MathematicsMedium244×since 2002Q4538Let A=(mnpq),d=∣A∣≠0A=\left(\begin{array}{cc}\mathrm{m} & \mathrm{n} \\ \mathrm{p} & \mathrm{q}\end{array}\right), \mathrm{d}=|\mathrm{A}| \neq 0A=(mpnq),d=∣A∣=0 and ∣A−d(AdjA)∣=0\mathrm{|A-d(A d j A)|=0}∣A−d(AdjA)∣=0. ThenA1+d2=m2+q21+\mathrm{d}^{2}=\mathrm{m}^{2}+\mathrm{q}^{2}1+d2=m2+q2B1+d2=(m+q)21+d^{2}=(m+q)^{2}1+d2=(m+q)2C(1+d)2=m2+q2(1+d)^{2}=m^{2}+q^{2}(1+d)2=m2+q2D(1+d)2=(m+q)2(1+d)^{2}=(m+q)^{2}(1+d)2=(m+q)2Check answerSkip