If P = 1 0 1 2 1 , then P 50 is:

If

P = [\begin{array}{l} 1 & 0 \\ \frac{1}{2} & 1 \end{array}],

then

P^{50}

is:

$[\begin{array}{l} 1 & 0 \\ 25 & 1 \end{array}]$
$[\begin{array}{l} 1 & 50 \\ 0 & 1 \end{array}]$
$[\begin{array}{l} 1 & 25 \\ 0 & 1 \end{array}]$
$[\begin{array}{l} 1 & 0 \\ 50 & 1 \end{array}]$

Solution

$P = [\begin{array}{l} 1 & 0 \\ \frac{1}{2} & 1 \end{array}]$

$P^{2} = [\begin{matrix} 1 & 0 \\ \frac{1}{2} & 1 \end{matrix}] [\begin{array}{l} 1 & 0 \\ \frac{1}{2} & 1 \end{array}] = [\begin{array}{l} 1 & 0 \\ 1 & 1 \end{array}]$

$P^{3} = [\begin{array}{l} 1 & 0 \\ 1 & 1 \end{array}] [\begin{matrix} 1 & 0 \\ \frac{1}{2} & 1 \end{matrix}] = [\begin{matrix} 1 & 0 \\ \frac{3}{2} & 1 \end{matrix}]$

$P^{4} = [\begin{array}{l} 1 & 0 \\ \frac{1}{2} & 1 \end{array}] [\begin{array}{l} 1 & 0 \\ \frac{3}{2} & 1 \end{array}] = [\begin{array}{l} 1 & 0 \\ 2 & 1 \end{array}]$

$⋮$

$∴ P^{50} = [\begin{matrix} 1 & 0 \\ \frac{1}{2} + 49 (\frac{1}{2}) & 1 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 25 & 1 \end{matrix}]$

Asked in: JEE Main 2021 (25 Jul Shift 2)

Practice more Matrices questions on Aicharya