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Matrices and Determinants question

2019 · 10 Jan · Shift 1 · Q32
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  5. /2019 · 10 Jan · Shift 1 · Q32

Matrices and Determinants question

2019 · 10 Jan · Shift 1 · Q32

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let d ∈\in∈ R, and A=[−24+d(sin⁡θ)−21(sin⁡θ)+2d5(2sin⁡θ)−d(−sin⁡θ)+2+2d],θ∈[0,2π]A = \left[ {\begin{matrix} { - 2} & {4 + d} & {\left( {\sin \theta } \right) - 2} \\ 1 & {\left( {\sin \theta } \right) + 2} & d \\ 5 & {\left( {2\sin \theta } \right) - d} & {\left( { - \sin \theta } \right) + 2 + 2d} \\ \end{matrix} } \right],\theta \in \left[ {0,2\pi } \right]A=​−215​4+d(sinθ)+2(2sinθ)−d​(sinθ)−2d(−sinθ)+2+2d​​,θ∈[0,2π] If the minimum value of det(A) is 8, then a value of d is -
  1. A
    −-− 7
  2. B
    2(2+2)2\left( {\sqrt 2 + 2} \right)2(2​+2)
  3. C
    −-− 5
  4. D
    2(2+1)2\left( {\sqrt 2 + 1} \right)2(2​+1)
View written solutionFree

Correct answer: C

  1. Write the matrix and simplify notation

Let s=sin⁡θ,s∈[−1,1].s=\sin\theta,\qquad s\in[-1,1].s=sinθ,s∈[−1,1]. Then

-2 & 4+d & s-2\\ 1 & s+2 & d\\ 5 & 2s-d & -s+2+2d \end{pmatrix}.$$ We need $\det(A)$ as a function of $s$ and $d$. --- 2. **Compute the determinant** Using expansion along the first row,

\det(A)=(-2)\begin{vmatrix}s+2 & d\ 2s-d & -s+2+2d\end{vmatrix} -(4+d)\begin{vmatrix}1 & d\ 5 & -s+2+2d\end{vmatrix} +(s-2)\begin{vmatrix}1 & s+2\ 5 & 2s-d\end{vmatrix}.

Nowevaluateeachminor: Now evaluate each minor: Nowevaluateeachminor:

\begin{vmatrix}s+2 & d\ 2s-d & -s+2+2d\end{vmatrix} =(s+2)(-s+2+2d)-d(2s-d).

Expanding,Expanding,Expanding,

(s+2)(-s+2+2d)=-s^2+4+2ds+4d,

sososo

M_1=-s^2+4+2ds+4d-(2ds-d^2)=-s^2+d^2+4d+4.

ThusfirsttermisThus first term isThusfirsttermis

(-2)M_1=2s^2-2d^2-8d-8.

Secondminor: Second minor:Secondminor:

\begin{vmatrix}1 & d\ 5 & -s+2+2d\end{vmatrix} =(-s+2+2d)-5d=-s+2-3d.

SosecondtermisSo second term isSosecondtermis

-(4+d)(-s+2-3d)=(4+d)(s-2+3d).

Thirdminor: Third minor:Thirdminor:

\begin{vmatrix}1 & s+2\ 5 & 2s-d\end{vmatrix} =(2s-d)-5(s+2)=-3s-d-10.

SothirdtermisSo third term isSothirdtermis

(s-2)(-3s-d-10).

Nowexpandeverything:Secondterm: Now expand everything: Second term:Nowexpandeverything:Secondterm:

(4+d)(s-2+3d)=4s+ds-8-2d+12d+3d^2=4s+ds-8+10d+3d^2.

Thirdterm: Third term:Thirdterm:

(s-2)(-3s-d-10)=-3s^2-ds-10s+6s+2d+20=-3s^2-ds-4s+2d+20.

Addingallthreeparts: Adding all three parts:Addingallthreeparts:

\det(A)=\bigl(2s^2-2d^2-8d-8\bigr)+\bigl(4s+ds-8+10d+3d^2\bigr)+\bigl(-3s^2-ds-4s+2d+20\bigr).

Now combine like terms: - $s^2$: $2s^2-3s^2=-s^2$ - $ds$: $ds-ds=0$ - $s$: $4s-4s=0$ - $d^2$: $-2d^2+3d^2=d^2$ - $d$: $-8d+10d+2d=4d$ - constants: $-8-8+20=4$ Hence

\det(A)=d^2+4d+4-s^2=(d+2)^2-\sin^2\theta.

--- 3. **Find the minimum value with respect to $\theta$** Since $\sin^2\theta\in[0,1]$, we have

\det(A)=(d+2)^2-\sin^2\theta.

For fixed $d$, the minimum occurs when $\sin^2\theta=1$. Thus

\min \det(A)=(d+2)^2-1.

Given that this minimum value is $8$, we get

(d+2)^2-1=8.

SoSoSo

(d+2)^2=9

whichgiveswhich giveswhichgives

d+2=\pm 3.

HenceHenceHence

d=1 \quad \text{or} \quad d=-5.

--- 4. **Match with the options** Options are: - A: $-7$ - B: $2(\sqrt2+2)$ - C: $-5$ - D: $2(\sqrt2+1)$ Among these, only $$d=-5$$ is present. So the correct option is **C**. --- 5. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They agree.
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