Prove the following identities :\(\begin{vmatrix} b+c & a-b& a \\[0.3e

1.	Prove the following identities :\(\begin{vmatrix} b+c & a-b& a \\[0.3em] c+a & b-c & b \\[0.3em] a+b & c-a & c \end{vmatrix}\) = 3abc - a3 - b3 - c3
Answer» L.H.S = \(\begin{vmatrix} b+c & a-b& a \\[0.3em] c+a & b-c & b \\[0.3em] a+b & c-a & c \end{vmatrix}\) As, \|A\| = \|A\|^T So, \(\begin{vmatrix} b+c & c + a& a+b \\[0.3em] a-b & b-c & c-a \\[0.3em] a &b & c \end{vmatrix}\) If any two rows or columns of the determinant are interchanged, then determinant changes its sign \(-\begin{vmatrix}a &b & c \\[0.3em] a-b & b-c & c-a \\[0.3em]b+c & c + a& a+b \end{vmatrix}\) Apply C₁→C₁ + C₂ + C₃ \(-\begin{vmatrix}a+b+c & b& c \\[0.3em] 0 & b-c & c-a \\[0.3em] 2(a+b+c) &c+a & a+b \end{vmatrix}\) Taking (a + b + c) common from C₁ we get, = - (a + b + c)\(\begin{vmatrix}1 & b& c \\[0.3em] 0 & b-c & c-a \\[0.3em] 2&c+a & a+b \end{vmatrix}\) Applying, R₃→R₃ – 2R₁ = - (a + b + c)\(\begin{vmatrix}1 & b& c \\[0.3em] 0 & b-c & c-a \\[0.3em] 0&c+a-2b & a+b-2c \end{vmatrix}\) = – (a + b + c)[(b – c)(a + b – 2c) – (c – a)(c + a – 2b)] = 3abc – a³ – b³– c³ As, L.H.S = R.H.S, hence proved.

Prove the following identities :\(\begin{vmatrix} b+c & a-b& a \\[0.3em] c+a & b-c & b \\[0.3em] a+b & c-a & c \end{vmatrix}\) = 3abc - a3 - b3 - c3

Answer»

L.H.S = \(\begin{vmatrix} b+c & a-b& a \\[0.3em] c+a & b-c & b \\[0.3em] a+b & c-a & c \end{vmatrix}\)

As,

|A| = |A|^T

So,

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

If any two rows or columns of the determinant are interchanged, then determinant changes its sign

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

Apply C₁→C₁ + C₂ + C₃

\(-\begin{vmatrix}a+b+c & b& c \\[0.3em] 0 & b-c & c-a \\[0.3em] 2(a+b+c) &c+a & a+b \end{vmatrix}\)

Taking (a + b + c) common from C₁ we get,

= - (a + b + c)\(\begin{vmatrix}1 & b& c \\[0.3em] 0 & b-c & c-a \\[0.3em] 2&c+a & a+b \end{vmatrix}\)

Applying,

R₃→R₃ – 2R₁

= - (a + b + c)\(\begin{vmatrix}1 & b& c \\[0.3em] 0 & b-c & c-a \\[0.3em] 0&c+a-2b & a+b-2c \end{vmatrix}\)

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

= 3abc – a³ – b³– c³

As,

L.H.S = R.H.S,

hence proved.