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Matrices and Determinants question

2019 · 9 Jan · Shift 2 · Q33
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  5. /2019 · 9 Jan · Shift 2 · Q33

Matrices and Determinants question

2019 · 9 Jan · Shift 2 · Q33

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If A=[ete−tcos⁡te−tsin⁡tet−e−tcos⁡t−e−tsin⁡t−e−tsin⁡t+e−tcosolimitstet2e−tsin⁡t−2e−tcos⁡t]A = \left[ {\begin{matrix} {{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \\ {{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s} olimits} t} \\ {{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \\ \end{matrix} } \right]A=​etetet​e−tcost−e−tcost−e−tsint2e−tsint​e−tsint−e−tsint+e−tcosolimitst−2e−tcost​​ then A is :
  1. A
    invertible for all t ∈\in∈ R.
  2. B
    invertible only if t =π=\pi=π
  3. C
    not invertible for any t ∈\in∈ R
  4. D
    invertible only if t =π2={\pi \over 2}=2π​.
View written solutionFree

Correct answer: A

  1. Write the matrix clearly

Given

A=[ete−tcos⁡te−tsin⁡tet−e−tcos⁡t−e−tsin⁡t−e−tsin⁡t+e−tcos⁡tet2e−tsin⁡t−2e−tcos⁡t].A=\begin{bmatrix} e^t & e^{-t}\cos t & e^{-t}\sin t\\[4pt] e^t & -e^{-t}\cos t-e^{-t}\sin t & -e^{-t}\sin t+e^{-t}\cos t\\[4pt] e^t & 2e^{-t}\sin t & -2e^{-t}\cos t \end{bmatrix}.A=​etetet​e−tcost−e−tcost−e−tsint2e−tsint​e−tsint−e−tsint+e−tcost−2e−tcost​​.

We need to determine when AAA is invertible. A matrix is invertible iff its determinant is nonzero.

  1. Factor common terms from columns

Observe:

  • first column has common factor ete^tet
  • second and third columns have common factor e−te^{-t}e−t

So,

det⁡(A)=et⋅e−t⋅e−tdet⁡[1cos⁡tsin⁡t1−cos⁡t−sin⁡t−sin⁡t+cos⁡t12sin⁡t−2cos⁡t].\det(A)=e^t\cdot e^{-t}\cdot e^{-t} \det\begin{bmatrix} 1 & \cos t & \sin t\\ 1 & -\cos t-\sin t & -\sin t+\cos t\\ 1 & 2\sin t & -2\cos t \end{bmatrix}.det(A)=et⋅e−t⋅e−tdet​111​cost−cost−sint2sint​sint−sint+cost−2cost​​.

Thus,

det⁡(A)=e−tdet⁡[1cos⁡tsin⁡t1−cos⁡t−sin⁡t−sin⁡t+cos⁡t12sin⁡t−2cos⁡t].\det(A)=e^{-t} \det\begin{bmatrix} 1 & \cos t & \sin t\\ 1 & -\cos t-\sin t & -\sin t+\cos t\\ 1 & 2\sin t & -2\cos t \end{bmatrix}.det(A)=e−tdet​111​cost−cost−sint2sint​sint−sint+cost−2cost​​.

Let

M=[1cos⁡tsin⁡t1−cos⁡t−sin⁡t−sin⁡t+cos⁡t12sin⁡t−2cos⁡t].M=\begin{bmatrix} 1 & \cos t & \sin t\\ 1 & -\cos t-\sin t & -\sin t+\cos t\\ 1 & 2\sin t & -2\cos t \end{bmatrix}.M=​111​cost−cost−sint2sint​sint−sint+cost−2cost​​.
  1. Simplify determinant using row operations

Apply:

R2→R2−R1,R3→R3−R1.R_2\to R_2-R_1,\qquad R_3\to R_3-R_1.R2​→R2​−R1​,R3​→R3​−R1​.

These operations do not change the determinant.

