Find the determinant of the following matrix: [ a h g h b f g f c ]

Consider | a h g h b f g f c | = a b f f c - h h f g c + g h b g f = a [ b c - f 2 ] - h [ h c - g f ] + g [ h f - g b ] = a b c - a f 2 - h 2 c + h g f + h g f - g 2 b = a b c - a f 2 - h 2 c + 2 h g f - g 2 b . Thus, the determinant of the given matrix is a b c - a f 2 - h 2 c + 2 h g f - g 2 b .

Find the determinant of the following matrix: [ a b c b c a c a b ]

Consider a b c b c a c a b = a c a a b - b b a c b + c b c c a = a ( c b - a 2 ) - b ( b 2 - c a ) + c ( b a - c 2 ) = a b c - a 3 - b 3 + a b c + a b c - c 3 = 3 a b c - a 3 - b 3 - c 3 Thus, the determinant of the given matrix is 3 a b c - a 3 - b 3 - c 3 .

Show that | b c b + c 1 c a c + a 1 a b a + b 1 | = ( a − b ) ( b − c ) ( c − a ) .

Consider, ∆ = b c b + c 1 c a a + c 1 a b a + b 1 = b c - c a b - a 0 c a a + c 1 a b a + b 1 (by a pplying R 1 → R 1 - R 2 ) = c ( b - a ) b - a 0 c a a + c 1 a b a + b 1 = ( b - a ) c 1 0 c a a + c 1 a b a + b 1 = ( b - a ) c 1 0 c a - a b c - b 0 a b a + b 1 (by a pplying R 2 → R 2 - R 3 ) = ( b - a ) c 1 0 a ( c - b ) c - b 0 a b a + b 1 = ( b - a ) ( c - b ) c 1 0 a 1 0 a b a + b 1 Expanding along C 3 , we get ∆ = ( b – a ) ( c – b ) [ 1 ( c ) ( 1 ) – ( a ) ( 1 ) ] ⇒ ∆ = ( b – a ) ( c – b ) ( c – a ) ⇒ ∆ = ( a – b ) ( b – c ) ( c – a ) Thus, we get b c b + c 1 c a a + c 1 a b a + b 1 = ( a – b ) ( b – c ) ( c – a ) .