JEE MATHEMATICS MCQ: MATRICES

1 Easy

1 of 1

Analyzing +4 | -1

Let

A

a 2 \times 2

matrix with real entries. Let

I

be the

2 \times 2

identity matrix. Denote bytr

(A)

, the sum of diagonal entries of

a

. Assume that

A^2=I

.

Statement-1 : If

A \neq I

and

A \neq-I

, then

\operatorname{det}(A)=-1

Statement-2 : If

A \neq I

and

A \neq-I

, then

\operatorname{tr}(A) \neq 0

View answer and explanation

Correct option: D

Statement -1 is true, Statement-2 is false

Option D (Statement -1 is true, Statement-2 is false) is correct because if A² = I and A is not I or -I, then det(A) = -1, but tr(A) can be 0.

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\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

a, b, c

a x+b y+c=0

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

2x + 3y + 1 = 0,3x + y - 2 = 0

ax + 2y - b = 0

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

\omega

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

A

2 \times 2

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

|A|