Then

M∼[1cos⁡tsin⁡t0−2cos⁡t−sin⁡t−2sin⁡t+cos⁡t02sin⁡t−cos⁡t−2cos⁡t−sin⁡t].M\sim \begin{bmatrix} 1 & \cos t & \sin t\\ 0 & -2\cos t-\sin t & -2\sin t+\cos t\\ 0 & 2\sin t-\cos t & -2\cos t-\sin t \end{bmatrix}.M∼​100​cost−2cost−sint2sint−cost​sint−2sint+cost−2cost−sint​​.

Hence,

det⁡(M)=det⁡[−2cos⁡t−sin⁡t−2sin⁡t+cos⁡t2sin⁡t−cos⁡t−2cos⁡t−sin⁡t].\det(M)= \det\begin{bmatrix} -2\cos t-\sin t & -2\sin t+\cos t\\ 2\sin t-\cos t & -2\cos t-\sin t \end{bmatrix}.det(M)=det[−2cost−sint2sint−cost​−2sint+cost−2cost−sint​].
  1. Compute the 2×22\times22×2 determinant

Let

a=−2cos⁡t−sin⁡t,a=-2\cos t-\sin t,a=−2cost−sint, b=−2sin⁡t+cos⁡t,b=-2\sin t+\cos t,b=−2sint+cost, c=2sin⁡t−cos⁡t.c=2\sin t-\cos t.c=2sint−cost.

Then

det⁡(M)=a2−bc.\det(M)=a^2-bc.det(M)=a2−bc.

Now,

bc=(−2sin⁡t+cos⁡t)(2sin⁡t−cos⁡t)=−(2sin⁡t−cos⁡t)2.bc=(-2\sin t+\cos t)(2\sin t-\cos t)=-(2\sin t-\cos t)^2.bc=(−2sint+cost)(2sint−cost)=−(2sint−cost)2.

So,

det⁡(M)=a2+(2sin⁡t−cos⁡t)2.\det(M)=a^2+(2\sin t-\cos t)^2.det(M)=a2+(2sint−cost)2.

That is,

det⁡(M)=(−2cos⁡t−sin⁡t)2+(2sin⁡t−cos⁡t)2.\det(M)=(-2\cos t-\sin t)^2+(2\sin t-\cos t)^2.det(M)=(−2cost−sint)2+(2sint−cost)2.

Expand:

(−2cos⁡t−sin⁡t)2=4cos⁡2t+sin⁡2t+4sin⁡tcos⁡t,(-2\cos t-\sin t)^2=4\cos^2 t+\sin^2 t+4\sin t\cos t,(−2cost−sint)2=4cos2t+sin2t+4sintcost, (2sin⁡t−cos⁡t)2=4sin⁡2t+cos⁡2t−4sin⁡tcos⁡t.(2\sin t-\cos t)^2=4\sin^2 t+\cos^2 t-4\sin t\cos t.(2sint−cost)2=4sin2t+cos2t−4sintcost.

Adding,

det⁡(M)=5cos⁡2t+5sin⁡2t=5.\det(M)=5\cos^2 t+5\sin^2 t=5.det(M)=5cos2t+5sin2t=5.

Therefore,

det⁡(A)=e−t⋅5=5e−t.\det(A)=e^{-t}\cdot 5=5e^{-t}.det(A)=e−t⋅5=5e−t.
  1. Check invertibility

Since

e−t>0for all t∈R,e^{-t}>0 \quad \text{for all } t\in\mathbb R,e−t>0for all t∈R,

we get

det⁡(A)=5e−t≠0for all t∈R.\det(A)=5e^{-t}\neq 0 \quad \text{for all } t\in\mathbb R.det(A)=5e−t=0for all t∈R.

Thus AAA is invertible for every real ttt.

  1. Evaluate the options
  • A: invertible for all t∈Rt\in\mathbb Rt∈R ✅
  • B: invertible only if t=πt=\pit=π ❌
  • C: not invertible for any t∈Rt\in\mathbb Rt∈R ❌
  • D: invertible only if t=π2t=\frac\pi2t=2π​ ❌

So the correct option is A.

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