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Matrices and Determinants question

2019 · 12 Jan · Shift 2 · Q38
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  5. /2019 · 12 Jan · Shift 2 · Q38

Matrices and Determinants question

2019 · 12 Jan · Shift 2 · Q38

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If A = [1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1]\left[ {\begin{matrix} 1 & {\sin \theta } & 1 \\ { - \sin \theta } & 1 & {\sin \theta } \\ { - 1} & { - \sin \theta } & 1 \\ \end{matrix} } \right]​1−sinθ−1​sinθ1−sinθ​1sinθ1​​; then for all θ∈\theta \inθ∈ (3π4,5π4)\left( {{{3\pi } \over 4},{{5\pi } \over 4}} \right)(43π​,45π​), det (A) lies in the interval :
  1. A
    (32,3]\left( {{3 \over 2},3} \right](23​,3]
  2. B
    (0,32]\left( {0,{3 \over 2}} \right](0,23​]
  3. C
    [52,4)\left[ {{5 \over 2},4} \right)[25​,4)
  4. D
    (1,52]\left( {1,{5 \over 2}} \right](1,25​]
View written solutionFree

Correct answer: A

  1. Write the matrix and simplify notation

Let

A=[1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1].A=\begin{bmatrix} 1 & \sin\theta & 1\\ -\sin\theta & 1 & \sin\theta\\ -1 & -\sin\theta & 1 \end{bmatrix}.A=​1−sinθ−1​sinθ1−sinθ​1sinθ1​​.

Set s=sin⁡θ.s=\sin\theta.s=sinθ. Then

A=[1s1−s1s−1−s1].A=\begin{bmatrix} 1 & s & 1\\ -s & 1 & s\\ -1 & -s & 1 \end{bmatrix}.A=​1−s−1​s1−s​1s1​​.

We need the range of det⁡(A)\det(A)det(A) for θ∈(3π4,5π4).\theta\in\left(\frac{3\pi}{4},\frac{5\pi}{4}\right).θ∈(43π​,45π​).

  1. Compute the determinant

Expand along the first row:

det⁡(A)=1∣1s−s1∣−s∣−ss−11∣+1∣−s1−1−s∣.\det(A)=1\begin{vmatrix}1&s\\-s&1\end{vmatrix}-s\begin{vmatrix}-s&s\\-1&1\end{vmatrix}+1\begin{vmatrix}-s&1\\-1&-s\end{vmatrix}.det(A)=1​1−s​s1​​−s​−s−1​s1​​+1​−s−1​1−s​​.

Now compute each minor:

  • First minor:
∣1s−s1∣=1+s2.\begin{vmatrix}1&s\\-s&1\end{vmatrix}=1+s^2.​1−s​s1​​=1+s2.
  • Second minor:
∣−ss−11∣=(−s)(1)−s(−1)=−s+s=0.\begin{vmatrix}-s&s\\-1&1\end{vmatrix}=(-s)(1)-s(-1)=-s+s=0.​−s−1​s1​​=(−s)(1)−s(−1)=−s+s=0.

So the middle term is −s⋅0=0-s\cdot 0=0−s⋅0=0.

  • Third minor:
∣−s1−1−s∣=(−s)(−s)−1(−1)=s2+1.\begin{vmatrix}-s&1\\-1&-s\end{vmatrix}=(-s)(-s)-1(-1)=s^2+1.​−s−1​1−s​​=(−s)(−s)−1(−1)=s2+1.

Hence,

det⁡(A)=(1+s2)+0+(1+s2)=2(1+s2).\det(A)=(1+s^2)+0+(1+s^2)=2(1+s^2).det(A)=(1+s2)+0+(1+s2)=2(1+s2).

So,

det⁡(A)=2(1+sin⁡2θ).\boxed{\det(A)=2\bigl(1+\sin^2\theta\bigr)}.det(A)=2(1+sin2θ)​.
  1. Find the range of sin⁡θ\sin\thetasinθ on the given interval

Given

θ∈(3π4,5π4),\theta\in\left(\frac{3\pi}{4},\frac{5\pi}{4}\right),θ∈(43π​,45π​),

this interval is centered at π\piπ and excludes the endpoints.

On this interval:

  • sin⁡θ\sin\thetasinθ is negative,
  • at θ=π\theta=\piθ=π, sin⁡θ=0\sin\theta=0sinθ=0,
  • as θ→3π4+\theta\to \frac{3\pi}{4}^+θ→43π​+ or θ→5π4−\theta\to \frac{5\pi}{4}^-θ→45π​−,
sin⁡θ→−22.\sin\theta\to -\frac{\sqrt2}{2}.sinθ→−22​​.

Therefore,

sin⁡2θ∈[0,12).\sin^2\theta\in\left[0,\frac12\right).sin2θ∈[0,21​).
  1. Find the range of the determinant

Using

det⁡(A)=2(1+sin⁡2θ),\det(A)=2(1+\sin^2\theta),det(A)=2(1+sin2θ),

and

sin⁡2θ∈[0,12),\sin^2\theta\in\left[0,\frac12\right),sin2θ∈[0,21​),

we get

det⁡(A)∈2(1+[0,12))=2[1,32)=[2,3).\det(A)\in 2\left(1+\left[0,\frac12\right)\right) =2\left[1,\frac32\right) =\left[2,3\right).det(A)∈2(1+[0,21​))=2[1,23​)=[2,3).

So the exact range is

[2,3).\boxed{[2,3)}.[2,3)​.
  1. Match with the options

Check which interval contains all values of [2,3)[2,3)[2,3):

  • A: (32,3]\left(\frac32,3\right](23​,3] contains every value of [2,3)[2,3)[2,3).
  • B: (0,32]\left(0,\frac32\right](0,23​] does not contain values near 222 or 333.
  • C: [52,4)\left[\frac52,4\right)[25​,4) misses values between 222 and 52\frac5225​.
  • D: (1,52]\left(1,\frac52\right](1,25​] misses values between 52\frac5225​ and 333.

Hence the correct option is

A.\boxed{A}.A​.
  1. Comparison with stored answer

Stored correct answer: A.

Our derived answer is also A, so they agree.

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