Let I = I3 be the identity matrix of order 3, and let f(x) = |A − xI|.

Question

Let I = I₃ be the identity matrix of order 3, and let f(x) = |A − xI|. (In the study of matrices, f(x) is the characteristic polynomial of A, and the zeros of f(x) are the characteristic values (eigenvalues) of A.)

A= [ 1 0 0

1 0 4

-1 1 -5]

(a) Find the polynomial f(x). 

f(x) = 

(b) Find the zeros of f(x). (Enter your answers as a comma-separated list.) 

Answer 1

Rather than x, I am used to seeing λ in the formulation.

We just put add -x to all the diagonal elements of the matrix and calculate the determinants, then find its zeroes.

1-x 0 0

1 -x 4

-1 1 -5-x

The 2 zeroes in the first row make this an easy determinant to calculate.

(1-x)[(-x)(-5-x) - 4*1)] = (1-x)(x^2+5x - 4) = x^2+5x - 4 - ( x^3+5x^2 - 4x) = -x^3 + 6x^2 + x - 4

The zeroes are 1, -5/2 + 1/2√41, -5/2 - 1/2√41

We solve this from the formula used to generate the polynomial, (1-x)(x^2+5x - 4) , as x = 1 solves the first and use the quadratic formula on the second term.