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If $p$ and

Q=PAP^T, then P^T Q²⁰¹⁵ P is :

Option 1)

Option 2)
$\begin{bmatrix} 2015 &1 \\ 0& 1 \end{bmatrix}$

Option 3)

Option 4)

Answers (2)

best_answer

As we have learned

Transpose of a Matrix -

The matrix obtained from any given matrix $A$ , by interchanging its rows and columns.

- wherein

Multiplication of matrices -

-

$P^TQ^{2015}P$

$= P^TQ\cdot Q\cdot Q\cdot Q\cdot Q........QP$

2015 times

Now , if $P= \begin{bmatrix} \sqrt3/2 &1/2 \\ -1/2 & \sqrt3/2 \end{bmatrix}$

so , $P^TP=$ $\begin{bmatrix} \sqrt3/2 &-1/2 \\ 1/2 & \sqrt3/2 \end{bmatrix}$ . $\begin{bmatrix} \sqrt3/2 &1/2 \\ -1/2 & \sqrt3/2 \end{bmatrix}$

$\begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix}= I$

$\rightarrow P^TQ^{2015}= A^{2015}$

$= \begin{bmatrix} 1 &1 \\ 0& 1 \end{bmatrix}\cdot \begin{bmatrix} 1 &1 \\ 0& 1 \end{bmatrix}.....\begin{bmatrix} 1 &1 \\ 0& 1 \end{bmatrix}$

$= \begin{bmatrix} 2 &1 \\ 0& 1 \end{bmatrix}........= \begin{bmatrix} 2015 &1 \\ 0& 1 \end{bmatrix}$

Option 1)

Option 2)

$\begin{bmatrix} 2015 &1 \\ 0& 1 \end{bmatrix}$

Option 3)

Option 4)

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gaurav

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