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Let $$A=\left(\begin{array}{cc}\mathrm{m} & \mathrm{n} \\ \mathrm{p} & \mathrm{q}\end{array}\right), \mathrm{d}=|\mathrm{A}| \neq 0$$ and $$\mathrm{|A-d(A d j A)|=0}$$. Then

JEE · Math · previous-year question

  1. A.$$1+\mathrm{d}^{2}=\mathrm{m}^{2}+\mathrm{q}^{2}$$
  2. B.$$1+d^{2}=(m+q)^{2}$$
  3. C.$$(1+d)^{2}=m^{2}+q^{2}$$
  4. D.$$(1+d)^{2}=(m+q)^{2}$$correct

Answer

D. $$(1+d)^{2}=(m+q)^{2}$$

Explanation

$$\left| {A - d\left( {\begin{matrix} q & { - n} \\ { - p} & m \\ \end{matrix} } \right)} \right| = 0$$ $$\left| {\begin{matrix} {m - qd} & {n(1 + d)} \\ {p(1 + d)} & {q - md} \\ \end{matrix} } \right| = 0$$ $$(m - qd)(q - md) = np{(1 + d)^2}$$ $$mq - ({q^2} + {m^2})d + qm{d^2} = np(1 + {d^2}) + 2npd$$ $${d^2}(mq - np) + 1(mq - np) = (2np + {m^2} + {q^2})d$$ $$({d^2} + 1)(mq - np) = (2np + m + a)d$$ $${d^2} + 1 = 2np + {m^2} + {q^2}$$ $$2d = 2mq - 2np$$ $$ \Rightarrow {(1 + d)^2} = {(m + q)^2}$$

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