Question

Evaluate: $\left|\begin{array}{ccc}ka& k^2+a^2& 1\\ kb& k^2+b^2& 1\\ kc& k^2+c^2& 1\end{array}\right|$

• A. $k(a+b)(b+c)(c+a)$
• B. $k(a-b)(b-c)(c-a)$
• C. $k^2(a-b)(b-c)(c-a)$
• D. $-k(a-b)(b-c)(c-a)$

Answer 1

The correct option is B $k(a-b)(b-c)(c-a)$

Let $A=\left|\begin{array}{ccc}ka& k^2+a^2& 1\\ kb& k^2+b^2& 1\\ kc& k^2+c^2& 1\end{array}\right|$
So, by row transformation
$r_1\to r_1-r_2$ and $r_3\to r_3-r_2$
So, $A=\left|\begin{array}{ccc}k(a-b)& (a^2-b^2)& 0\\ kb& k^2+b^2& 1\\ k(c-b)& c^2-b^2& 0\end{array}\right|$
$=-1\left[k\right(a-b\left)\right(c^2-b^2)-k(c-b\left)\right(a^2-b^2\left)\right]$
$=k(a-b)[b^2-c^2-(b-c\left)\right(a+b\left)\right]$
$=k(a-b)\left[\right(b-c\left)\right(c-a\left)\right]$