Question

please solve this question

Answer 1

Step-by-step explanation:

Just use the simple method of expansion of a determinant.

hope this answer helps.

Answer 2

Given :

$\left|\begin{array}{ccc}\sf1&\sf ab&\sf (a+b)c\\\sf1&\sf ca&\sf(c+a)b\\ \sf1 & \sf bc&\sf(b+c)a\end{array}\right| = \sf 0$

To prove :

LHS = RHS

Proof :

RHS = 0

$\implies \rm LHS =\left|\begin{array}{ccc}\sf1&\sf ab&\sf (a+b)c\\\sf1&\sf ca&\sf(c+a)b\\ \sf1 & \sf bc&\sf(b+c)a\end{array}\right| \\ \\ \implies \left|\begin{array}{ccc}\sf1&\sf ab&\sf ac+bc\\\sf1&\sf ca&\sf cb+ab\\ \sf1 & \sf bc&\sf ba+ca\end{array}\right|$

$\tt \: C_3 \rightarrow C_3+C_2 \\ \\ \implies \left|\begin{array}{ccc}\sf1&\sf ab&\sf ab + ac+bc\\\sf1&\sf ca&\sf ca + cb+ab\\ \sf1 & \sf bc&\sf bc + ba+ca\end{array}\right| \\ \\ \\ \implies \left|\begin{array}{ccc}\sf1&\sf ab&\sf ab + ac+bc\\\sf1&\sf ca&\sf ab + ac+bc\\ \sf1 & \sf bc&\sf ab + ac+bc\end{array}\right| \\ \\ \sf \: taking \: out \: (ab + ac+bc) \: from \: C_3 \: as \: common \: : \\ \\ \implies \sf (ab + ac+bc) \left|\begin{array}{ccc}\sf1&\sf ab&\sf 1\\\sf1&\sf ca&\sf 1\\ \sf1 & \sf bc&\sf 1\end{array}\right|$

We know that , if all the elements of rows or column of determinant are identical to the elements of some other row or column, then determinant is zero.

$\implies \sf (ab + ac+bc) \left|\begin{array}{ccc}\sf1&\sf ab&\sf 1\\\sf1&\sf ca&\sf 1\\ \sf1 & \sf bc&\sf 1\end{array}\right| = (ab + ac+bc) \times 0 \\ \\ \implies0$

Therefore, LHS = RHS

hence proved