Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Matrices and Determinants question

2021 · 26 Aug · Shift 1 · Q26
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Matrices and Determinants
  5. /2021 · 26 Aug · Shift 1 · Q26

Matrices and Determinants question

2021 · 26 Aug · Shift 1 · Q26

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let θ∈(0,π2)\theta \in \left( {0,{\pi \over 2}} \right)θ∈(0,2π​). If the system of linear equations (1+cos⁡2θ)x+sin⁡2θy+4sin⁡3 θz=0cos⁡2θx+(1+sin⁡2θ)y+4sin⁡3 θz=0cos⁡2θx+sin⁡2θy+(1+4sin⁡3 θ)z=0(1 + {\cos ^2}\theta )x + {\sin ^2}\theta y + 4\sin 3\,\theta z = 0{\cos ^2}\theta x + (1 + {\sin ^2}\theta )y + 4\sin 3\,\theta z = 0{\cos ^2}\theta x + {\sin ^2}\theta y + (1 + 4\sin 3\,\theta )z = 0(1+cos2θ)x+sin2θy+4sin3θz=0cos2θx+(1+sin2θ)y+4sin3θz=0cos2θx+sin2θy+(1+4sin3θ)z=0 has a non-trivial solution, then the value of θ\thetaθ is :
  1. A
    4π9{{4\pi } \over 9}94π​
  2. B
    7π18{{7\pi } \over {18}}187π​
  3. C
    π18{\pi \over {18}}18π​
  4. D
    5π18{{5\pi } \over {18}}185π​
View written solutionFree

Correct answer: B

We need the homogeneous system to have a non-trivial solution, so the determinant of its coefficient matrix must be zero.

1. Write the coefficient matrix

The system is

(1+cos⁡2θ)x+sin⁡2θ y+4sin⁡3θ z=0(1+\cos^2\theta)x+\sin^2\theta\,y+4\sin 3\theta\,z=0(1+cos2θ)x+sin2θy+4sin3θz=0 cos⁡2θ x+(1+sin⁡2θ)y+4sin⁡3θ z=0\cos^2\theta\,x+(1+\sin^2\theta)y+4\sin 3\theta\,z=0cos2θx+(1+sin2θ)y+4sin3θz=0 cos⁡2θ x+sin⁡2θ y+(1+4sin⁡3θ)z=0\cos^2\theta\,x+\sin^2\theta\,y+(1+4\sin 3\theta)z=0cos2θx+sin2θy+(1+4sin3θ)z=0

So the coefficient matrix is

A=(1+cos⁡2θsin⁡2θ4sin⁡3θcos⁡2θ1+sin⁡2θ4sin⁡3θcos⁡2θsin⁡2θ1+4sin⁡3θ).A=\begin{pmatrix} 1+\cos^2\theta & \sin^2\theta & 4\sin 3\theta\\ \cos^2\theta & 1+\sin^2\theta & 4\sin 3\theta\\ \cos^2\theta & \sin^2\theta & 1+4\sin 3\theta \end{pmatrix}.A=​1+cos2θcos2θcos2θ​sin2θ1+sin2θsin2θ​4sin3θ4sin3θ1+4sin3θ​​.

Let a=cos⁡2θ,b=sin⁡2θ,c=4sin⁡3θ.a=\cos^2\theta,\quad b=\sin^2\theta,\quad c=4\sin 3\theta.a=cos2θ,b=sin2θ,c=4sin3θ. Then a+b=1a+b=1a+b=1, and

A=(1+abca1+bcab1+c).A=\begin{pmatrix} 1+a & b & c\\ a & 1+b & c\\ a & b & 1+c \end{pmatrix}.A=​1+aaa​b1+bb​cc1+c​​.

2. Compute the determinant

Observe that

A=I+(abcabcabc).A=I+\begin{pmatrix}a&b&c\\a&b&c\\a&b&c\end{pmatrix}.A=I+​aaa​bbb​ccc​​.

The added matrix has all rows equal, say row vector (a,b,c)(a,b,c)(a,b,c) repeated three times.

