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

2019 · Shift 1 · Q20
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 Advanced
  3. /Mathematics
  4. /Matrices and Determinants
  5. /2019 · Shift 1 · Q20

Matrices and Determinants question

2019 · Shift 1 · Q20

JEE AdvancedMathematicsMatrices and DeterminantsMCQ+3 / −1
Let M=[sin⁡4θ1+cos⁡2θ−1−sin⁡2θcos⁡4θ]=αI+βM−1M = \left[ {\begin{matrix} {{{\sin }^4}\theta } \\ {1 + {{\cos }^2}\theta } \\ \end{matrix} \begin{matrix} { - 1 - {{\sin }^2}\theta } \\ {{{\cos }^4}\theta } \\ \end{matrix} } \right] = \alpha I + \beta {M^{ - 1}}M=[sin4θ1+cos2θ​−1−sin2θcos4θ​]=αI+βM−1, where α\alphaα=α\alphaα(θ\thetaθ) and β\betaβ=β\betaβ(θ\thetaθ) are real numbers, and I is the 2 ×\times× 2 identity matrix. If α\alphaα* is the minimum of the set {}α\alphaα(θ\thetaθ) : θ∈\theta \inθ∈ [0, 2 π\piπ)} and {β\betaβ(θ\thetaθ) : θ∈\theta \inθ∈ [0, 2 π\piπ{)}, then the value of α\alphaα* + β\betaβ* is
  1. A
    −1716- {{17} \over {16}}−1617​
  2. B
    −3116- {{31} \over {16}}−1631​
  3. C
    −3716- {{37} \over {16}}−1637​
  4. D
    −2916- {{29} \over {16}}−1629​
View written solutionFree

Correct answer: D

Let M=(sin⁡4θ−1−sin⁡2θ1+cos⁡2θcos⁡4θ).M=\begin{pmatrix}\sin^4\theta & -1-\sin^2\theta\\[4pt] 1+\cos^2\theta & \cos^4\theta\end{pmatrix}.M=(sin4θ1+cos2θ​−1−sin2θcos4θ​).

We are given M=αI+βM−1.M=\alpha I+\beta M^{-1}.M=αI+βM−1.

We need the minimum values α∗\alpha^*α∗ and β∗\beta^*β∗ of α(θ)\alpha(\theta)α(θ) and β(θ)\beta(\theta)β(θ) respectively, and then compute α∗+β∗\alpha^*+\beta^*α∗+β∗.

1. Use the characteristic relation for a 2×22\times 22×2 matrix

From M=αI+βM−1,M=\alpha I+\beta M^{-1},M=αI+βM−1, multiply both sides by MMM: M2=αM+βI.M^2=\alpha M+\beta I.M2=αM+βI.

Now for any 2×22\times 22×2 matrix, Cayley–Hamilton gives M2−(tr⁡M)M+(det⁡M)I=0,M^2-(\operatorname{tr} M)M+(\det M)I=0,M2−(trM)M+(detM)I=0, so M2=(tr⁡M)M−(det⁡M)I.M^2=(\operatorname{tr} M)M-(\det M)I.M2=(trM)M−(detM)I.

Comparing with M2=αM+βI,M^2=\alpha M+\beta I,M2=αM+βI, we get α=tr⁡M,β=−det⁡M.\alpha=\operatorname{tr} M,\qquad \beta=-\det M.α=trM,β=−detM.

So we only need trace and determinant of MMM.

2. Compute α(θ)=tr⁡M\alpha(\theta)=\operatorname{tr}Mα(θ)=trM

The trace is α=sin⁡4θ+cos⁡4θ.\alpha=\sin^4\theta+\cos^4\theta.α=sin4θ+cos4θ.

Use sin⁡4θ+cos⁡4θ=(sin⁡2θ+cos⁡2θ)2−2sin⁡2θcos⁡2θ=1−2sin⁡2θcos⁡2θ.\sin^4\theta+\cos^4\theta=(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta=1-2\sin^2\theta\cos^2\theta.sin4θ+cos4θ=(sin2θ+cos2θ)2−2sin2θcos2θ=1−2sin2θcos2θ.

Also, sin⁡2θcos⁡2θ≤14.\sin^2\theta\cos^2\theta\le \frac14.sin2θcos2θ≤41​.

Hence α≥1−2⋅14=12.\alpha\ge 1-2\cdot\frac14=\frac12.α≥1−2⋅41​=21​.

Therefore, α∗=12.\alpha^*=\frac12.α∗=21​.

3. Compute β(θ)=−det⁡M\beta(\theta)=-\det Mβ(θ)=−detM

Let s=sin⁡2θ,c=cos⁡2θ,s=\sin^2\theta,\qquad c=\cos^2\theta,s=sin2θ,c=cos2θ, so that s+c=1.s+c=1.s+c=1.

