Prove the following identities – |(a+b+c,-c,-b)(-c,a+b+c,-c)(-b,-a,a+b+c)| = 2(a + b)(b + c)(c + a) u200b

Question

Prove the following identities –

(\begin{vmatrix} a+b+c& -c & -b \[0.3em] -c& a+b+c & -c \[0.3em] -b &-a &a+b+c \end{vmatrix})=2(a + b)(b + c)(c + a)

Answer 1

Let Δ = (\begin{vmatrix}
a+b+c& -c & -b \[0.3em]
-c& a+b+c & -c \[0.3em]
-b &-a &a+b+c
\end{vmatrix})

Recall that the value of a determinant *** same if we apply the operation Ri→ Ri + kRj or Ci→ Ci + kCj.

Applying R1→ R1 + R2, we get

Taking (a + b) and (c + a) common from C2 and C3, we get

Expanding the determinant along C3, we have

Δ = (a + b)(c + a)[(–c + a)(–2) – (–b – a)(2)]

⇒ Δ = (a + b)(c + a)[2c – 2a + 2a + 2b]

⇒ Δ = (a + b)(c + a)(2b + 2c)

∴ Δ = 2(a + b)(b + c)(c + a)

Thus,

(\begin{vmatrix}
a+b+c& -c & -b \[0.3em]
-c& a+b+c & -c \[0.3em]
-b &-a &a+b+c
\end{vmatrix})

= 2(a + b) (b + c) (c + a)