Now perform row operations on AAA:

  • R1→R1−R2R_1 \to R_1-R_2R1​→R1​−R2​
  • R2→R2−R3R_2 \to R_2-R_3R2​→R2​−R3​

Then

R1=(1,−1,0),R2=(0,1,−1),R3=(a,b,1+c).R_1=(1,-1,0),\qquad R_2=(0,1,-1),\qquad R_3=(a,b,1+c).R1​=(1,−1,0),R2​=(0,1,−1),R3​=(a,b,1+c).

So

det⁡A=∣1−1001−1ab1+c∣.\det A= \begin{vmatrix} 1&-1&0\\ 0&1&-1\\ a&b&1+c \end{vmatrix}.detA=​10a​−11b​0−11+c​​.

Expand along the first row:

det⁡A=1⋅∣1−1b1+c∣−(−1)⋅∣0−1a1+c∣.\det A=1\cdot \begin{vmatrix} 1&-1\\ b&1+c \end{vmatrix} -(-1)\cdot \begin{vmatrix} 0&-1\\ a&1+c \end{vmatrix}.detA=1⋅​1b​−11+c​​−(−1)⋅​0a​−11+c​​.

Thus

det⁡A=(1+c+b)+(a)=1+a+b+c=1+1+c=2+c.\det A=(1+c+b)+(a)=1+a+b+c=1+1+c=2+c.detA=(1+c+b)+(a)=1+a+b+c=1+1+c=2+c.

Since c=4sin⁡3θc=4\sin 3\thetac=4sin3θ,

det⁡A=2+4sin⁡3θ=2(1+2sin⁡3θ).\det A=2+4\sin 3\theta=2(1+2\sin 3\theta).detA=2+4sin3θ=2(1+2sin3θ).

For a non-trivial solution,

det⁡A=0  ⟹  2+4sin⁡3θ=0  ⟹  sin⁡3θ=−12.\det A=0 \implies 2+4\sin 3\theta=0 \implies \sin 3\theta=-\frac12.detA=0⟹2+4sin3θ=0⟹sin3θ=−21​.

3. Solve for θ\thetaθ

Given θ∈(0,π2),\theta\in\left(0,\frac\pi2\right),θ∈(0,2π​), we have 3θ∈(0,3π2).3\theta\in\left(0,\frac{3\pi}{2}\right).3θ∈(0,23π​). In this interval, sin⁡3θ=−12\sin 3\theta=-\frac12sin3θ=−21​ when 3θ=7π6.3\theta=\frac{7\pi}{6}.3θ=67π​. (The other standard value 11π6\frac{11\pi}{6}611π​ is not in (0,3π/2)(0,3\pi/2)(0,3π/2).)

Hence

θ=13⋅7π6=7π18.\theta=\frac{1}{3}\cdot \frac{7\pi}{6}=\frac{7\pi}{18}.θ=31​⋅67π​=187π​.

4. Check options

  • A: 4π9\frac{4\pi}{9}94π​
  • B: 7π18\frac{7\pi}{18}187π​
  • C: π18\frac{\pi}{18}18π​
  • D: 5π18\frac{5\pi}{18}185π​

Therefore, the correct option is 7π18.\boxed{\frac{7\pi}{18}}.187π​​.

PreviousNext

More from Matrices and Determinants

  • If A=(5​1​5​−2​​5​2​5​1​​), B=(1i​01​)…2021 · MCQ
  • Let A=​101​010​010​​. Then A2025 − A2020 is equal to :2021 · MCQ
  • Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of A2 is 1, then the possible number of such matrices is :2021 · MCQ
  • The value of ​(a+1)(a+2)(a+2)(a+3)(a+3)(a+4)​a+2a+3a+4​111​​ is :2021 · MCQ
  • Consider the following system of equations : x + 2y − 3z = a 2x + 6y − 11z = b x − 2y + 7z = c, where a, b and c are real constants. Then the system of equations :2021 · MCQ
  • If the matrix A=​103​020​00−1​​ satisfies the equation A20+αA19+βA=​100​040​001​​…2021 · Numerical
  • If the matrix A=(0K​2−1​) satisfies A(A3+3I)=2I, then the value of K is :2021 · MCQ
  • If the system of linear equations 2x + y − z = 3 x − y − z =α 3x + 3y +β z = 3 has infinitely many solution, then α+β−αβ is equal to ​.2021 · Numerical