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

2023 · 25 Jan · Shift 2 · Q36
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. /2023 · 25 Jan · Shift 2 · Q36

Matrices and Determinants question

2023 · 25 Jan · Shift 2 · Q36

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A=[110310−310110]A = \left[ {\begin{matrix} {{1 \over {\sqrt {10} }}} & {{3 \over {\sqrt {10} }}} \\ {{{ - 3} \over {\sqrt {10} }}} & {{1 \over {\sqrt {10} }}} \\ \end{matrix} } \right]A=[10​1​10​−3​​10​3​10​1​​] and B=[1−i01]B = \left[ {\begin{matrix} 1 & { - i} \\ 0 & 1 \\ \end{matrix} } \right]B=[10​−i1​], where i=−1i = \sqrt { - 1}i=−1​. If M=ATBA\mathrm{M=A^T B A}M=ATBA, then the inverse of the matrix AM2023AT\mathrm{AM^{2023}A^T}AM2023AT is
  1. A
    [1−2023i01]\left[ {\begin{matrix} 1 & { - 2023i} \\ 0 & 1 \\ \end{matrix} } \right][10​−2023i1​]
  2. B
    [102023i1]\left[ {\begin{matrix} 1 & 0 \\ {2023i} & 1 \\ \end{matrix} } \right][12023i​01​]
  3. C
    [12023i01]\left[ {\begin{matrix} 1 & {2023i} \\ 0 & 1 \\ \end{matrix} } \right][10​2023i1​]
  4. D
    [10−2023i1]\left[ {\begin{matrix} 1 & 0 \\ { - 2023i} & 1 \\ \end{matrix} } \right][1−2023i​01​]
View written solutionFree

Correct answer: C

  1. Given matrices
A=[110310−310110],B=[1−i01]A=\begin{bmatrix} \frac1{\sqrt{10}} & \frac3{\sqrt{10}}\\[4pt] \frac{-3}{\sqrt{10}} & \frac1{\sqrt{10}} \end{bmatrix}, \qquad B=\begin{bmatrix}1 & -i\\0 & 1\end{bmatrix}A=[10​1​10​−3​​10​3​10​1​​],B=[10​−i1​]

We need the inverse of

AM2023ATAM^{2023}A^TAM2023AT

where

M=ATBA.M=A^TBA.M=ATBA.
  1. First observe that AAA is orthogonal

Let

A=110[13−31].A=\frac1{\sqrt{10}}\begin{bmatrix}1&3\\-3&1\end{bmatrix}.A=10​1​[1−3​31​].

Now,

A^TA= rac1{10} \begin{bmatrix}1&-3\\3&1\end{bmatrix} \begin{bmatrix}1&3\\-3&1\end{bmatrix} = rac1{10} \begin{bmatrix}10&0\\0&10\end{bmatrix} =egin{bmatrix}1&0\\0&1\end{bmatrix}=I.

Hence,

AT=A−1.A^T=A^{-1}.AT=A−1.

So AAA is orthogonal.

  1. Simplify AM2023ATAM^{2023}A^TAM2023AT using similarity

Since

M=ATBA=A−1BA,M=A^TBA=A^{-1}BA,M=ATBA=A−1BA,

MMM is similar to BBB. Therefore,

M2023=(ATBA)2023=ATB2023A.M^{2023}=(A^TBA)^{2023}=A^TB^{2023}A.M2023=(ATBA)2023=ATB2023A.

Now multiply by AAA on the left and ATA^TAT on the right:

AM2023AT=A(ATB2023A)AT.AM^{2023}A^T=A(A^TB^{2023}A)A^T.AM2023AT=A(ATB2023A)AT.

Using AAT=IAA^T=IAAT=I and AAT=IAA^T=IAAT=I again,

AM^{2023}A^T=(AA^T)B^{2023}}(AA^T)=B^{2023}.

So we only need

(AM2023AT)−1=(B2023)−1=B−2023.\left(AM^{2023}A^T\right)^{-1}=(B^{2023})^{-1}=B^{-2023}.(AM2023AT)−1=(B2023)−1=B−2023.
  1. Compute B2023B^{2023}B2023

Write

B=I+N,quadN=[0−i00].B=I+N, quad N=\begin{bmatrix}0&-i\\0&0\end{bmatrix}.B=I+N,quadN=[00​−i0​].

Since

N2=0,N^2=0,N2=0,

we use

(I+N)n=I+nN.(I+N)^n=I+nN.(I+N)n=I+nN.

Therefore,

B2023=I+2023N=[1−2023i01].B^{2023}=I+2023N =\begin{bmatrix}1&-2023i\\0&1\end{bmatrix}.B2023=I+2023N=[10​−2023i1​].
  1. Find its inverse

For a matrix of the form

[1a01]−1=[1−a01],\begin{bmatrix}1&a\\0&1\end{bmatrix}^{-1} =\begin{bmatrix}1&-a\\0&1\end{bmatrix},[10​a1​]−1=[10​−a1​],

so with a=−2023ia=-2023ia=−2023i,

(B2023)−1=[12023i01].(B^{2023})^{-1}=\begin{bmatrix}1&2023i\\0&1\end{bmatrix}.(B2023)−1=[10​2023i1​].

Hence,

(AM2023AT)−1=[12023i01]\boxed{\left(AM^{2023}A^T\right)^{-1}=\begin{bmatrix}1&2023i\\0&1\end{bmatrix}}(AM2023AT)−1=[10​2023i1​]​
  1. Match with options

This is Option C.

  1. Comparison with stored answer

Stored correct answer: C

Derived answer: C

So they agree.

PreviousNext

More from Matrices and Determinants

  • Let α and β be real numbers. Consider a 3 × 3 matrix A such that A2=3A+αI. If A4=21A+βI, then2023 · MCQ
  • Consider the following system of equations αx+2y+z=12αx+3y+z=13x+αy+2z=β for some α,β∈R. Then which of the following is NOT correct.2023 · MCQ
  • Let A be a symmetric matrix such that ∣A∣=2 and [23​123​​]A=[1α​2β​]. If the…2023 · Numerical
  • Let the system of linear equations x+y+kz=22x+3y−z=13x+4y+2z=k have infinitely many solutions. Then the system (k+1)x+(2k−1)y=7(2k+1)x+(k+5)y=10 has :2023 · MCQ
  • For α,β∈R, suppose the system of linear equations ​x−y+z=52x+2y+αz=83x−y+4z=β​ has infinitely many solutions. Then α and β are…2023 · MCQ
  • For the system of linear equations x+y+z=6αx+βy+7z=3x+2y+3z=14 which of the following is NOT true ?2023 · MCQ
  • Let A=​100​0412​0−1−3​​. Then the sum of the diagonal elements of the matrix (A+I)11 is equal to :2023 · MCQ
  • The number of values of α for which the system of equations : x + y + z =αα x + 2 α y + 3z = − 1 x + 3 α y + 5z = 4 is inconsistent, is2022 · MCQ