By using properties of determinants, show that : 1 1 1 a b c a 3 b 3 c 3 = ( a - b ) ( b - c ) ( c - a ) ( a + b + c )

Let, Δ = 1 1 1 a b c a 3 b 3 c 3 Applying C 2 → C 2 - C 3 , C 3 → C 3 - C 1 , = 1 0 0 a b - c c - a a 3 b 3 - c 3 c 3 - a 3 = b - c c - a 1 0 0 a 1 1 a 3 b 2 + b c + c 2 c 2 + c a + a 2 ∵ x 3 - y 3 = ( x - y ) ( x 2 + x y + y 2 ) = ( b - c ) ( c - a ) 1 c 2 + c a + a 2 - 1 b 2 + b c + c 2 - 0 + 0 [expanding along R 1 ] = ( b - c ) ( c - a ) c a + a 2 - b 2 - b c = ( b - c ) ( c - a ) c a - b + a + b a - b = ( b - c ) ( c - a ) ( a - b ) ( a + b + c ) Hence, proved.

Verify A adj A = A I for the matrix A = 2 0 - 1 5 1 0 0 1 3 .

Given that A = 2 0 - 1 5 1 0 0 1 3 Cofactors of the above matrix are: C 11 = - 1 1 + 1 1 0 1 3 = 3 - 0 = 3 C 12 = - 1 1 + 2 5 0 0 3 = - 15 - 0 = - 15 C 13 = - 1 1 + 3 5 1 0 1 = 5 - 0 = 5 C 21 = - 1 2 + 1 0 - 1 1 3 = - 0 + 1 = - 1 C 22 = - 1 2 + 2 2 - 1 0 3 = 6 - 0 = 6 C 23 = - 1 2 + 3 2 0 0 1 = - 2 - 0 = - 2 C 31 = - 1 3 + 1 0 - 1 1 0 = 0 + 1 = 1 C 32 = - 1 3 + 2 2 - 1 5 0 = - 0 + 5 = - 5 C 33 = - 1 3 + 3 2 0 5 1 = 2 - 0 = 2 adj A = 3 - 15 5 - 1 6 - 2 1 - 5 2 T = 3 - 1 1 - 15 6 - 5 5 - 2 2 Now, A adj A = 2 0 - 1 5 1 0 0 1 3 3 - 1 1 - 15 6 - 5 5 - 2 2 = 6 - 0 - 5 - 2 + 0 + 2 2 + 2 - 2 15 - 15 + 0 - 5 + 6 + 0 5 - 5 + 0 0 - 15 + 15 0 + 6 - 6 0 - 5 + 6 = 1 0 0 0 1 0 0 0 1 Now, A = 2 0 - 1 5 1 0 0 1 3 = 2 3 - 0 - 0 - 1 5 - 0 = 6 - 5 = 1 Now, A I = 1 1 0 0 0 1 0 0 0 1 = 1 0 0 0 1 0 0 0 1 Hence, proved.

Determine the inverse of the matrix A = 2 0 − 1 5 1 0 0 1 3 using elementary operations.

We have, A = I A or 2 0 − 1 5 1 0 0 1 3 = 1 0 0 0 1 0 0 0 1 A Applying R 1 → R 1 2 ⇒ 1 0 − 1 2 5 1 0 0 1 3 = 1 2 0 0 0 1 0 0 0 1 A Applying R 2 → R 2 - 5 R 1 ⇒ 1 0 − 1 2 0 1 5 2 0 1 3 = 1 2 0 0 - 5 2 1 0 0 0 1 A Applying R 3 → R 3 - R 2 ⇒ 1 0 − 1 2 0 1 5 2 0 0 1 2 = 1 2 0 0 - 5 2 1 0 5 2 - 1 1 A Applying R 3 → 2 R 3 ⇒ 1 0 − 1 2 0 1 5 2 0 0 1 = 1 2 0 0 - 5 2 1 0 5 - 2 2 A Applying R 1 → R 1 + 1 2 R 3 ⇒ 1 0 0 0 1 5 2 0 0 1 = 3 - 1 1 - 5 2 1 0 5 - 2 2 A Applying R 2 → R 2 - 5 2 R 3 ⇒ 1 0 0 0 1 0 0 0 1 = 3 - 1 1 - 15 6 - 5 5 - 2 2 A Hence, A - 1 = 3 − 1 1 − 15 6 − 5 5 − 2 2

