Find the $100^{th}$ power of the matrix $\left( \begin{matrix} 1& 1\\ -2& 4\end{matrix} \right)$.

Question

Find the $100^{th}$ power of the matrix $\left( \begin{matrix} 1& 1\\ -2& 4\end{matrix} \right)$.

Can you give a hint/method?

Answer 1

The characteristic polynomial of the given matrix say $A$ is $x^2-5x+6$, the zeroes of whose are $2,3$. Thus there exist an invertible matrix $P$ such that $A=P\begin{pmatrix} 2 & 0\\ 0 & 3\end{pmatrix}P^{-1}$. Hence $A^{100}=P\begin{pmatrix} 2^{100} & 0\\ 0 & 3^{100}\end{pmatrix}P^{-1}$, where $P = \begin{pmatrix}1 &1\\ 2 & 1 \end{pmatrix}$

Answer 2

You want to find the eigen decomposition for the matrix and transform it into a diagonal matrix (for which computing powers is easy).

The eigenvectors of the matrix $M$ given by

$$ M = \left( \begin{matrix} 1& 1\\ -2& 4\end{matrix} \right) $$

are:

$$ v_1 = \left ( \begin{matrix} 1\\ 2\end{matrix} \right) \qquad v_2 = \left ( \begin{matrix} 1\\ 1\end{matrix} \right) $$

giving us the change of base matrix $B$ as:

$$ B = \left( \begin{matrix} 1& 1\\ 2& 1\end{matrix} \right) $$

The matrix, $\Gamma = B^{-1}MB$, is a diagonal matrix as such:

$$ \begin{align} \Gamma &= B^{-1}MB \\ &= \left( \begin{matrix} -1& 1\\ 2& -1\end{matrix} \right) \left( \begin{matrix} 1& 1\\ -2& 4\end{matrix} \right) \left( \begin{matrix} 1& 1\\ 2& 1\end{matrix} \right) \\ &= \left( \begin{matrix} 3& 0\\ 0& 2\end{matrix} \right) \\ \end{align} $$

Now we compute $\Gamma^{100}$ which is easy since it is diagonalizable:

$$ \Gamma^{100} = \left( \begin{matrix} 3^{100} & 0\\ 0& 2^{100}\end{matrix} \right) $$

Now we convert it back to the standard basis:

$$ \begin{align} M^{100} &= B\Gamma^{100}B^{-1} \\ &= \left( \begin{matrix} 1& 1\\ 2& 1\end{matrix} \right) \left( \begin{matrix} 3^{100}& 0\\ 0 & 2^{100}\end{matrix} \right) \left( \begin{matrix} -1& 1\\ 2& -1\end{matrix} \right) \\ &= \left( \begin{matrix} -3^{100} + 2^{101}& 3^{100} - 2^{100}\\ -2\cdot 3^{100} + 2^{101} & 2\cdot 3^{100} - 2^{100}\end{matrix} \right) \\ \end{align} $$

Answer 3

Since $A=PDP^{-1}$ then $A^{100}=PD^{100}P^{-1}$

Then we know that $M=\begin{bmatrix} 1& 1\\ -2& 4\end{bmatrix}$ has eigenvalues $\lambda=(3,2)$ and eigenvector matrix P=$\begin{bmatrix} 1&1\\2&1\end{bmatrix}$ therefore $D=\begin{bmatrix} 3&0\\0&2\end{bmatrix}$

Now apply $A^{100}=PD^{100}P^{-1}$

$D^{100}=\begin{bmatrix} 3^{100}&0\\0&2^{100}\end{bmatrix}$

$P^{-1}=\begin{bmatrix} -1&1\\2&-1 \end{bmatrix}$

Then $M^{100}=(-515377520732011328501159929309162469708701111249, 515377520732011329768810529537391871205404316625, -1030755041464022659537621059074783742410808633250, 1030755041464022660805271659303013143907511838626)$

Answer 4

Let $$P= \begin {pmatrix} 1& 1\\ -2 & 4 \end {pmatrix} $$The key is to write the matrix $P $ in terms of a diagonal matrix $D $. Thus we wish to write $$P=ADA^{-1} $$ where $D $ is a diagonal matrix. Thus, $$P^n =AD^n A^{-1} $$ $D^n $ being very easy to compute, we find the values of $D $ and $A $ using eigenvalues and eigenvectors.

---

For more, you can see here. Hope it helps.