Engineering
Mathematics
Introduction to Determinants
Question

First row of the matrix A is [1   3   2]. If adj(A)=24a1213a52 then a possible value of det(A) is

1

2

– 1

– 2

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Solution
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|A| = a11A11 + a12A21 + a13A31

|A| = 1(– 2) + 3(– 1) + 2(3α)
|A| = – 5 + 6α
adj A=[24α1223α52]
|adj A| = – 2(– 4 + 5) – 4(2 – 3α) + α(5 – 6α)
= – 2 – 8 + 12α + 5α – 6α2
|A|2 = – 6α2 + 17α – 10
(6α – 5)2 = – 6α2 + 17α – 10
36α2 + 25 – 60α = – 6α3 + 17α – 10
42α2 – 77α + 35 = 0
2 – 11α + 5 = 0
(6α – 5) (α – 1) = 0
α=56 or α = 1
|A| = – 5 + 6α
When, α=56,|A|=0
When, α = 1, |A| = 1
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