If A= [ array cc 1 & 0 0 & -1 array ], P= [ array ll 1 & 1 0 & 1 array ] and X=A P A^T , then A^T X^ 50 A=

Correct answer: D. [ array cc 1 & 50 0 & 1 array ] Since A A^T=I therefore matrix A is orthogonal matrix. Now, A^T X^ 50 A=A^T X^ 49 (A P A^T) A=A^T X^ 49 A P(A^T A)=A^T X^ 49 A P=A^T X^ 48 (A P A^T) A P=A^T X^ 48 A P^2 . =A^T A P^ 50 =I P^ 50 =P^ 50 P=[ array ll 1 & 1 0 & 1 array ] P^2= [ array ll 1 & 2 0 & 1 array ] P^3=[ array ll 1 & 3 0 & 1 array ] . . P^ 50 =[ array cc 1 & 50 0 & 1 array ] So, A^T X^ 50 A=P^ 50 =[ array cc 1 & 50 0 & 1 array ]

BITSAT 2024Paper 1

Exam: BITSAT
Year: 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Matrices

BITSAT 2024 · Paper 1 · Mathematics · Matrices

If A= [ array cc 1 & 0 0 & -1 array ], P= [ array ll 1 & 1 0 & 1...

A=

\left[

\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}

\right], P=

\left[

\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}

\right]

and

X=A P A^T

, then

A^T X^{50} A=

Options

A $\left[$ $\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}$ $\right]$
B $\left[$ $\begin{array}{cc}2 & 1 \\ 0 & -1\end{array}$ $\right]$
C $\left[$ $\begin{array}{cc}25 & 1 \\ 1 & -25\end{array}$ $\right]$
D $\left[$ $\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}$ $\right]$

Correct answer

D. $\left[$ $\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}$ $\right]$

Step-by-step solution

Since

A A^T=I

therefore matrix

A

is orthogonal matrix. Now,

A^T X^{50} A=A^T X^{49}\left(A P A^T\right) A=A^T X^{49} A P\left(A^T A\right)=A^T X^{49} A P=A^T X^{48}\left(A P A^T\right) A P=A^T X^{48} A P^2 \ldots \ldots

=A^T A P^{50}=I P^{50}=P^{50}\because \quad P=\left[

\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}

\right] \Rightarrow P^2=

\left[

\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}

\right]\Rightarrow P^3=\left[

\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}

\right] \ldots.

.

\Rightarrow \quad P^{50}=\left[

\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}

\right]

So,

A^T X^{50} A=P^{50}=\left[

\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}

\right]

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