The value of determinant $\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$ is :

0

The correct answer is Option 4: 0

$P = \left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$

so $c_1 \rightarrow c_1+c_2+c_3$

$P=\left|\begin{array}{lll}
a-b+b-c+c-a & b-c & c-a \\
b-c+c-a+a-b & c-a & a-b \\
c-a+a-b+b-c & a-b & b-c
\end{array}\right|$

$=\left|\begin{array}{ccc}0 & b-c & c-a \\ 0 & c-a & a-b \\ 0 & a-b & b-c\end{array}\right|=0$

$P=0$

Alternatively,

In the given determinant,

• Row 1 = (a−b,  b−c,  c−a)
• Row 2 = (b−c,  c−a,  a−b)
• Row 3 = (c−a,  a−b,  b−c)

If we add the elements of any row: (a−b)+(b−c)+(c−a)=0

Similarly, the sum of elements of every row is 0.

Therefore, all rows are linearly dependent. A determinant whose rows (or columns) are linearly dependent has value 0.

Hence, the determinant is equal to 0.