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Question

Mathematics Question on Integration by Partial Fractions

Let
A =(12 −2−5)\begin{pmatrix}1 & 2\\\ -2 & -5 \end{pmatrix}
Let α, β ∈ R be such that αA2+βA = 2I. Then α+β is equal to

A

–10

B

–6

C

6

D

10

Answer

10

Explanation

Solution

The correct answer is (D):
A2 = [12 −2−5]\begin{bmatrix} 1&2 \\\ -2 &-5 \end{bmatrix} [12 −2−5]\begin{bmatrix} 1&2 \\\ -2 & -5 \end{bmatrix} = [−3−8 821]\begin{bmatrix} -3 &-8 \\\ 8 &21 \end{bmatrix}
αA2 + βA =[−3α8α 8α21α]\begin{bmatrix} -3α &8α \\\ 8α & 21α \end{bmatrix} [β2β −2β−5β]\begin{bmatrix} β &2β \\\ -2β & -5β \end{bmatrix}
=[−3α+β−8α+2β 8α−2β21α−5β]\begin{bmatrix} -3α+β &-8α+2β \\\ 8α-2β & 21α-5β \end{bmatrix} [20 02]\begin{bmatrix} 2 &0 \\\ 0 & 2 \end{bmatrix}
On Comparing
8α = 2β, -3α+β = 2,21α-5β = 2
⇒ α = 2,β = 8
So, α + β = 10