 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2016-resit/7/solution

For a diagonal matrix, each diagonal entry separately obeys

d^{2} + 2 d - 15 = (d - 3) (d + 5) = 0

. Hence all four possibilities are

diag (3, 3), diag (3, - 5), diag (- 5, 3), diag (- 5, - 5) .

For the second form,

X^{2} = I

gives

(a I + b X)^{2} + 2 (a I + b X) = (a^{2} + b^{2} + 2 a) I + 2 b (a + 1) X .

Because

I, X

are linearly independent and

b \neq = 0

, the

X

a = - 1

; the

I

coefficient then gives

b^{2} = 16

. Thus

A = - I \pm 4 X = (- 1 \pm 4 \pm 4 - 1) .

Both constants are nonzero, as required.

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