Matrix to power $2012$

Question

How to calculate $A^{2012}$?

$A = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$

How can one calculate this? It must be tricky or something, cause there was only 1 point for solving this.

Answer 1

Observe that $A^2$ is $A$.

So $A^{\large2012}$ is $A$ too.

Answer 2

Write $A$ as: $$A=C^{-1}BC$$

where $B$ is a diagonal matrix with eigenvalues.

Then do the computation.

Answer 3

Hints:

• Diagonalize (Jordan Normal Form) the matrix $A = P J P^{-1}$
• Take advantage of the diagonalization, such that $A^{2012} = P J^{2012} P^{-1}$

Answer 4

This particular matrix satisfies $A^2 = A$. Thus $A^{2012} = A$ as well.

Answer 5

$$A = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$$ $$A^2=\left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]\cdot \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$$

$$A^2 = \left[\begin{array}{ccc}{9-2-4}&{-3+2}&{-6+2+2}\\{6-4}&{-2+2}&{-4+2}\\{6-2}&{-2+1}&{-4+2+1}\end{array}\right]$$ $$A^2 = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$$ $$A^2=A$$

$$A=A^2=A^3=\cdots =A^n,n\in N$$

so $$A^{2012}=A$$

Answer 6

This particular example, as noted above, is easy.

More generally, you could use a technique called "exponentiation via squaring." (Since the first step of this operation is to square $A$, it would lead you immediately to see that $A^2=A$, so you wouldn't have to do the entire operation.)

Or you could write out the characteristic polynomial $p(x)=\det(x I-A)$. Then divide the polynomial $x^{2012}$ by $p(x)$ to get a polynomial of degree $\leq 2$. That is:

$$x^{2012} = p(x)q(x) + r(x)$$

Then $A^{2012} = r(A)$. So you only have to compute $A^2$ to and $r(x)$ to figure out $A^{2012}$.

(The polynomial division might seem like it requires more operations than the "exponentiation by squaring," but it's actually pretty much equivalent, I believe - there is a way to compute $r(x)$ that is equivalent to "exponentiation by squaring.")