2005 Mu Matrices And Vectors Solutions



(C) A singular matrix has determinant equal to zero. Going through each of the answer choices, we see that the only singular matrix is the one in choice C.

1. (C) Setting some of the corresponding entries equal to each other, we get [pic] and [pic]. Solving these two equations simultaneously yields x = 1 and y = –1, so [pic].

2. (D) Perform the following sequence of row operations in order to get the matrix into row-echelon form: [pic], [pic], [pic], and [pic]. This should produce the matrix

[pic],

which has two pivots (the first two columns), and therefore, has rank 2.

3. (B) If [pic], then [pic], or [pic]. The larger root is 4.

4. (C) The element in the second row and first column of [pic] is [pic]. We take this value and from it subtract the value in the second row and first column of [pic], which is –3. So the answer is 8 – (–3) = 11.

5. (C) For a product of two matrices to be defined, the number of columns of the first matrix must equal the number of rows of the second matrix. Note that ABC is not defined unless n = m and a = b. If [pic] is defined, then a = b, which isn’t necessarily true; same thing with [pic]. Checking choice C, we see that it is indeed defined; the result is a b ( a matrix.

6. (A) Say the matrices are A and B, then [pic] and [pic]. Add the two equations together to get [pic], or [pic]; consequently, [pic]. Thus, [pic], making [pic].

7. (D) Statements I, III, and IV are standard theorems. Statement II is easy to disprove by finding a counterexample, say, [pic] and [pic]. Note that tr(A)tr(B) = –24, but tr(AB) = 0.

8. (A) The characteristic polynomial is [pic]. The smaller root of this polynomial is –5.

9. (B) Since [pic], the vector in choice B is an eigenvector (with a corresponding eigenvalue of 2). It’s easy to check that the vectors in the other answer choices do not yield a scalar multiple of the vector when multiplied with the given matrix.

10. (C) The system will not have a unique solution if the determinant of the coefficient matrix is equal to 0. Thus, we have

[pic].

The solution to this equation is, of course, –7.

11. (A) Let [pic]. Notice that [pic] is a symmetric matrix and [pic] is a skew-symmetric matrix. The sum of the above matrices is

[pic].

So choice A is true. Now, to disprove the other choices. In choice B, the given summation is the entry in the ith row and jth column of [pic], not AB. It’s easy to find a counter-example for choice D; just multiply a 3 ( 2 matrix with a 2 ( 3 matrix (in that order) to obtain a 3 ( 3 matrix, for instance. For choice C, just take [pic] and [pic], for example. It’s clear that [pic], but A and B aren’t inverses of each other.

12. (C) By the standard matrix formula, the area is equal to [pic].

13. (B) By the inverse “trick” for 2 ( 2 matrices,

[pic].

Left-multiplying both sides of the given equation by this inverse yields

[pic].

The sum of the entries of B is – 6 – 7 + 24 + 20 = 31.

14. (B) Since a ( c = –7, b ( c = –36, and the dot product is distributive, we have

[pic].

15. (D) Notice that [pic], a counter-clockwise rotation of [pic] radians. Thus,

[pic]

The product of the entries is 0.

16. (B) The two vectors will be orthogonal when their dot product is 0. Thus

[–7, 3, 1] ( [13, 37, [pic]] = [pic].

The smaller solution to this equation is c = 4.

17. (C) First, form the augmented matrix [pic]. Perform the following sequence of row operations on both sides of the line to turn the left side into the identity matrix; the outcome of the right side is the desired inverse: [pic], [pic], and [pic]. The right side should equal [pic], which is precisely the inverse.

18. (B) Using the standard formula, the distance is [pic].

19. (B) Using properties of the cross product, we have

[pic].

20. (A) For each determinant d in the sum, we can find a corresponding determinant whose matrix is the same as the matrix of d, but with the first two rows switched. Since switching rows changes the sign of the determinant, d and its negative counterpart will cancel each other out in the sum, leaving us with a sum of 0.

21. (A) Finding powers of A can be a pain, so instead, let’s see what A does to a certain vector:

[pic].

We now see that A is simply a vector-permutation matrix that shifts the components downward, with the lowest entry moving to the top. Thus, it takes six applications of A in order to revert a vector into its original form—in other words, [pic]. So we have [pic]. We can conclude that the third row of [pic] is the same as the third row of A.

22. (D) The characteristic polynomial of A is [pic]. Factoring [pic], we find that [pic]. By the Cayley-Hamilton theorem, [pic] (that is, the characteristic polynomial of a matrix evaluated at the matrix is equal to the zero matrix), and it follows that the determinant of [pic] is 0.

23. (D) Let a = [1, 9, –5] and b = [1, –2, 2]. By the standard formula, the projection of a along b is given by

[pic].

24. (C) If we let [pic] while a and b, equal the number of rows (or columns) that A, and B have, respectively, then the equation |2A| + |3B| = |C| becomes [pic]. Some guess-and-checking quickly gives the solution (a, b, c) = (4, 2, 5), and so A is a 4 ( 4 matrix. Now we show that this solution is unique. Taking the equation [pic] modulo 3, we get [pic], or [pic]. Since the perfect squares in modulo 3 are only 0 and 1, this must mean that a is even, say, a = 2x. Plugging this back into the equation, rearranging, and factoring produces [pic]. Thus, [pic] and [pic] for some integers m and n, where m + n = b. Subtracting the second equation from the first to get rid of c, we have [pic]. The left-hand side of this equation has no factor of 3, and so n must equal 0. Thus, [pic]. Notice that the right-hand side of the equation is a product of powers of 2 that are 2 apart from each other, which means that [pic] and [pic], making m = 2. If m = 2, then x = 2 and a = 4. It follows that b = 2 and c = 5.

25. (B) Instead of working out the entire matrix product and inverse, we use the fact that the determinant of a product is the product of the determinants. So we have

[pic].

Directly substituting 0 in for x gives the limit, [pic].

26. (D) Let’s try to find a pattern with the powers of (. Observe that

[pic], [pic], [pic], [pic], and [pic].

So in general, [pic]; this fact can be proven more rigorously using induction. We then have

[pic]

Recalling the cosine Maclaurin series for real numbers, [pic], we can simplify the above matrix as [pic], and its determinant is [pic].

27. (A) Since [pic], the slope of the tangent line to the original graph at (0, 1) is 5. The slope of the new tangent line to C is calculated by transforming a tangent vector at (0, 1) using the matrix. Taking the tangent vector to be [1, 5], we get [pic], and so the new slope is [pic]. The point that corresponds to (0, 1) on C is [pic], or [pic]. Thus, the equation of the tangent line is [pic], or 24y – 13x = 7.

28. (A) The (signed) scale factor in which the old area is multiplied is precisely the determinant of the transformation matrix. Thus, the answer is the absolute value of the determinant of [pic], or |(2)(3) – (8)(1)| = 2.

29. (A) Let’s find several values for [pic] as well as corresponding values for [pic] and try to find a pattern. Note that

[pic], [pic], [pic], [pic], and [pic].

Thus, we have [pic], [pic], [pic], [pic], and [pic]. Now the pattern is a lot clearer: [pic], the (n + 3)rd Fibonacci Number, where [pic] and [pic] for all n > 2. Thus, the desired limit is simply the ratio of the (n + 4)th Fibonacci number to the (n + 3)rd as [pic]. This limit, which we’ll call L, can be computed as follows:

[pic].

Solving the equation [pic] for the positive root, we get [pic], the Golden Ratio.

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