answer:We see that [begin{pmatrix} sqrt{3} & -1 1 & sqrt{3} end{pmatrix} = 2 begin{pmatrix} sqrt{3}/2 & -1/2 1/2 & sqrt{3}/2 end{pmatrix} = 2 begin{pmatrix} cos frac{pi}{6} & -sin frac{pi}{6} sin frac{pi}{6} & cos frac{pi}{6} end{pmatrix}.]Note that begin{pmatrix} cos frac{pi}{6} & -sin frac{pi}{6} sin frac{pi}{6} & cos frac{pi}{6} end{pmatrix} corresponds to a rotation of frac{pi}{6} around the origin. In general, for a rotation matrix, begin{pmatrix} costheta & -sintheta sintheta & costheta end{pmatrix}^k = begin{pmatrix} cos ktheta & -sin ktheta sin ktheta & cos ktheta end{pmatrix}. Hence, begin{pmatrix} sqrt{3} & -1 1 & sqrt{3} end{pmatrix}^6 = 2^6 begin{pmatrix} cos frac{pi}{6} & -sin frac{pi}{6} sin frac{pi}{6} & cos frac{pi}{6} end{pmatrix}^6 = 2^6 begin{pmatrix} cos {pi} & -sin {pi} sin {pi} & cos {pi}end{pmatrix} = begin{pmatrix} -64 & 0 0 & -64 end{pmatrix} .

answer:The distance between two points on different spheres depends on the distance between the centers of the spheres and the radii of the spheres. To find the distance between the centers, I can use the distance formula for three dimensions: d=sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}. Plugging in the coordinates of the centers, I get d=sqrt{(12-(-2))^2+(8-(-10))^2+(-16-5)^2}=sqrt{196+324+441}=sqrt{961}=31. Now, the largest possible distance between two points on different spheres occurs when the points are on the same line as the centers, and on opposite sides of the centers. In that case, the distance between the points is the sum of the distance between the centers and the radii of the spheres. So, the largest possible distance is 31+19+87=137.

answer:I recognize that the left-hand side of the equation is related to the trigonometric identity sin x + cos x = sqrt{2} cos (x - frac{pi}{4}). So I can rewrite the equation as sqrt{2} cos (x - frac{pi}{4}) = sqrt{2}. Dividing both sides by sqrt{2} gives me cos (x - frac{pi}{4}) = 1. The cosine function is 1 when the angle is a multiple of 2 pi, so I can write x - frac{pi}{4} = 2 pi k for some integer k. Adding frac{pi}{4} to both sides gives me x = 2 pi k + frac{pi}{4}. But I also need to consider the given interval 0 le x < 2 pi. So I only want values of k that make x lie in this interval. I can try some values of k and see if they work. If k = 0, then x = frac{pi}{4}, which is in the interval. If k = 1, then x = frac{9 pi}{4}, which is larger than 2 pi and not in the interval. If k = -1, then x = frac{-7 pi}{4}, which is negative and not in the interval. So the only solution in the given interval is x = frac{pi}{4}.

answer:Note that begin{align*} mathbf{A}^2 &= begin{pmatrix} 4 & 1 -9 & -2 end{pmatrix} begin{pmatrix} 4 & 1 -9 & -2 end{pmatrix} &= begin{pmatrix} 7 & 2 -18 & -5 end{pmatrix} &= 2 begin{pmatrix} 4 & 1 -9 & -2 end{pmatrix} - begin{pmatrix} 1 & 0 0 & 1 end{pmatrix} &= 2 mathbf{A} - mathbf{I}. end{align*}Then mathbf{A}^2 - 2 mathbf{A} + mathbf{I} = 0, so [(mathbf{A} - mathbf{I})^2 = mathbf{A}^2 - 2 mathbf{A} + mathbf{I} = mathbf{0}.]Thus, let [mathbf{B} = mathbf{A} - mathbf{I} = begin{pmatrix} 4 & 1 -9 & -2 end{pmatrix} - begin{pmatrix} 1 & 0 0 & 1 end{pmatrix} = begin{pmatrix} 3 & 1 -9 & -3 end{pmatrix}.]Then mathbf{B}^2 = mathbf{0}, and mathbf{A} = mathbf{B} + mathbf{I}, so by the Binomial Theorem, begin{align*} mathbf{A}^{100} &= (mathbf{B} + mathbf{I})^{100} &= mathbf{B}^{100} + binom{100}{1} mathbf{B}^{99} + binom{100}{2} mathbf{B}^{98} + dots + binom{100}{98} mathbf{B}^2 + binom{100}{99} mathbf{B} + mathbf{I} &= 100 mathbf{B} + mathbf{I} &= 100 begin{pmatrix} 3 & 1 -9 & -3 end{pmatrix} + begin{pmatrix} 1 & 0 0 & 1 end{pmatrix} &= begin{pmatrix} 301 & 100 -900 & -299 end{pmatrix}. end{align*}Note: We can expand (mathbf{B} + mathbf{I})^{100} using the Binomial Theorem because the matrices mathbf{B} and mathbf{I} commute, i.e. mathbf{B} mathbf{I} = mathbf{I} mathbf{B}. In general, expanding a power of mathbf{A} + mathbf{B} is difficult. For example, [(mathbf{A} + mathbf{B})^2 = mathbf{A}^2 + mathbf{A} mathbf{B} + mathbf{B} mathbf{A} + mathbf{B}^2,]and without knowing more about mathbf{A} and mathbf{B}, this cannot be simplified.