Question

1
Using properties of determinant, find the value of​

Answer 1

Answer :

$\tt{\left|\begin{array}{ccc}1 & 1 & 1\\a & b & c\\b+c & c+a & a+b\end{array}\right|}$

Applying : R3 = R3 + R2

$\tt{=\left|\begin{array}{ccc}1 & 1 & 1\\a & b & c\\a+b+c & a+b+c & a+b+c\end{array}\right|}$

Taking common : (a+b+c) from R3

$\tt{=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1\\a & b & c\\1 & 1 & 1\end{array}\right|}$

If two columns or rows of a determinant are identical, then value of determinant is zero.

Hence,

$\tt{=(a+b+c)0}$

$\tt{=0}$

Answer 2

Applying : R3 = R3 + R2

$\begin{gathered}\tt{=\left|\begin{array}{ccc}1 & 1 & 1\\a & b & c\\a+b+c & a+b+c & a+b+c\end{array}\right|}\end{gathered}$

Taking common : (a+b+c) from R3

$\begin{gathered}\tt{=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1\\a & b & c\\1 & 1 & 1\end{array}\right|}\end{gathered}$

= (a+b+c)

If two columns or rows of a determinant are identical, then value of determinant is zero.

Hence,

$</p><p>\tt{=(a+b+c)0}=(a+b+c)0$

$</p><p>\tt{=0}=0$