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Joint Entrance Examination
Mathematics
Matrices and Determinants
Question 310
Question 310 of 472
Easy
MCQ
If
A
=
(
a
b
b
a
)
A=\begin{pmatrix}a&b\\b&a\end{pmatrix}
A
=
(
a
b
b
a
)
and
A
2
=
(
α
β
β
α
)
A^2=\begin{pmatrix}\alpha&\beta\\\beta&\alpha\end{pmatrix}
A
2
=
(
α
β
β
α
)
, then:
A
α
=
2
a
b
,
β
=
a
2
+
b
2
\alpha=2ab,\ \beta=a^2+b^2
α
=
2
ab
,
β
=
a
2
+
b
2
B
α
=
a
2
+
b
2
,
β
=
a
b
\alpha=a^2+b^2,\ \beta=ab
α
=
a
2
+
b
2
,
β
=
ab
C
α
=
a
2
+
b
2
,
β
=
2
a
b
\alpha=a^2+b^2,\ \beta=2ab
α
=
a
2
+
b
2
,
β
=
2
ab
D
α
=
a
2
+
b
2
,
β
=
a
2
−
b
2
\alpha=a^2+b^2,\ \beta=a^2-b^2
α
=
a
2
+
b
2
,
β
=
a
2
−
b
2
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Solutions · 1
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Official solution
\[A^2=
(
a
b
b
a
)
\begin{pmatrix}a&b\\b&a\end{pmatrix}
(
a
b
b
a
)
(
a
b
b
a
)
\begin{pmatrix}a&b\\b&a\end{pmatrix}
(
a
b
b
a
)
=
(
a
2
+
b
2
2
a
b
2
a
b
a
2
+
b
2
)
\begin{pmatrix}a^2+b^2&2ab\\2ab&a^2+b^2\end{pmatrix}
(
a
2
+
b
2
2
ab
2
ab
a
2
+
b
2
)
.\] So
α
=
a
2
+
b
2
,
β
=
2
a
b
\alpha=a^2+b^2,\ \beta=2ab
α
=
a
2
+
b
2
,
β
=
2
ab
.
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