+p{\alpha }^2)=0$. That implies $q^2=pr$ or $p(\alpha {)}^2+2q(\alpha )+r=0$

Therefore either p,q,r are G.P. or $\alpha$ satisfies the equation $p{x}^2+2qx+r=0$

Hence either p,q,r are in G.P. or, $\alpha$ is a root of the equation $p{x}^2+2qx+r=0$