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Mathematics Question on Determinants

Let A=(mn pq),d=∣A∣≠0A= \begin{pmatrix} m & n \\\ p & q\end{pmatrix}, d=|A| \neq 0 and ∣A−d(Adj⁡A)∣=0|A-d(\operatorname{Adj} A)|=0 Then

A

1+d2=m2+q21+d^2=m^2+q^2

B

1+d2=(m+q)21+d^2=(m+q)^2

C

(1+d)2=m2+q2(1+d)^2=m^2+q^2

D

(1+d)2=(m+q)2(1+d)^2=(m+q)^2

Answer

(1+d)2=(m+q)2(1+d)^2=(m+q)^2

Explanation

Solution

The correct answer is (D) : (1+d)2=(m+q)2(1+d)^2=(m+q)^2
A=[mp​nq​],∣A−d(adjA)∣=0
⇒∣A−d(adjA)∣=∣∣​[mp​nq​]−d[q−p​−nm​]∣∣​
=∣∣​m−qdp(1+d)​n(1+d)q−md​∣∣​=0
⇒(m−qd)(q−md)−np(1+d)2=0
⇒mq−m2d−q2d+mqdd2−np(1+d)2=0
⇒(mq−np)+d2(mq−np)−d(m2+q2+2np)=0
⇒d+d3−d((m+q)2−2d)=0
⇒1+d2=(m+q)2−2d
⇒(1+d)2=(m+q)2