If a ≠ 0 , then solve the equation x + a x x x x + a x x x x + a = 0

Let △ = x + a x x x x + a x x x x + a Applying C 1 → C 1 + C 2 + C 3 , we get: ⇒ △ = 3 x + a x x 3 x + a x + a x 3 x + a x x + a Taking 3 x + a common from C 1 , we get: ⇒ △ = 3 x + a 1 x x 1 x + a x 1 x x + a Applying R 2 → R 2 − R 1 , R 3 → R 3 − R 1 , we get: ⇒ Δ = ( 3 x + a ) 1 x x 0 a 0 0 0 a ⇒ Δ = a 2 ( 3 x + a ) For solving the equation putting △ = 0 ⇒ a 2 ( 3 x + a ) = 0 Now, a ≠ 0 ⇒ a 2 ≠ 0 ⇒ 3 x + a = 0 ⇒ x = - a 3

Find the area of the triangle whose vertices are ( 2 , 7 ) , ( 1 , 1 ) and ( 10 , 8 ) .

Given ( 2 , 7 ) , ( 1 , 1 ) and ( 10 , 8 ) are the vertices of the triangle. We know that if vertices of a triangle are x 1 , y 1 , x 2 , y 2 and x 3 , y 3 , then the area of the triangle is given by: Δ = 1 2 x 1 y 1 1 x 2 y 2 1 x 3 y 3 1 Now, substituting given value in above formula ⇒ Δ = 1 2 2 7 1 1 1 1 10 8 1 Expanding along R 1 ⇒ Δ = 1 2 2 1 1 8 1 - 7 1 1 10 1 + 1 1 1 10 8 ⇒ Δ = 1 2 [ 2 ( - 7 ) - 7 ( - 9 ) + 1 ( - 2 ) ] ⇒ Δ = 1 2 [ - 14 + 63 - 2 ] ⇒ Δ = 47 2 square units Thus, area of triangle is 47 2 square units .

If A = 1 − 1 2 3 0 − 2 1 0 3 , then verify that A adj A = A I . Also find A - 1 .

We have, A = 1 − 1 2 3 0 − 2 1 0 3 Now, A = 1 - 1 2 3 0 - 2 1 0 3 ⇒ A = 1 0 + 1 9 + 2 - 2 0 ⇒ A = 11 ≠ 0 Let C i j be the co-factors of elements a i j in A = a i j . Then, C 11 = - 1 1 + 1 0 - 2 0 3 = 0 C 12 = - 1 1 + 2 3 - 2 1 3 = - 9 + 2 = - 11 C 13 = - 1 1 + 3 3 0 1 0 = 0 C 21 = - 1 2 + 1 - 1 2 0 3 = - - 3 = 3 C 22 = - 1 2 + 2 1 2 1 3 = 3 - 2 = 1 C 23 = - 1 2 + 3 1 - 1 1 0 = - 1 = - 1 C 31 = - 1 3 + 1 - 1 2 0 - 2 = 2 C 32 = - 1 3 + 2 1 2 3 - 2 = - - 2 - 6 = 8 C 33 = - 1 3 + 3 1 - 1 3 0 = 3 Thus, adj A = 0 - 11 0 3 1 - 1 2 8 3 T = 0 3 2 - 11 1 8 0 - 1 3 Consider, A adj A = 1 − 1 2 3 0 − 2 1 0 3 0 3 2 − 11 1 8 0 − 1 3 ⇒ A adj A = 1 ⋅ 0 + ( − 1 ) ( − 11 ) + 2 ⋅ 0 1 ⋅ 3 + ( − 1 ) ⋅ 1 + 2 ( − 1 ) 1 ⋅ 2 + ( − 1 ) ⋅ 8 + 2 ⋅ 3 3 ⋅ 0 + 0 ⋅ ( − 11 ) + ( − 2 ) ⋅ 0 3 ⋅ 3 + 0 ⋅ 1 + ( − 2 ) ( − 1 ) 3 ⋅ 2 + 0 ⋅ 8 + ( − 2 ) ⋅ 3 1 ⋅ 0 + 0 ⋅ ( − 11 ) + 3 ⋅ 0 1 ⋅ 3 + 0 ⋅ 1 + 3 ( − 1 ) 1 ⋅ 2 + 0 ⋅ 8 + 3 ⋅ 3 ⇒ A ( adj ⁡ A ) = 11 0 0 0 11 0 0 0 11 ⇒ A ( adj ⁡ A ) = 11 1 0 0 0 1 0 0 0 1 ⇒ A ( adj ⁡ A ) = 11 I ⇒ A ( adj ⁡ A ) = A I Hence, proved. Thus, A - 1 = 1 A adj A = 1 11 0 3 2 − 11 1 8 0 − 1 3

