Find all the positive values of 2 × 2 determinants whose entries are from the set { - 1 , 0 , 1 } .

Given, { - 1 , 0 , 1 } For positive values, determinant > 0 ⇒ a b c d > 0 ⇒ a d - b c > 0 ⇒ a d > b c Case (i): a d = 1 and b c = 0 or b c = - 1 Case (ii): a d = 0 & b c = - 1 Therefore, possible positive values of determinant are 1 , 2 Therefore, total positive values of 2 × 2 determinants whose entries are from the given set are 2 .

Find all the positive values of 2 × 2 determinants whose entries are from the set { - 1 , 0 , 1 } .

Given, { - 1 , 0 , 1 } For positive values, determinant > 0 ⇒ a b c d > 0 ⇒ a d - b c > 0 ⇒ a d > b c Case (i): a d = 1 and b c = 0 or b c = - 1 Case (ii): a d = 0 & b c = - 1 Therefore, possible positive values of determinant are 1 , 2 Therefore, total positive values of 2 × 2 determinants whose entries are from the given set are 2 .

Prove that a - b - c 2 a 2 a 2 b b - c - a 2 b 2 c 2 c c - a - b = a + b + c 3 .

Let LHS = a - b - c 2 a 2 a 2 b b - c - a 2 b 2 c 2 c c - a - b Operating R 1 → R 1 + ( R 2 + R 3 ) , we get = a + b + c a + b + c a + b + c 2 b b - c - a 2 b 2 c 2 c c - a - b Taking out ( a + b + c ) common from R 1 , we get = a + b + c 1 1 1 2 b b - c - a 2 b 2 c 2 c c - a - b Operating C 2 → C 2 – C 1 C 3 → C 3 – C 1 , we get = a + b + c 1 0 0 2 b - b - c - a 0 2 c 0 - c - a - b Expanding in terms of R 1 , we get = ( a + b + c ) [ ( – a – b – c ) 2 ] = ( a + b + c ) 3 =RHS Hence proved

Without expanding show that a x + a - x 2 a x - a - x 2 2 b x + b - x 2 b x - b - x 2 2 c x + c - x 2 c x - c - x 2 2 = 0 .

LHS = a x + a - x 2 a x - a - x 2 2 b x + b - x 2 b x - b - x 2 2 c x + c - x 2 c x - c - x 2 2 Applying C 1 → C 1 - C 2 , we get = 4 a x - a - x 2 2 4 b x - b - x 2 2 4 c x - c - x 2 2 [Using a + b 2 - a - b 2 = 4 a b ] = 4 × 2 1 a x - a - x 2 1 1 b x - b - x 2 1 1 c x - c - x 2 1 = 8 × 0 = 0 ∵ C 1 and C 3 are identical = RHS Hence, proved.

Prove that a 2 + 1 a b a c a b b 2 + 1 b c c a c b c 2 + 1 = 1 + a 2 + b 2 + c 2

Let L.H.S= ∆ = a 2 + 1 a b a c a b b 2 + 1 b c a c b 2 c 2 + 1 Applying R 1 a , R 2 b , R 3 c we get, = a + 1 a b c a b + 1 b c a b c + 1 c Applying a C 1 , b C 2 , C C 3 we get, = a 2 + 1 b 2 c 2 a 2 b 2 + 1 c 2 a 2 b 2 c 2 + 1 Applying C 1 → C 1 + C 2 + C 3 we get, = 1 + a 2 + b 2 + c 2 b 2 c 2 1 + a 2 + b 2 + c 2 b 2 + 1 c 2 1 + a 2 + b 2 + c 2 b 2 c 2 + 1 Taking 1 + a 2 + b 2 + c 2 common = 1 + a 2 + b 2 + c 2 1 b 2 c 2 1 b 2 + 1 c 2 1 b 2 c 2 + 1 Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 we get ∆ = 1 + a 2 + b 2 + c 2 1 b 2 c 2 0 1 0 0 0 1 = 1 + a 2 + b 2 + c 2 1 0 0 1 = 1 + a 2 + b 2 + c 2 =R.H.S Hence proved.

Using matrix method solve the following system of linear equations : x - y + z = 4 , 2 x + y - 3 z = 0 & x + y + z = 2

