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Need RD Sharma solution for Maths Class 12 Chapter 5 Determinants Exercise Very short question Question 7 for maths textbook solution.

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Answer: 0

Hint: Here we use basic concept of determinant of  matrix

Given: \left|\begin{array}{lll} a & 1 & b+c \\ b & 1 & c+a \\ c & 1 & a+b \end{array}\right|

Solution :

Let's perform some column operations

\begin{aligned} &C_{1} \rightarrow C_{1}+C_{3} \\ &\left|\begin{array}{lll} a+b+c & 1 & b+c \\ a+b+c & 1 & c+a \\ a+b+c & 1 & a+b \end{array}\right| \end{aligned}

\rightarrow(a+b+c) take common from colomn (1)

(a+b+c)\left|\begin{array}{lll} 1 & 1 & b+c \\ 1 & 1 & c+a \\ 1 & 1 & a+b \end{array}\right|

\begin{aligned} &=(a+b+c) \times 0 \\ &=0 \end{aligned}

\therefore [Because two rows are similar it's determinant must be zero according to property of determinant]

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