Find the adjoint of the matrix A = 1 2 3 - 5 and verify the result A ( adj A ) = ( adj A ) A = | A | · I .

Given that, the matrix is A = 1 2 3 - 5 Adjoint of a 2 × 2 matrix is obtained by interchanging diagonal elements and adding negative sign in front of other elements. ∴ Adj A = - 5 - 2 - 3 1 Now, A ( adj A ) = 1 2 3 - 5 - 5 - 2 - 3 1 = - 5 - 6 - 2 + 2 - 15 + 15 - 6 - 5 = - 11 0 0 - 11 Similarly, ( adj A ) A = - 5 - 2 - 3 1 1 2 3 - 5 = - 5 - 6 - 10 + 10 - 3 + 3 - 6 - 5 = - 11 0 0 - 11 Also, A · I = - 11 1 0 0 1 = - 11 0 0 - 11 Hence, A ( adj A ) = ( adj A ) A = | A | · I

Find the adjoint of the matrix A = 1 2 3 - 5 and verify the result A ( adj A ) = ( adj A ) A = | A | · I .

Given that, the matrix is A = 1 2 3 - 5 Adjoint of a 2 × 2 matrix is obtained by interchanging diagonal elements and adding negative sign in front of other elements. ∴ Adj A = - 5 - 2 - 3 1 Now, A ( adj A ) = 1 2 3 - 5 - 5 - 2 - 3 1 = - 5 - 6 - 2 + 2 - 15 + 15 - 6 - 5 = - 11 0 0 - 11 Similarly, ( adj A ) A = - 5 - 2 - 3 1 1 2 3 - 5 = - 5 - 6 - 10 + 10 - 3 + 3 - 6 - 5 = - 11 0 0 - 11 Also, A · I = - 11 1 0 0 1 = - 11 0 0 - 11 Hence, A ( adj A ) = ( adj A ) A = | A | · I

For what values of μ the system of homogeneous equations x + y + 3 z = 0 ; 4 x + 3 y + μ z = 0 ; 2 x + y + 2 z = 0 have : only trivial solution.

Given system of homogenous equations are: x + y + 3 z = 0 4 x + 3 y + μ z = 0 2 x + y + 2 z = 0 We can write, given equations as A X = B , where A = 1 1 3 4 3 μ 2 1 2 , X = x y z and B = 0 0 0 Augmented matrix A | B = 1 1 3 0 4 3 μ 0 2 1 2 0 We can have infinite solutions or a trivial solution as given equations are passing through origin. In order to have trivial solution i.e., x , y , z = 0 , 0 , 0 , we should get ρ ( A ) = ρ ( A ∣ B ) = Number of variables i.e, ρ A = ρ A | B = 3 Rank of augmented matrix is same as rank of A here, as B is null matrix. So, we can check condition directly for A . ⇒ A ≠ 0 ⇒ A = 1 1 3 4 3 μ 2 1 2 ≠ 0 ⇒ 6 - μ - 1 8 - 2 μ + 3 4 - 6 ≠ 0 ⇒ μ + 6 - 6 - 8 ≠ 0 ⇒ μ ≠ 8 So given system of homogeneous equations have only trivial solution for all real values of μ except 8 .