$4^\text{th}$ power of a $2\times 2$ matrix

$$A = \left(\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right)$$ is given as a matrix. What is the result of $$ad + bc \text{ if } A^4=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$

Note that $A^4$ is the $4^\text{th}$ power of the matrix $A$.

I tried to use some trigonometric expressions but it gets very complicated and couldn't solve it.

• Hint: can you describe the geometric effect of multiplying by your matrix as a transformation of the plane? – Mark Bennet Aug 18 '13 at 15:45
• This is a high school question of the topic matrix. I dont think that it involves transformation of a plane. – guest Aug 18 '13 at 15:48
• Most high schools curricula don't even mention matrices, but it would be an extremely strange one that did so without explaining their relation to geometry. – Henning Makholm Aug 18 '13 at 15:52
• Also, are you completely sure that what is being asked for isn't $ad-bc$ rather than $ad+bc$? – Henning Makholm Aug 18 '13 at 15:53
• Yes it is a.d+b.c , not a.d-b.c. I think in that case we could first find the determinant of A and then calculate the 4th power of that value. – guest Aug 18 '13 at 15:56

All you need to do here is to recognise that $A$ represents an anti-clockwise rotation about the origin by $x$. If you square the matrix then you repeat the rotation a second time, i.e. rotate by $2x$. Similarly, the cube $A^3$ can be thought of as performing the rotation three times. Hopefully you can see that $A^4$ is an anti-clockwise rotation about the origin by $4x$. As a matrix, this looks like $$A^4 = \left[ \begin{array} 1\cos(4x) & -\sin(4x) \\ \sin(4x) & \cos(4x) \end{array}\right]$$

If you want $ad+bc$ then you have $$\cos(4x)\cos(4x)-\sin(4x)\sin(4x) \equiv \cos(4x+4x) \equiv \cos(8x)$$ Here I used the double angle formula. If, however, you actually wanted the determinant, i.e. $ad-bc$ then you have $$\cos(4x)\cos(4x)+\sin(4x)\sin(4x) \equiv \cos(4x-4x) \equiv 1$$

Note that the last result is obvious. A rotation preserves volume and so its matrix must have determinant one. (If the determinant had been $2$, say, then all areas would double.)

• Altough your solution is a bit complicated for me,the answer is cos(8x). Thank you for your explanation. – guest Aug 18 '13 at 16:10
• @guest You're welcome. What level are you working at? Would you like me to explain anything further? Just let me know and I can edit the post. – Fly by Night Aug 18 '13 at 16:11

Here is a hint:

We have

$A = \left(\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right)$

Use matrix multiplication to compute (I'll do it for you) ...

$B=A^2 = \left(\begin{array}{cc}\cos^2 x -\sin^2 x& -2\sin x \cos x \\ 2\sin x \cos x & \cos^2 x -\sin^2 x\end{array}\right)$

Now simplify using the trigonometry you know, and compute $B^2=A^4$.

• I found the result as cos(8x) and it is correct. Thanks to you and to@Fly by Night. – guest Aug 18 '13 at 16:11

Here's my way of looking at this: let $J$ be the $2 \times 2$ matrix given by

$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$;

then $J^2 = -I$, where $I$ is the $2 \times 2$ identity matrix:

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

Now observe that

$A(x) = \cos x I + \sin x J$,

and that since $J^2 = -I$, $J$ behaves algebraically exactly like $i = \sqrt {-1} \in \mathbb{C}$, $\mathbb{C}$ being the ordinary complex numbers. In particular we have, for $a, b$ real

$(aI + bJ)^2 = (a^2 - b^2)I + 2abJ$,

and if $c,d$ are real as well,

$(aI + bJ)(cI + dJ) = (ac - bd)I + (ad + bc)J$.

These formulas are easy to verify simply using elementary matrix algebra, one really doesn't even need to look at specific matrix entries, just maneuver expressions as if they were polynomials in $J$, and reduce using $J^2 = -I$. Furthermore, they show that de Moivre's classic formula

$(\cos \theta + i\sin \theta)^n = \cos n \theta + i\sin n \theta$

holds for matrices of the form $\cos \theta I + \sin \theta J$:

$(\cos \theta I + \sin\theta J)^n = \cos n \theta I + \sin n \theta J$.

