Let [P=[[arraycc √32 & (1)/(2) -(1)/(2) & √32 array]], A=[[arrayll 1 & 1 0 & 1 array]]] an…

Question

Let $$P=\left[\[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{array}$$\right], A=\left[$$\begin{array}{ll} 1 & 1 \ 0 & 1 \end{array}$$\right]\] and $Q=P A P^{T}$. If $$P^{T} Q^{2007} P=\left[\[\begin{array}{ll} a & b \ c & d \end{array}$$\right]\], then $2 a+b-3 c-4 d$ equal to :

Accepted Answer

$$
\text { Here, } P=\left[\[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{array}$$\right], A=\left[$$\begin{array}{ll} 1 & 1 \ 0 & 1 \end{array}$$\right]
\]

$$
\text { Here, } \mathrm{PP}^{\mathrm{T}}=\left[\[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{array}$$\right]\left[$$\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{-1}{2} \ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array}$$\right]
\]

$$
=\left[\[\begin{array}{cc} \frac{3}{4}+\frac{1}{4} & -\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \ \frac{-\sqrt{3}}{4}+\frac{\sqrt{3}}{4} & \frac{1}{4}+\frac{3}{4} \end{array}$$\right]=\left[$$\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}$$\right]=\mathrm{I}
\]

Similarly $P^T P=1$

$$ \because $$ $Q=P A P^{T}$

Now, $Q^{2007}=\left(P A P^T\right)\left(P A P^T\right) \ldots 2007$ times $=P A^{2007} P^T$

$$
\begin{aligned}
& \text { As, } A=\left[\[\begin{array}{ll} 1 & 1 \ 0 & 1 \end{array}\end{aligned}$$\right] \\
\& \Rightarrow A\^2=\left[$$\begin{array}{ll} 1 & 1 \ 0 & 1 \end{array}$$\right]\left[$$\begin{array}{ll} 1 & 1 \ 0 & 1 \end{array}$$\right]=\left[$$\begin{array}{ll} 1+0 & 1+1 \ 0+0 & 0+1 \end{array}$$\right]=\left[$$\begin{array}{ll} 1 & 2 \ 0 & 1 \end{array}$$\right]

\]

$$
A^3=\left[\[\begin{array}{ll} 1 & 3 \ 0 & 1 \end{array}$$\right]
\]

.

$$
A^{2007}=\left[\[\begin{array}{cc} 1 & 2007 \ 0 & 1 \end{array}$$\right]
\]

$$
\begin{aligned}
& \text { Hence, } \mathrm{P}^{\mathrm{T}} \mathrm{Q}^{2007} \mathrm{P}=\mathrm{A}^{2007}=\left[\[\begin{array}{cc} 1 & 2007 \ 0 & 1 \end{array}\end{aligned}$$\right] \\
\& \Rightarrow\left[$$\begin{array}{ll} a & b \ c & d \end{array}$$\right]=\left[$$\begin{array}{cc} 1 & 2007 \ 0 & 1 \end{array}$$\right]

\]

$$
\begin{aligned}
& \Rightarrow a=1, b=2007, c=0, d=1 \\
& \therefore 2 a+b-3 c-4 d=2(1)+2007-3(0)-4(1) \\
& =2+2007-4=2005
\end{aligned}
$$

Solutions · 1