Question-51

matrix

If \(\displaystyle A=\begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}\) what is \(\displaystyle A^{37}\)?

\(\displaystyle 2^{37} A\)
\(\displaystyle 2^{36} A\)
\(\displaystyle A\)
\(\displaystyle A^{2}\)

Hint

Compute \(A^2\) and \(A^3\) and observe the pattern. Generalize.

Answer

\(\displaystyle 2^{37} A\)
\(\displaystyle 2^{36} A\)
\(\displaystyle A\)
\(\displaystyle A^{2}\)

Solution

If \(\displaystyle v=\begin{bmatrix} 1\\ 1 \end{bmatrix}\), then \(\displaystyle A=vv^{T}\). We see that:

\[ \begin{equation*} \begin{aligned} A^{2} & =\left( vv^{T}\right)\left( vv^{T}\right)\\ & =\left( v^{T} v\right)\left( vv^{T}\right)\\ & =\left( v^{T} v\right) A \end{aligned} \end{equation*} \]

A couple of more rounds will show the following pattern:

\[ \begin{equation*} A^{n} =\left( v^{T} v\right)^{n-1} A \end{equation*} \]

We have \(\displaystyle v^{T} v=2\) and \(\displaystyle n=37\). Therefore, \(\displaystyle A^{37} =2^{36} A\).