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

2016 · 9 Apr · 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. /2016 · 9 Apr · Shift 1 · Q26

Matrices and Determinants question

2016 · 9 Apr · Shift 1 · Q26

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If P = [3212−1232],A=[1101]   \left[ {\begin{matrix} {{{\sqrt 3 } \over 2}} & {{1 \over 2}} \\ { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \\ \end{matrix} } \right],A = \left[ {\begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} } \right]\,\,\,[23​​−21​​21​23​​​],A=[10​11​] Q = PAPT, then PT Q2015 P is :
  1. A
    [0201500]\left[ {\begin{matrix} 0 & {2015} \\ 0 & 0 \\ \end{matrix} } \right][00​20150​]
  2. B
    [2015102015]\left[ {\begin{matrix} {2015} & 1 \\ 0 & {2015} \\ \end{matrix} } \right][20150​12015​]
  3. C
    [2015012015]\left[ {\begin{matrix} {2015} & 0 \\ 1 & {2015} \\ \end{matrix} } \right][20151​02015​]
  4. D
    [1201501]\left[ {\begin{matrix} 1 & {2015} \\ 0 & 1 \\ \end{matrix} } \right][10​20151​]
View written solutionFree

Correct answer: D

  1. Given matrices
P=[3212−1232],A=[1101]P=\begin{bmatrix} \frac{\sqrt 3}{2} & \frac12\\[4pt] -\frac12 & \frac{\sqrt 3}{2} \end{bmatrix}, \qquad A=\begin{bmatrix} 1&1\\ 0&1 \end{bmatrix}P=[23​​−21​​21​23​​​],A=[10​11​]

and

Q=PAPT.Q=PAP^T.Q=PAPT.

We need to find PTQ2015PP^TQ^{2015}PPTQ2015P.

  1. Use the expression for QQQ

Since

Q=PAPT,Q=PAP^T,Q=PAPT,

we get

Q2015=(PAPT)2015.Q^{2015}=(PAP^T)^{2015}.Q2015=(PAPT)2015.

Now note that PPP is an orthogonal matrix, because its columns are orthonormal, so

PTP=PPT=I.P^TP=PP^T=I.PTP=PPT=I.

Hence,

(PAPT)n=PAnPT.(PAP^T)^n=PA^nP^T.(PAPT)n=PAnPT.

Therefore,

Q2015=PA2015PT.Q^{2015}=PA^{2015}P^T.Q2015=PA2015PT.

Now multiply by PTP^TPT on the left and PPP on the right:

PTQ2015P=PT(PA2015PT)P=(PTP)A2015(PTP)=IA2015I=A2015.P^TQ^{2015}P=P^T(PA^{2015}P^T)P=(P^TP)A^{2015}(P^TP)=IA^{2015}I=A^{2015}.PTQ2015P=PT(PA2015PT)P=(PTP)A2015(PTP)=IA2015I=A2015.

So the problem reduces to finding A2015A^{2015}A2015.

  1. Find A2015A^{2015}A2015

Write

A=[1101]=I+N,N=[0100].A=\begin{bmatrix}1&1\\0&1\end{bmatrix}=I+N, \qquad N=\begin{bmatrix}0&1\\0&0\end{bmatrix}.A=[10​11​]=I+N,N=[00​10​].

Now,

N2=[0100]2=[0000].N^2=\begin{bmatrix}0&1\\0&0\end{bmatrix}^2=\begin{bmatrix}0&0\\0&0\end{bmatrix}.N2=[00​10​]2=[00​00​].

So by binomial expansion,

An=(I+N)n=I+nNA^n=(I+N)^n=I+nNAn=(I+N)n=I+nN

because all higher powers of NNN vanish.

Thus,

A2015=I+2015N=[1001]+2015[0100]=[1201501].A^{2015}=I+2015N =\begin{bmatrix}1&0\\0&1\end{bmatrix}+2015\begin{bmatrix}0&1\\0&0\end{bmatrix} =\begin{bmatrix}1&2015\\0&1\end{bmatrix}.A2015=I+2015N=[10​01​]+2015[00​10​]=[10​20151​].

Therefore,

PTQ2015P=[1201501].P^TQ^{2015}P=\begin{bmatrix}1&2015\\0&1\end{bmatrix}.PTQ2015P=[10​20151​].
  1. Match with options

This is exactly Option D:

[1201501].\begin{bmatrix}1&2015\\0&1\end{bmatrix}.[10​20151​].
PreviousNext

More from Matrices and Determinants

  • If A = [−43​−11​], then the determinant of the matrix (A2016 − 2A2015 − A2014) is :2016 · MCQ
  • Let A be a 3 × 3 matrix such that A2 − 5A + 7I = 0 Statement - I : A − 1 = 71​(5I − A). Statement - II : The polynomial A3 − 2A2 − 3A + I can be reduced to 5(A − 4I). Then :2016 · MCQ
  • The system of linear equations x+λy−z=0λx−y−z=0x+y−λz=0​ has a non-trivial solution for :2016 · MCQ
  • If A=[5a3​−b2​] and A adj A=AAT, then 5a+b is equal to :2016 · MCQ
  • If A=​12a​212​2−2b​​ is a matrix satisfying the equation AAT=9I, where I is 3×3 identity matrix, then the…2015 · MCQ
  • The set of all values of λ for which the system of linear equations: 2x1​−2x2​+x3​=λx1​2x1​−3x2​+2x3​=λx2​−x1​+2x2​=λx3​​…2015 · MCQ
  • If A is a 3×3 non-singular matrix such that AA′=A′A and B=A−1A′, then BB′ equals:2014 · MCQ
  • If α,βe0, and f(n)=αn+βn and ​31+f(1)1+f(2)​1+f(1)1+f(2)1+f(3)​1+f(2)1+f(3)1+f(4)​​…2014 · MCQ