Proving the above equation merely relies on a simple induction on the exponent $n \in \Bbb{Z}$, $n \ge 0$: if

$(\cos \theta I + \sin\theta J)^k = \cos k \theta I + \sin k \theta J$,

then multiplying through by $\cos \theta I + \sin \theta J$ yields

$(\cos \theta I + \sin\theta J)^{k + 1} = (\cos \theta I + \sin \theta J)(\cos k \theta I + \sin k \theta J)$,

and we have using the above formulas

$(\cos \theta I + \sin \theta J)(\cos k \theta I + \sin k \theta J)$

$= (\cos \theta \cos k \theta - \sin \theta \sin k \theta)I + (\sin \theta \cos k \theta + \cos \theta \sin k \theta)J$

$=\cos(k + 1) \theta I + \sin(k + 1) \theta J$,

this last equality relying on the standard addition formulas for $\sin (x + y)$ and $\cos (x + y)$. For more information on this derivation, see this wikipedia page.

The OP's specific concern is neatly wrapped up by observing that since

$A = \cos x I + \sin x J$,

it follows from what we have seen that

$A^4 = \cos 4x I + \sin 4x J$,

and we have $ad + bc$ (using now our OP guest's $a, b, c, d$) as

$ad + bc = \cos^2 4x - \sin^2 4x = \cos 8x$,

as our friend Fly by Night pointed out in his answer to this question.

What I like about the present approach is, that once we have

$(\cos \theta I + \sin\theta J)^n = \cos n \theta I + \sin n \theta J$,

it is really not much extra work to calculate all sorts of stuff about $A^n$ for any $n \in \Bbb{Z}$, $n \ge 0$; for example

$A^{17} = \cos 17 x I + \sin 17 x J$,

and we have

$ad + bc = \cos 34 x$

in this case.

Hope I haven't been too long winded, or gone too far afield. Cheeer-i-o, Ladies and Gents!

• Did you notice that the OP said that my "solution is a bit complicated for" him? – Fly by Night Aug 18 '13 at 21:12
• @Fly by Night: No, in fact, I did not. I confess that I am none too diligent about reading comments that aren't directed at me or directly relate to one of my posts. Not that I don't want to/enjoy doing so, just busy and somewhat preoccupied with other matters. I can only hope that, if my answer overshot the mark, it is communicative enough that guest or some other SE member finds in it at least some of the satisfaction I attained in it's composition. And if someone learns something from it, even if they can't grasp the whole thing, that is very good by me. Thanks for the feedback! – Robert Lewis Aug 18 '13 at 21:31

\begin{eqnarray*} A & = & \left(% \begin{array}{rr} \cos\left(x\right) & -\sin\left(x\right) \\ \sin\left(x\right) & \cos\left(x\right) \end{array} \right) = \cos\left(x\right) - {\rm i}\sin\left(x\right)\,\sigma_{y} \\ A^{2} & = & \cos^{2}\left(x\right) - \sin^{2}\left(x\right) - 2{\rm i}\sin\left(x\right)\cos\left(x\right)\,\sigma_{y} = \cos\left(2x\right) - {\rm i}\sin\left(2x\right)\,\sigma_{y} \\&&\mbox{Then} \\ A^{4} & = & \cos\left(4x\right) - {\rm i}\sin\left(4x\right)\,\sigma_{y} = \left(% \begin{array}{rr} \cos\left(4x\right) & -\sin\left(4x\right) \\ \sin\left(4x\right) & \cos\left(4x\right) \end{array} \right) \quad\Longrightarrow\quad {\rm det}\,\left(A^{4}\right) = 1 \\[5mm]&&\mbox{} \end{eqnarray*}

$$ad + bc = \cos^{2}\left(4x\right) - \sin^{2}\left(4x\right) = \cos\left(8x\right)$$

$\sigma_{y}$ is a Pauli Matrix: http://en.wikipedia.org/wiki/Pauli_matrices

Indeed, ${\rm det}\,\left(A^{4}\right) =\left({\rm det}\,A\right)^{4} = 1^{4} = 1$. Also,

$$A' = -\sin\left(x\right) - {\rm i}\cos\left(x\right)\,\sigma_{y} = -{\rm i}\left\lbrack\cos\left(x\right) - {\rm i}\sin\left(x\right)\,\sigma_{y}\right\rbrack = -{\rm i}A \ \Longrightarrow\ A = {\rm e}^{-{\rm i}\,x\,\sigma_{y}}$$ since $A'' + A = 0$ with $\left.A\right\vert_{x\ =\ 0} = 1$ and $\left.A'\right\vert_{x\ =\ 0} = -{\rm i}\,\sigma_{y}$. That means $$A^{n} = {\rm e}^{-{\rm i}\,n\,x\,\sigma_{y}} =\left(% \begin{array}{rr} \cos\left(nx\right) & -\sin\left(nx\right) \\ \sin\left(nx\right) & \cos\left(nx\right) \end{array} \right)\,, \quad A\ \mbox{is a Rotation Matrix}$$