Mark the tick against the correct answer in the following: $\left|\begin{array}{lll} \mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{ca} & \mathrm{c}+\mathrm{a} & 1 \\ \mathrm{ab} & \mathrm{a}+\mathrm{b} & 1 \end{array}\right|=?$ A. $(a-b)(b-c)(c-a)$ B. $-(a-b)(b-c)(c-a)$ C. $(a+b)(b+c)(c+a)$ D. None of these

Mark the tick against the correct answer in the following: $\left|\begin{array}{lll} \mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{ca} & \mathrm{c}+\mathrm{a} & 1 \\ \mathrm{ab} & \mathrm{a}+\mathrm{b} & 1 \end{array}\right|=?$
A. $(a-b)(b-c)(c-a)$
B. $-(a-b)(b-c)(c-a)$
C. $(a+b)(b+c)(c+a)$
D. None of these

CBSE | Class 12 | Determinants | Maths | RS Aggarwal

Solution:

Option(A)
To find: Value of $\left|\begin{array}{lll}\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{ca} & \mathrm{a}+\mathrm{c} & 1 \\ \mathrm{ab} & \mathrm{a}+\mathrm{b} & 1\end{array}\right|$
We have, $\left|\begin{array}{lll}\text { bc } & \text { b+c } & 1 \\ \text { ca } & \text { a+c } & 1 \\ \text { ab } & \text { a+b } & 1\end{array}\right|$
Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$
$\Rightarrow\left|\begin{array}{ccc} b c-c a & b-a & 0 \\ c a & a+c & 1 \\ a b & a+b & 1 \end{array}\right|$
$\Rightarrow\left|\begin{array}{ccc} c(b-a) & b-a & 0 \\ c a & a+c & 1 \\ a b & a+b & 1 \end{array}\right|$
Taking (b – a) common
$\Rightarrow(b-a)\left|\begin{array}{ccc} c & 1 & 0 \\ c a & a+c & 1 \\ a b & a+b & 1 \end{array}\right|$
Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$
$\begin{array}{l} \Rightarrow(b-a)\left|\begin{array}{ccc} c & 1 & 0 \\ c a-a b & c-b & 0 \\ a b & a+b & 1 \end{array}\right| \\ \Rightarrow(b-a)\left|\begin{array}{ccc} c(c-b) & c-b & 0 \\ a b & a+b & 1 \end{array}\right| \end{array}$
Taking $(c-b)$ common
$\Rightarrow(b-a)(c-b)\left|\begin{array}{ccc} c & 1 & 0 \\ a & 1 & 0 \\ a b & a+b & 1 \end{array}\right|$
Expanding along $C_{3}$
$\Rightarrow(b-a)(c-b)[1\{(c)(1)-(a)(1)\}]$
$\begin{array}{l} \Rightarrow(b-a)(c-b)(c-a) \\ \Rightarrow(a-b)(b-c)(c-a) \end{array}$