Download Math. 3013: Sample Test Solutions - Linear Algebra - Prof. Mahdi Asgari and more Exams Linear Algebra in PDF only on Docsity!
- (Exs. 17–20, §1.1) Draw parallels (dotted lines) to u and v. Then r ≈ − 0 .6 and s ≈ − 0 .4. u
v
x
r^ s
- a) [− 2 , 2 , 9]; b)
[
]
- (Ex. 24, §1.1) Suppose there is a scalar r such that [c, c, c^3 ] = r[1, 1 , 4]. Then c = r and c^3 = 4r = 4c. That allows c = 0, ±2. Since c = 0 gives the zero vector, the answer is c = ±2, both of which lead to parallel vectors [− 2 , − 2 , −8] and [2, 2 , 8].
- (Ex. 34, §1.2) The distance is
‖[1, 2 , − 1 , 4] − [1, − 1 , 1 , −2]‖ = ‖[0, 3 , − 2 , 6]‖ =
02 + 3^2 + 2^2 + 6^2 = 7
- (Ex. 43, §1.2; compare with ex. 46) Compute the dot product of v − w and v + w, using properties of the dot product:
(v − w) · (v + w) = v · v + v · w − w · v + w · w = v · v − w · w = ‖v‖^2 − ‖w‖^2
Therefore, v − w and v + w are perpendicular if and only if their dot product is 0, which by the above calculation is true if and only if ‖v‖^2 = ‖w‖^2. Since norms are nonnegative, the last statement is true if and only if ‖v‖ = ‖w‖.
- (a)
(b) AT^ = −A. (We say a matrix A = [aij ] is skew-symmetric if AT^ = −A. That means aji = −aij for all i and j. In this case, we have aji = j − i and aij = i − j, and therefore it is clear that aji = −aij. That’s a proof.)
- (§1.7: exs. 13–18) Compute the square:
T 2 =
Then the cube is
T 2 =
= all +’s
Hence T is regular, and the smallest power that is all positive is T 3.
- (§1.7: exs. 28, 29, 33)
(a) The columns sum up to 1.
T =
(b) This is the 31 entry of T 2 (two transitions). That is [0,. 1 , 0] · [. 6 ,. 4 , 0] =. 1 · .4 = 0 .04, or 4%. (c) The linear system is T s = s. The third row gives 0. 1 s 2 = s 3. The second one then gives by back substitution s 2 =. 4 s 1 +. 6 s 2 +. 8 s 3 =. 4 s 1 +. 6 s 2 +. 08 s 2 , or s 1 =. 8 s 2 , after simplifying. Since s 1 + s 2 + s 3 = 1, we get. 8 s 2 + s 2 +. 1 s 2 = 1, or
- 9 s 2 = 1, or s 2 = 1/ 1. 9
= .53. Then s 1
= .42 and s 3
- (Similar to §1.4: exs. 30–37) Write out the matrix equation as
[ x 1 + x 2 x 1 + 2x 2 x 3 + x 4 x 3 + 2x 4
]
[
x 1 + x 3 x 2 + x 4 x 1 + 2x 3 x 2 + 2x 4
]
These give four linear equations:
x 1 + x 2 = x 1 + x 3 x 1 + 2x 2 = x 2 + x 4 x 3 + x 4 = x 1 + 2x 3 x 3 + 2x 4 = x 2 + 2x 4
- (a) Row-echelon form only requires pivots moving from left to right as we go down, and all 0’s below the pivot. This is true for A, C, D, but not B (look at the last column). (b) Reduced row-echelon form requires all the pivots to be 1, and all entries above a pivot to be 0, too. This isn’t true for any of the matrices. (c) Consistency merely requires that no pivots occur in the last column. That happens only for D. (d) (Similar to Exs. 7–12, §1.4) The equations are x 1 − x 3 + 2x 4 = 1 x 3 + 2x 4 = 0 x 4 = 0 Working from the bottom up, we see x 4 = 0, then x 3 = 0, then x 1 = 1. x 2 is a free variable. Thus, the set of solutions is all vectors [1, r, 0 , 0] for all scalars r.
- The righthand matrix is obtained from the left by the row operation of subtracting 2 times the first row from the second. Thus, the elementary matrix E is the result of the same row operation applied to the 3 × 3 identity matrix. That gives
E =
- The righthand matrix is obtained from the left by the row operations of subtracting the first row from the second and from the third. Thus, the matrix C is the product
C =
- We have to convert the matrix below to RREF.
1 a b 1 0 0 0 1 c 0 1 0 0 0 1 0 0 1
Add −a times row 2 to row 1 to get
1 0 b − ac 1 −a 0 0 1 c 0 1 0 0 0 1 0 0 1
Now add −c times row 3 to row 2 and ac − b times row 3 to row 1
1 0 0 1 −a ac − b 0 1 0 0 1 −c 0 0 1 0 0 1
The inverse is the 3 × 3 matrix on the right. Following the row operations, we have
1 −a ac − b 0 1 −c 0 0 1
1 0 ac − b 0 1 0 0 0 1
0 1 −c 0 0 1
1 −a 0 0 1 0 0 0 1
