Given, A=[mpnq],∣A−d(adjA)∣=0
Here, d=∣A∣=mq−np
⇒∣A−d(adjA)∣=∣[mpnq]−d[q−p−nm]∣=0
⇒∣m−qdp(1+d)n(1+d)q−md∣=0
⇒(m−qd)(q−md)−np(1+d)2=0
⇒mq−m2d−q2d+mqd2−np(1+2d+d2)=0
⇒mq−m2d−q2d+mqd2−np−2npd−npd2=0
⇒(mq−np)+d2(mq−np)−d(m2+q2+2np)=0
⇒(mq−np)(1+d2)−d(m+q)2=0
⇒1+d2=(mq−np)d(m+q)2
⇒1+d2=(mq−np)(mq−np)(m+q)2[∵d=mq−np]
⇒(1+d)2=(m+q)2
Hence, option (A) is correct.