Let A = bmatrix 1 & 0 & 0 0 & & 1 1 & 0 & 3 bmatrix . If A⁻¹ = A^2 + A + I and 6 + 3 + = 1 6 , then the value of 10 is equal to :

Step-by-step solution

The characteristic equation of matrix A is given by |A - xI| = 0 . vmatrix 1-x & 0 & 0 0 & -x & 1 1 & 0 & 3-x vmatrix = 0 (1-x)[( -x)(3-x) - 0] = 0 (1-x)(x^2 - ( +3)x + 3 ) = 0 x^3 - ( +4)x^2 + (4 +3)x - 3 = 0 By the Cayley-Hamilton theorem, A satisfies its own characteristic equation: A^3 - ( +4)A^2 + (4 +3)A - 3 I = O Multiplying both sides by A⁻¹ gives: A^2 - ( +4)A + (4 +3)I - 3 A⁻¹ = O A⁻¹ = 1 3 A^2 - +4 3 A + 4 +3 3 I Comparing this with A⁻¹ = A^2 + A + I , we get: = 1 3 , = - +4 3 , = 4 +3 3 Given 6 + 3 + =