Then M=(s2−(1+s)1+cc2).M=\begin{pmatrix}s^2 & -(1+s)\\ 1+c & c^2\end{pmatrix}.M=(s21+c​−(1+s)c2​).

Its determinant is det⁡M=s2c2−(−(1+s)(1+c))=s2c2+(1+s)(1+c).\det M=s^2c^2-\bigl(-(1+s)(1+c)\bigr)=s^2c^2+(1+s)(1+c).detM=s2c2−(−(1+s)(1+c))=s2c2+(1+s)(1+c).

Since c=1−sc=1-sc=1−s, (1+s)(1+c)=(1+s)(2−s)=2+s−s2.(1+s)(1+c)=(1+s)(2-s)=2+s-s^2.(1+s)(1+c)=(1+s)(2−s)=2+s−s2.

Also, s2c2=s2(1−s)2=s2−2s3+s4.s^2c^2=s^2(1-s)^2=s^2-2s^3+s^4.s2c2=s2(1−s)2=s2−2s3+s4.

Thus det⁡M=s4−2s3+2+s.\det M=s^4-2s^3+2+s.detM=s4−2s3+2+s.

So β=−det⁡M=−(s4−2s3+s+2),s∈[0,1].\beta=-\det M=-(s^4-2s^3+s+2), \qquad s\in[0,1].β=−detM=−(s4−2s3+s+2),s∈[0,1].

To minimize β\betaβ, we must maximize f(s)=s4−2s3+s+2.f(s)=s^4-2s^3+s+2.f(s)=s4−2s3+s+2.

4. Maximize f(s)=s4−2s3+s+2f(s)=s^4-2s^3+s+2f(s)=s4−2s3+s+2 on [0,1][0,1][0,1]

Differentiate: f′(s)=4s3−6s2+1.f'(s)=4s^3-6s^2+1.f′(s)=4s3−6s2+1.

Factor: 4s3−6s2+1=(2s−1)(2s2−2s−1).4s^3-6s^2+1=(2s-1)(2s^2-2s-1).4s3−6s2+1=(2s−1)(2s2−2s−1).

In [0,1][0,1][0,1], the only valid critical point is s=12,s=\frac12,s=21​, since roots of 2s2−2s−1=02s^2-2s-1=02s2−2s−1=0 are outside [0,1][0,1][0,1] except one negative and one greater than 1.

Now check values: f(0)=2,f(0)=2,f(0)=2, f(1)=2,f(1)=2,f(1)=2, f(12)=116−2⋅18+12+2=116−14+12+2=3716.f\left(\frac12\right)=\frac1{16}-2\cdot\frac18+\frac12+2=\frac1{16}-\frac14+\frac12+2=\frac{37}{16}.f(21​)=161​−2⋅81​+21​+2=161​−41​+21​+2=1637​.

So maximum of fff is 3716.\frac{37}{16}.1637​.

Hence β∗=−3716.\beta^*=-\frac{37}{16}.β∗=−1637​.

5. Compute α∗+β∗\alpha^*+\beta^*α∗+β∗

α∗+β∗=12−3716=8−3716=−2916.\alpha^*+\beta^*=\frac12-\frac{37}{16}=\frac{8-37}{16}=-\frac{29}{16}.α∗+β∗=21​−1637​=168−37​=−1629​.

6. Compare with options

The value is −2916,-\frac{29}{16},−1629​, which matches Option D.

7. Verification with stored answer

Stored correct answer: D

Our derived answer: D

So the stored answer is correct.

PreviousNext

More from Matrices and Determinants

  • Let M=​013​12b​a31​​ and adj M=​−18−5​1−63​−12−1​​…2019 · Multiple correct
  • Let x ∈ R and let P=​100​120​123​​, Q=​20x​x4x​x06​​…2019 · Multiple correct
  • P1​=I=​100​010​001​​,P2​=​100​001​010​​,P3​=​010​100​001​​,P4​=​001​100​010​​,P5​=​010​001​100​​,P6​=​001​010​100​​…2019 · Multiple correct
  • Suppose det ​k=0∑n​kk=0∑n​nCk​.k​k=0∑n​nCk​k2k=0∑n​nCk​3k​​=0 holds…2019 · Numerical
  • Let S be the set of all column matrices ​b1​b2​b3​​​ such that b1​,b2​,b3​∈R and the system of equations (in real variables) ​−x+2y+5z=b1​2x−4y+3z=b2​x−2y+2z=b3​​…2018 · Multiple correct
  • Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {− 1, 0, 1}. Then, the maximum possible value of the determinant of P is ............ .2018 · Numerical
  • Which of the following is(are) NOT the square of a 3 × 3 matrix with real entries?2017 · Multiple correct
  • For a real number α, if the system ​1αα2​α1α​α2α1​​​xyz​​=​1−11​​…2017 · Numerical