SATHEE: Matrices and Determinants 1 Question 8

Matrices and Determinants 1 Question 8

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8. If $P = \begin{array}{cc} \sqrt{3} / 2 & 1 / 2 - 1 / 2 & \sqrt{3} / 2 \end{array}, A = \begin{array}{ll} 1 & 1 0 & 1 \end{array}$ and $Q = P A P^{T}$ , then $P^{T} Q^{2005} P$ is

======= ####8. If $P = [\begin{array}{cc} \sqrt{3} / 2 & 1 / 2 \\ - 1 / 2 & \sqrt{3} / 2 \end{array}], A = [\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}]$ and $Q = P A P^{T}$ , then $P^{T} Q^{2005} P$ is

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed

(2005, 1M)

(a) $[\begin{array}{cc} 1 & 2005 \\ 0 & 1 \end{array}]$

(b) $[\begin{array}{cc} 1 & 2005 \\ 2005 & 1 \end{array}]$

(d) $[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$

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Answer:

Correct Answer: 8. (a)

Solution:

Now,

$P^{T} P = [\begin{array}{cc} \sqrt{3} / 2 & - 1 / 2 \\ 1 / 2 & \sqrt{3} / 2 \end{array}] [\begin{array}{cc} \sqrt{3} / 2 & 1 / 2 \\ - 1 / 2 & \sqrt{3} / 2 \end{array}]$

$\begin{aligned} \Rightarrow P^{T} P = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] \\ \Rightarrow P^{T} P = I \\ \Rightarrow P^{T} = P^{- 1} \end{aligned}$

Since, $Q = P A P^{T}$

$\begin{aligned} ∴ P^{T} Q^{2005} P = P^{T} [(P A P^{T}) (P A P^{T}) \dots 2005 times] P \\ = \underset{2005 times}{\underset{⏟}{(P^{T} P) A (P^{T} P) A (P^{T} P) \dots (P^{T} P) A (P^{T} P)}} \\ = I A^{2005} = A^{2005} \\ ∴ A = [\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}] \\ A^{2} = [\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}] [\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}] = [\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}] \\ A^{3} = [\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}] [\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}] = [\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}] \end{aligned}$

$\begin{aligned} A^{2005} & = [\begin{array}{cc} 1 & 2005 \\ 0 & 1 \end{array}] \\ P^{T} Q^{2005} P & = [\begin{array}{cc} 1 & 2005 \\ 0 & 1 \end{array}] \end{aligned}$