If A = ( \beginmatrix 1 \over \sqrt 5 & 2 \over \sqrt 5 - 2 \over \sqrt 5 & 1 \over \sqrt 5 \endmatrix ) , B = ( \beginmatrix 1 & 0 i & 1 \endmatrix ) , i = \sqrt - 1 , and Q = ATBA, then the inverse of the matrix A Q2021 AT is equal to :

( \beginmatrix 1 & 0 - 2021i & 1 \endmatrix )

( \beginmatrix 1 \over \sqrt 5 & - 2021 2021 & 1 \over \sqrt 5 \endmatrix )

( \beginmatrix 1 & 0 2021i & 1 \endmatrix )

( \beginmatrix 1 & - 2021i 0 & 1 \endmatrix )

Matrices and Determinants PYQ — JEE Main 2021 · 26 Aug · Shift 1 · Q32 with Solution

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Correct answer: B

Interpret the expression carefully

Given

A=\begin{pmatrix} \frac1{\sqrt5} & \frac2{\sqrt5}\\[4pt] -\frac2{\sqrt5} & \frac1{\sqrt5} \end{pmatrix}, \qquad B=\begin{pmatrix}1&0\\ i&1\end{pmatrix}

and $Q=A^TBA.$

We need the inverse of the matrix $AQ^{2021}A^T$ (the printed text appears as $A Q2021 A^T$ , which is naturally read as $AQ^{2021}A^T$ ).

Use the definition of $Q$

Since $Q=A^TBA,$ we get $Q^{2021}=(A^TBA)^{2021}.$

Now consider $AQ^{2021}A^T=A(A^TBA)^{2021}A^T.$

Check that $A$ is orthogonal

Let

with $a=\frac1{\sqrt5},\qquad b=\frac2{\sqrt5}.$ Then $a^2+b^2=\frac15+\frac45=1.$ Hence $A^TA=I=AA^T.$ So $A^{-1}=A^T$ .

Simplify $AQ^{2021}A^T$

Using $A A^T=I$ repeatedly,

A(A^TBA)A^T=(AA^T)B(AA^T)=B.

Similarly,

A(A^TBA)^2A^T=B^2,

and in general,

A(A^TBA)^nA^T=B^n.

Therefore, $AQ^{2021}A^T=B^{2021}.$ So the required inverse is

(AQ^{2021}A^T)^{-1}=(B^{2021})^{-1}=B^{-2021}.

Compute $B^{2021}$

Write

\qquad N=\begin{pmatrix}0&0\\ i&0\end{pmatrix}.

Now $N^2=0.$ Hence by binomial expansion,

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

for any positive integer $n$ . Thus,

B^{2021}=I+2021N= \begin{pmatrix} 1&0\\ 2021i&1 \end{pmatrix}.

Find its inverse

For a matrix of the form $\begin{pmatrix}1&0\\x&1\end{pmatrix},$ its inverse is $\begin{pmatrix}1&0\\-x&1\end{pmatrix}.$ Therefore,

(B^{2021})^{-1}= \begin{pmatrix} 1&0\\ -2021i&1 \end{pmatrix}.

Match with the options

This is exactly Option B.

$\boxed{\begin{pmatrix}1&0\\-2021i&1\end{pmatrix}}$

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View written solutionFree

Correct answer: B

Interpret the expression carefully

Given

A=\begin{pmatrix} \frac1{\sqrt5} & \frac2{\sqrt5}\\[4pt] -\frac2{\sqrt5} & \frac1{\sqrt5} \end{pmatrix}, \qquad B=\begin{pmatrix}1&0\\ i&1\end{pmatrix}

and $Q=A^TBA.$

We need the inverse of the matrix $AQ^{2021}A^T$ (the printed text appears as $A Q2021 A^T$ , which is naturally read as $AQ^{2021}A^T$ ).

Use the definition of $Q$

Since $Q=A^TBA,$ we get $Q^{2021}=(A^TBA)^{2021}.$

Now consider $AQ^{2021}A^T=A(A^TBA)^{2021}A^T.$

Check that $A$ is orthogonal

Let

with $a=\frac1{\sqrt5},\qquad b=\frac2{\sqrt5}.$ Then $a^2+b^2=\frac15+\frac45=1.$ Hence $A^TA=I=AA^T.$ So $A^{-1}=A^T$ .

Simplify $AQ^{2021}A^T$

Using $A A^T=I$ repeatedly,

A(A^TBA)A^T=(AA^T)B(AA^T)=B.

Similarly,

A(A^TBA)^2A^T=B^2,

and in general,

A(A^TBA)^nA^T=B^n.

Therefore, $AQ^{2021}A^T=B^{2021}.$ So the required inverse is

(AQ^{2021}A^T)^{-1}=(B^{2021})^{-1}=B^{-2021}.

Compute $B^{2021}$

Write

\qquad N=\begin{pmatrix}0&0\\ i&0\end{pmatrix}.

Now $N^2=0.$ Hence by binomial expansion,

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

for any positive integer $n$ . Thus,

B^{2021}=I+2021N= \begin{pmatrix} 1&0\\ 2021i&1 \end{pmatrix}.

Find its inverse

For a matrix of the form $\begin{pmatrix}1&0\\x&1\end{pmatrix},$ its inverse is $\begin{pmatrix}1&0\\-x&1\end{pmatrix}.$ Therefore,

(B^{2021})^{-1}= \begin{pmatrix} 1&0\\ -2021i&1 \end{pmatrix}.

Match with the options

This is exactly Option B.

$\boxed{\begin{pmatrix}1&0\\-2021i&1\end{pmatrix}}$

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