Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

The value of $\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|$ is

Options:

0

$(a+b+c)$

$(a-b)(b-c)(c-a)$

$a^2+b^2+c^2$

Correct Answer:

0

Explanation:

It $\Delta=\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & a+c \\ 1 & c & a+b\end{array}\right|$

⇒  $c_3 \rightarrow c_3+c_2 \quad \Delta=\left|\begin{array}{lll}1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c\end{array}\right|$

taking (a + b + c) common factor out

$\Rightarrow(a+b+c)\left|\begin{array}{lll}1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1\end{array}\right|=(a+b+c) \times 0=0$

(two columns have same values here value is zero)