If a = i + j , b = j + k , c = i + k , then [( b c ) ( c a )( c a ) ( a b )( a b ) ( b c )] [ b + c c + a a + b ][ b c a a b ]

Step-by-step solution

Since, ( b c ) ( c a )=[ a b c ] c aligned & ( c a ) ( a b )=[ a b c ] a & ( a b ) ( b c )=[ a b c ] b & and [ b + c c + a a + b ]=2[ a b c ] & and [ b c ~ c a a b ]=[ a b c ]^2 & [( b c ) ( c a )( c a ) ( a b ) & So, ( a b ) ( b c )] [ b + c c + a a + b ][ b c a a b ] & = [[ a b c ] c [ a b c ] a [ a b c ] b ] 2[ a b c ][ a b ]^2 & = [ a b c ]^3[ a b c ] 2[ a b c ]^3 = [ a b c ] 2 =1 & [ a b c ]= | array lll 1 & 1 & 0 0 & 1 & 1 1 & 0 & 1 array |=2 & aligned