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

2024 · 8 Apr · Shift 1 · Q58
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. /2024 · 8 Apr · Shift 1 · Q58

Matrices and Determinants question

2024 · 8 Apr · Shift 1 · Q58

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
Let A=[2−111]A=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]A=[21​−11​]. If the sum of the diagonal elements of A13A^{13}A13 is 3n3^n3n, then nnn is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 7

  1. We need the sum of the diagonal elements of A13A^{13}A13, i.e. the trace of A13A^{13}A13.

Given

A=[2−111]A=\begin{bmatrix}2 & -1\\ 1 & 1\end{bmatrix}A=[21​−11​]

we want

tr⁡(A13).\operatorname{tr}(A^{13}).tr(A13).
  1. First find the characteristic polynomial of AAA:
det⁡(A−λI)=∣2−λ−111−λ∣=(2−λ)(1−λ)+1.\det(A-\lambda I)=\begin{vmatrix}2-\lambda & -1\\ 1 & 1-\lambda\end{vmatrix} =(2-\lambda)(1-\lambda)+1.det(A−λI)=​2−λ1​−11−λ​​=(2−λ)(1−λ)+1.

Expanding,

(2−λ)(1−λ)+1=2−3λ+λ2+1=λ2−3λ+3.(2-\lambda)(1-\lambda)+1=2-3\lambda+\lambda^2+1=\lambda^2-3\lambda+3.(2−λ)(1−λ)+1=2−3λ+λ2+1=λ2−3λ+3.

So the characteristic equation is

λ2−3λ+3=0.\lambda^2-3\lambda+3=0.λ2−3λ+3=0.
  1. Let the eigenvalues be α,β\alpha,\betaα,β. Then
α+β=3,αβ=3.\alpha+\beta=3, \qquad \alpha\beta=3.α+β=3,αβ=3.

Also,

tr⁡(A13)=α13+β13.\operatorname{tr}(A^{13})=\alpha^{13}+\beta^{13}.tr(A13)=α13+β13.

Let

Sk=αk+βk.S_k=\alpha^k+\beta^k.Sk​=αk+βk.

Since α,β\alpha,\betaα,β satisfy

x2=3x−3,x^2=3x-3,x2=3x−3,

we get the recurrence

Sk=3Sk−1−3Sk−2.S_k=3S_{k-1}-3S_{k-2}.Sk​=3Sk−1​−3Sk−2​.
  1. Compute initial values:
S0=α0+β0=2,S_0=\alpha^0+\beta^0=2,S0​=α0+β0=2, S1=α+β=3.S_1=\alpha+\beta=3.S1​=α+β=3.

Now use the recurrence:

S2=3S1−3S0=3⋅3−3⋅2=3S_2=3S_1-3S_0=3\cdot 3-3\cdot 2=3S2​=3S1​−3S0​=3⋅3−3⋅2=3 S3=3S2−3S1=3⋅3−3⋅3=0S_3=3S_2-3S_1=3\cdot 3-3\cdot 3=0S3​=3S2​−3S1​=3⋅3−3⋅3=0 S4=3S3−3S2=0−9=−9S_4=3S_3-3S_2=0-9=-9S4​=3S3​−3S2​=0−9=−9 S5=3S4−3S3=−27S_5=3S_4-3S_3=-27S5​=3S4​−3S3​=−27 S6=3S5−3S4=−81+27=−54S_6=3S_5-3S_4=-81+27=-54S6​=3S5​−3S4​=−81+27=−54 S7=3S6−3S5=−162+81=−81S_7=3S_6-3S_5=-162+81=-81S7​=3S6​−3S5​=−162+81=−81 S8=3S7−3S6=−243+162=−81S_8=3S_7-3S_6=-243+162=-81S8​=3S7​−3S6​=−243+162=−81 S9=3S8−3S7=−243+243=0S_9=3S_8-3S_7=-243+243=0S9​=3S8​−3S7​=−243+243=0 S10=3S9−3S8=0+243=243S_{10}=3S_9-3S_8=0+243=243S10​=3S9​−3S8​=0+243=243 S11=3S10−3S9=729S_{11}=3S_{10}-3S_9=729S11​=3S10​−3S9​=729 S12=3S11−3S10=2187−729=1458S_{12}=3S_{11}-3S_{10}=2187-729=1458S12​=3S11​−3S10​=2187−729=1458 S13=3S12−3S11=4374−2187=2187.S_{13}=3S_{12}-3S_{11}=4374-2187=2187.S13​=3S12​−3S11​=4374−2187=2187.

Thus

tr⁡(A13)=S13=2187=37.\operatorname{tr}(A^{13})=S_{13}=2187=3^7.tr(A13)=S13​=2187=37.
  1. Therefore,
n=7.n=7.n=7.
  1. Comparison with stored answer: Stored correct answer = 777, which matches our result.
PreviousNext

More from Matrices and Determinants

  • If αeqa,βeqb,γeqc and ​αaa​bβb​ccγ​​=0…2024 · MCQ
  • If the system of equations x+4y−z=λ,7x+9y+μz=−3,5x+y+2z=−1 has infinitely many solutions, then (2μ+3λ) is equal to :2024 · MCQ
  • Let λ,μ∈R. If the system of equations ​3x+5y+λz=37x+11y−9z=297x+155y−189z=μ​ has infinitely many solutions, then μ+2λ is equal to…2024 · MCQ
  • Let A be a non-singular matrix of order 3. If det(3adj(2adj((detA)A)))=3−13⋅2−10 and det(3adj(2A))=2m⋅3n…2024 · Numerical
  • Let B=[11​35​] and A be a 2×2 matrix such that AB−1=A−1. If BCB−1=A and C4+αC2+βI=O, then 2β−α is equal to2024 · MCQ
  • Consider the matrices : A=[23​−5m​],B=[20m​] and X=[xy​]. Let the set of all m, for which…2024 · Numerical
  • Consider the matrix f(x)=​cosxsinx0​−sinxcosx0​001​​. Given below are two statements : Statement I : f(−x) is the inverse of the matrix f(x)…2024 · MCQ
  • Let A=​211​010​101​​,B=[B1​,B2​,B3​], where B1​,B2​,B3​ are column matrics, and AB1​=​100​​,AB2​=​230​​,AB3​=​321​​…2024 · Numerical