Given that, x - y + z = 4 , 2 x + y - 3 z = 0 & x + y + z = 2 . This can be written as A X = B where A = 1 - 1 1 2 1 - 3 1 1 1 , X = x y z , B = 4 0 2 Here, | A | = 1 ( 1 + 3 ) + 1 ( 2 + 3 ) + 1 ( 2 - 1 ) = 10 Since, | A | ≠ 0 Hence, the system of equations is consistent and has a unique solution given by X = A - 1 B A - 1 = adj A | A | and adj A = C T C 11 = ( - 1 ) 1 + 1 1 - 3 1 1 = 1 + 3 = 4 C 12 = ( - 1 ) 1 + 2 2 - 3 1 1 = - 2 + 3 = - 5 C 13 = ( - 1 ) 1 + 3 2 1 1 1 = 2 - 1 = 1 C 21 = ( - 1 ) 2 + 1 - 1 1 1 1 = 2 C 22 = ( - 1 ) 2 + 2 1 1 1 1 = 0 C 23 = ( - 1 ) 2 + 3 1 - 1 1 1 = - 2 C 31 = ( - 1 ) 3 + 1 - 1 1 1 - 3 = 2 C 32 = ( - 1 ) 3 + 2 1 1 2 - 3 = 5 C 33 = ( - 1 ) 3 + 3 1 - 1 2 1 = 3 Therefore, cofactor matrix is C = 4 - 5 1 2 0 - 2 2 5 3 ⇒ adj A = C T = 4 2 2 - 5 0 5 1 - 2 3 ⇒ A - 1 = adj A | A | = 1 10 4 2 2 - 5 0 5 1 - 2 3 Solution is given by X = A - 1 B x y z = 1 10 4 2 2 - 5 0 5 1 - 2 3 4 0 2 = 1 10 16 + 4 - 20 + 10 4 + 6 = 1 10 20 - 10 10 = 2 - 1 1 Therefore, x = 2 , y = - 1 , z = 1

Prove that a 2 + 1 a b a c a b b 2 + 1 b c c a c b c 2 + 1 = 1 + a 2 + b 2 + c 2

Let L.H.S= ∆ = a 2 + 1 a b a c a b b 2 + 1 b c a c b 2 c 2 + 1 Applying R 1 a , R 2 b , R 3 c we get, = a + 1 a b c a b + 1 b c a b c + 1 c Applying a C 1 , b C 2 , C C 3 we get, = a 2 + 1 b 2 c 2 a 2 b 2 + 1 c 2 a 2 b 2 c 2 + 1 Applying C 1 → C 1 + C 2 + C 3 we get, = 1 + a 2 + b 2 + c 2 b 2 c 2 1 + a 2 + b 2 + c 2 b 2 + 1 c 2 1 + a 2 + b 2 + c 2 b 2 c 2 + 1 Taking 1 + a 2 + b 2 + c 2 common = 1 + a 2 + b 2 + c 2 1 b 2 c 2 1 b 2 + 1 c 2 1 b 2 c 2 + 1 Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 we get ∆ = 1 + a 2 + b 2 + c 2 1 b 2 c 2 0 1 0 0 0 1 = 1 + a 2 + b 2 + c 2 1 0 0 1 = 1 + a 2 + b 2 + c 2 =R.H.S Hence proved.

Without expanding show that a x + a - x 2 a x - a - x 2 2 b x + b - x 2 b x - b - x 2 2 c x + c - x 2 c x - c - x 2 2 = 0 .

LHS = a x + a - x 2 a x - a - x 2 2 b x + b - x 2 b x - b - x 2 2 c x + c - x 2 c x - c - x 2 2 Applying C 1 → C 1 - C 2 , we get = 4 a x - a - x 2 2 4 b x - b - x 2 2 4 c x - c - x 2 2 [Using a + b 2 - a - b 2 = 4 a b ] = 4 × 2 1 a x - a - x 2 1 1 b x - b - x 2 1 1 c x - c - x 2 1 = 8 × 0 = 0 ∵ C 1 and C 3 are identical = RHS Hence, proved.

Prove that a - b - c 2 a 2 a 2 b b - c - a 2 b 2 c 2 c c - a - b = a + b + c 3 .

Let LHS = a - b - c 2 a 2 a 2 b b - c - a 2 b 2 c 2 c c - a - b Operating R 1 → R 1 + ( R 2 + R 3 ) , we get = a + b + c a + b + c a + b + c 2 b b - c - a 2 b 2 c 2 c c - a - b Taking out ( a + b + c ) common from R 1 , we get = a + b + c 1 1 1 2 b b - c - a 2 b 2 c 2 c c - a - b Operating C 2 → C 2 – C 1 C 3 → C 3 – C 1 , we get = a + b + c 1 0 0 2 b - b - c - a 0 2 c 0 - c - a - b Expanding in terms of R 1 , we get = ( a + b + c ) [ ( – a – b – c ) 2 ] = ( a + b + c ) 3 =RHS Hence proved

Show that x x 2 y + z y y 2 z + x z z 2 x + y = ( y - z ) ( z - x ) ( x - y ) ( x + y + z )

LHS = x x 2 y + z y y 2 x + z z z 2 x + y Applying C 2 = C 2 - x C 1 = x 0 y + z y y ( - x + y ) x + z z z ( - x + z ) x + y Applying C 3 = C 3 - y + z x C 1 = x 0 0 y y ( - x + y ) ( x - y ) ( x + y + z ) x z z ( - x + z ) ( x - z ) ( x + y + z ) x Expanding along row 1 , = ( y ( - x + y ) ) · ( x - z ) ( x + y + z ) x - ( x - y ) ( x + y + z ) x · ( z ( - x + z ) ) = - x 2 y + x 2 z + y 3 - z 3 - y 3 z x + y z 3 x = - ( x - y ) ( x - z ) ( y - z ) ( x + y + z ) = ( y - z ) ( z - x ) ( x - y ) ( x + y + z ) = RHS Hence, showed.