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1. What constraints on b1, b2, and b3 are necessary in order for the following system to have any solutions? [5]
2x - y + 3z = b1
3x - (3/2)y + (9/2)z = b2
-4x + 2y - 6z = b3
b2 = (3/2) b1 and
b3 = -2 b1 (or equivalent)
2. Calculate the determinant of the following matrix: [6]
 2 0 1 0 4 1 0 -2 1 2 2 0 0 -1 0 1

det = 5.
One way is to do: 2R4+R2, then -(1/2)R3+R1, then -R2+R1, then you get a lower-triangular matrix. Multiply down the diagonal: (-5/2)(-1)(2)(1) = 5.
3. A surface is being rendered in a 3D graphics system. If the surface faces away from the camera, "backface culling" means the surface does not need to be rendered, saving processing time. The vector from the surface to the camera is u = (3, -2, 8). The normal vector to the surface is (-1/2, 0, 0.2). Is the surface facing toward or away from the camera? [4]
Dotproduct = -1.5 + 0 + 1.6 = 0.1 > 0,
so it faces toward the camera.
4. Let A be a skew-symmetric n x n matrix; i.e., AT = -A.
1. Find the trace tr(A). [3]
For the diagonal entries aii, we must have aii = -aii, which is only possible if they are zero. Hence all the diagonal entries are zero, so the trace is zero.
2. If n is odd, prove that det(A) = 0. [5]
det(A) = det(AT) = det(-A) = (-1)ndet(A).
If n is odd, then det(A) = -det(A), so it must be zero.
5. Let u and v be any non-collinear (non-parallel) vectors in 3-space. Let w = v × (u × v).
1. Draw a sketch illustrating how u, v, and w are oriented relative to each other. [4]
All three of u, v, w are in the same plane, and w is perpendicular to v.
2. Find the dot product (v ⋅ w). [2]
Since w is perpendicular to v, v ⋅ w is zero.
3. What can you say about the dot product (u ⋅ w)? (Simplify) [5]
||u × v|| = ||u|| ||v|| sinθ.
Since the angle between v and (u × v) is 90deg,
||w|| = ||v|| (||u × v||) sin(90deg)
= ||v|| (||u × v||)
= ||u|| ||v||2 sinθ.
Now, because u, v, w are all in the same plane and v and w are perpendicular, the angle φ between u and w and the angle θ between u and v add up to 90deg. Hence, cosφ = sinθ.
So u ⋅ w = ||u|| ||w|| cosφ
= ||u|| ( ||u|| ||v||2 sinθ ) cosφ
= ||u||2 ||v||2 sin2θ
= ||u × v||2.
6. Let y = f(x) be a cubic polynomial in x, i.e., y = c1x3 + c2x2 + c3x + c4. The graph of y has a horizontal tangent when it goes through the point (0,1), and goes through the point (2,3) with a tangent slope of -2. Find y = f(x) (this is called a cubic Hermite spline). [6]
f(0) = 1 ⇒ c4 = 1
f'(0) = 0 ⇒ c3 = 0
f(2) = 3 ⇒ 8c1 + 4c2 + 2c3 + c4 = 3
f'(2) = -2 ⇒ 12c1 + 4c2 + c3 = -2
⇒ 8c1 + 4c2 + 1 = 3 and 12c1 + 4c2 = -2
⇒ c1 = -1, c2 = 5/2
⇒ y = - x3 + (5/2) x2 + 1