Question

If a , b and c are real numbers, and , Show that either a + b + c = 0 or a = b = c .

Answer 1

Three real numbers are a, b and c.

It is given that the determinant is 0.

Δ = | b+c c + a a + b c+a a + b b + c a + b b + c c + a | = 0

This means that either a+b+c = 0 or a = b=c.

Apply row operation R 1 → R 1 +R 2 +R 3 on the above determinant.

Here, R 1 is the first row, R 2 is the second row and R 3 is the third row.

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

Apply column operation C 2 → C 2 − C 1 on the above determinant.

= 2( a+b+c )  | 1 1−1 1 c+a a+b−c−a b+c a+b  b+c−a−b c+a | = 2( a+b+c )  | 1 0 1 c+a b−c b+c a+b  c−a c+a |

Apply column operation C 3 → C 3 − C 1 on the above determinant.

= 2( a+b+c )  | 1 0 1−1 c+a b−c b+c−c−a a+b  c−a c+a−a−b | = 2( a+b+c )  | 1 0 0 c+a b−c b−a a+b  c−a c−b |

The determinant has to be expanded along R 1 as,

= 2( a+b+c )( 1| b−c b−a  c−a  c−b |−0| c+a b−a a+b  c−b |+0| c+a b−c a+b  c−a | ) = 2( a+b+c )( 1| b−c b−a  c−a  c−b |−0+0 ) =2( a+b+c )( ( c−b )( b−c )−( b−a )( c−a ) )

Further simplify the above equation.

Δ=2( a+b+c )( −( a 2 + b 2 + c 2 )+ab+bc+ac )

The determinant value is given as 0.

Substitute the value of determinant in the above equation.

 2( a+b +c )( −( a 2 + b 2 + c 2 )+ab +bc+ac )=0

From the above equation, two parts can be written as,

2( a+b+c )=0 and −( a 2 + b 2 + c 2 )+ab+bc+ac=0

The first part can be written as ( a+b+c )=0.

Further simplify the second part.

( a 2 + b 2 + c 2 )−ab −bc −ac =0 ( 2 2 )( ( a 2 + b 2 + c 2 )−ab−bc−ac )=0 ( ( 2 a 2 +2 b 2 +2 c 2 )−2ab−2bc−2ac )=0 ( a 2 + b 2 + c 2 + a 2 + b 2 + c 2 −2ab −2bc −2ac )=0

( a−b ) 2 + ( b−c ) 2 + ( c−a ) 2 =0 a−b = 0,b−c = 0,c−a = 0 a =b,b = c,c = a a=b=c

Hence, from the first and second parts, it can be concluded that either ( a+b+c )=0 or a=b=c.