QJEEM08Ma6a86: JEE Main 2008 Mathematics - Class 12 JEEM08Ma6a86 Solution: Let be matrix with real entries. Let be the identity matrix.…

Question

Verified Solution • 2025–26 Session moderate

Let

A

be

a\,2 \times 2

matrix with real entries. Let

I

be the

2 \times 2

identity matrix. Denote by tr

(A)

, the sum of diagonal entries of

a

. Assume that

{a^2} = I.

Statement-1 : If

A \ne I

and

A \ne - I

, then det

(A)=-1

Statement- 2 : If

A \ne I

and

A \ne - I

, then tr

(A)

\ne 0

.

A statement - 1 is false, statement -2 is true
B statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1.
C statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1.
D statement - 1 is true, statement - 2 is false.

Accepted Answer

Let

A = [\begin{matrix} a & b \\ c & d \end{matrix}]

\,\,\,

then

{A^2} = 1

\Rightarrow {a^2} + bc = 1\,\,\,\,ab + bd = 0

ac + cd = 0\,\,\,\,bc + {d^2} = 1

From these four relations,

{a^2} + bc = bc + {d^2} \Rightarrow {a^2} = {d^2}

and

\,\,b\left( {a + d} \right) = 0 = c\left( {a + d} \right)

\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = - d

We can take

a = 1,b = 0,c = 0,d = - 1

as one possible set of values, then

A = [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}]

Clearly

A \ne I\,\,\,

and

\,\,\,\,A \ne - I\,\,

and

\,\,\,A = - 1

\therefore

\,\,\,\,\,

Statement

1

is true.

Also if

A \ne I\,\,\,\,\,tr\left( A \right) = 0

\therefore

\,\,\,\,\,

Statement

2

is false.

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