 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/5/c/solution

The diagonalization in part (b) gives

M^{k} = P diag (a^{k}, b^{k}, c^{k}) P^{- 1}

. Hence

M^{k} \to 0 ⟺ ∣ a ∣ < 1, ∣ b ∣ < 1, ∣ c ∣ < 1.

Sufficiency follows from the scalar powers tending to zero; necessity also follows immediately from the diagonal entries of

M^{k}

. Under these conditions the matrix geometric series converges, and its sum is

(I - M)^{- 1} = P diag (\frac{1}{1 - a}, \frac{1}{1 - b}, \frac{1}{1 - c}) P^{- 1} .

A_{0} = 1/ (1 - a)

B_{0} = 1/ (1 - b)

and

C_{0} = 1/ (1 - c)

, all entries are explicitly

(I - M)^{- 1} = A_{0} 00 B_{0} - A_{0} B_{0} 0 C_{0} - B_{0} C_{0} - B_{0} C_{0} .

For example,

B_{0} - A_{0} = (b - a) / [(1 - a) (1 - b)]

and

C_{0} - B_{0} = (c - b) / [(1 - b) (1 - c)]

. These expressions remain valid when parameter values coincide.

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