Find the inverse of the matrix 4 5 2 3 by using elementary operations.

Given A = 4 5 2 3 We know that A = I A ⇒ 4 5 2 3 = 1 0 0 1 A Applying R 1 → R 1 4 ⇒ 1 5 4 2 3 = 1 4 0 0 1 A Applying R 2 → R 2 - 2 R 1 ⇒ 1 5 4 0 1 2 = 1 4 0 - 1 2 1 A Applying R 2 → 2 R 2 ⇒ 1 5 4 0 1 = 1 4 0 - 1 2 A Applying R 1 → R 1 - 5 4 R 2 ⇒ 1 0 0 1 = 3 2 - 5 2 - 1 2 A ⇒ I = A - 1 A Hence, A - 1 = 3 2 - 5 2 - 1 2

Find the values of x for which 3 x x 1 = 3 2 4 1

Given that both the determinants are equal, i.e. 3 x x 1 = 3 2 4 1 ⇒ 3 - x 2 = 3 - 8 ∵ a b c d = a d - b c ⇒ x 2 = 8 ⇒ x = ± 2 2 which are the required values of x .

Evaluate determinant: x y x + y y x + y x x + y x y .

Let the given determinant be ∆ = x y x + y y x + y x x + y x y Operating R 1 → R 1 + R 2 + R 3 , then we get: ∆ = 2 x + 2 y 2 x + 2 y 2 x + 2 y y x + y x x + y x y ⇒ ∆ = 2 x + 2 y 1 1 1 y x + y x x + y x y [Taking 2 x + 2 y From R 1 ] Operating C 2 → C 2 - C 1 and C 3 → C 3 - C 1 , then we get: ∆ = 2 x + y 1 0 0 y x x - y x + y - y - x Expanding along row one, we obtain as ∆ = 2 x + y 1 - x × x - - y × x - y - 0 + 0 ⇒ ∆ = 2 x + y - x 2 + y x - y ⇒ ∆ = 2 x + y - x 2 + y x - y 2 ⇒ ∆ = - 2 x + y x 2 - x y + y 2 ⇒ ∆ = - 2 x 3 - y 3 ⇒ ∆ = 2 y 3 - x 3 Hence, x y x + y y x + y x x + y x y = 2 y 3 - x 3 .

By using properties of determinants, show that : 1 1 1 a b c a 3 b 3 c 3 = ( a - b ) ( b - c ) ( c - a ) ( a + b + c )

Let, Δ = 1 1 1 a b c a 3 b 3 c 3 Applying C 2 → C 2 - C 3 , C 3 → C 3 - C 1 , = 1 0 0 a b - c c - a a 3 b 3 - c 3 c 3 - a 3 = b - c c - a 1 0 0 a 1 1 a 3 b 2 + b c + c 2 c 2 + c a + a 2 ∵ x 3 - y 3 = ( x - y ) ( x 2 + x y + y 2 ) = ( b - c ) ( c - a ) 1 c 2 + c a + a 2 - 1 b 2 + b c + c 2 - 0 + 0 [expanding along R 1 ] = ( b - c ) ( c - a ) c a + a 2 - b 2 - b c = ( b - c ) ( c - a ) c a - b + a + b a - b = ( b - c ) ( c - a ) ( a - b ) ( a + b + c ) Hence, proved.