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Joint Entrance Examination
Mathematics
Matrices and Determinants
Question 434
Question 434 of 472
Hard
MCQ
Let \[A=\left(\[
m
n
p
q
\begin{array}{|ccc|}\mathrm{m} & \mathrm{n} \\ \mathrm{p} & \mathrm{q}\end{array}
m
p
n
q
\]\right), \mathrm{d}=|\mathrm{A}| \neq 0\] and \[\mathrm{|A-d(A d j A)|=0}\]. Then
A
\[1+\mathrm{d}^{2}=\mathrm{m}^{2}+\mathrm{q}^{2}\]
B
\[1+d^{2}=(m+q)^{2}\]
C
\[(1+d)^{2}=m^{2}+q^{2}\]
D
\[(1+d)^{2}=(m+q)^{2}\]
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Official solution
\[\left| {A - d\left( {
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−
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m
)
\begin{pmatrix} q & { - n} \\ { - p} & m \\ \end{pmatrix}
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)
} \right)} \right| = 0\] \[\left| {
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p
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)
\begin{pmatrix} {m - qd} & {n(1 + d)} \\ {p(1 + d)} & {q - md} \\ \end{pmatrix}
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−
q
d
p
(
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n
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} \right| = 0\]
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=
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(m - qd)(q - md) = np{(1 + d)^2}
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\[mq - ({q^2} + {m^2})d + qm{d^2} = np(1 + {d^2}) + 2npd\] \[{d^2}(mq - np) + 1(mq - np) = (2np + {m^2} + {q^2})d\]
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=
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d
({d^2} + 1)(mq - np) = (2np + m + a)d
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=
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{d^2} + 1 = 2np + {m^2} + {q^2}
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=
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=
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2d = 2mq - 2np
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d
=
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n
p
\[ \Rightarrow {(1 + d)^2} = {(m + q)^2}\]
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