Additional Problems
For problem 1-2, let
Prove that
To prove:
That
Answer to Problem 1AP
Solution:
It is verified that
Explanation of Solution
Given:
Let
Calculation:
Suppose
Then,
Substitute
Thus, it is verified that
Conclusion:
Hence, it is verified that
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Chapter 4 Solutions
Differential Equations and Linear Algebra (4th Edition)
- In Problems 21–26, decompose v into two vectors v1 and v2 , where v1 is parallel to w, and v2 is orthogonal to w. 25. v = 3i + j, w = - 2i - jarrow_forwardIn Problem ,use the vectors in the figure at the right to graph each of the following vectors. u - varrow_forwardIn Problem,decompose v into two vectors v1 and v2 where v1 is parallel to w and v2 is orthogonal to w. v = 3i + j, w =2i - jarrow_forward
- If u and v are vectors in R^n, and if v≠0 ⃗, then u can be expressed in exactly one way in the form u=w_1+w_2, where w_1 is a scalar multiple of v and w_2 is orthogonal to v. The vector w_1 is called the _____.arrow_forwardthe component of vector a are ax=2.0 units and ay=3.0units while that of another vector b are Bx= 3.0 units and by=2.0 units. what is the smallest between vectors a and barrow_forwardwrite the general solution to x+y+z+w=1 x+y-z-w=1 as a vector in the form of y or w I've tried this so many times and can't get it right, please explain how you got it.arrow_forward
- From the system x' = -x + 5y og y' = -y show that the vector function (top of picture) is a solution of the system only if (bottom of picture) is true.arrow_forwardTwo vectors in the plane, i & j, have the following properties: i · i = 1, i · j = 0, j · j = 1a) Is there a vector k, that is not equal to i, such that: k · k = 1, k · j = 0? What is it? Are there many vectors with these properties?b) Is there a vector k such that: k · k = 1, k · j = 0, k · i = 0? Why not?c) If i and j were vectors in 3D, how would the answers to the above questions change?arrow_forwardIf A is invertible, i.e., for any given r.h.s. vector b, the system Ax = b has a solution. How can you compute the 3rd column of A^−1(what r.h.s. b would youchoose)?arrow_forward
- I reached this same conclusion " Tx=Ax1x2⇒Tx=0110x1x2⇒Tx=x2x1" but had difficulty explaining it in words because it is only a reflection with respect to the y=x line if there is a negative value in either the x or y original vector. eg. (-3,2) => (2,-3) reflects across the x and y axis however if both values of x,y are negative of positive the values stay in the same quadrant and the reflection occurs over a corresponding diagonal midline. As far as I can tell, the values are transpose vectors of one another, but transpose is not a linear transformation. I know I must be confusing some theorem or missing something, but I cannot find it. Is there something I am misinterpretting? Sometimes online classes are extra difficult when asking for clarification isn't immediate or easy, so sorry for the needed follow up.arrow_forwardSuppose column 5 of U has no pivot. Then x5 is a __ variable. The zero vector (is) (is not) the only solution to Ax = 0. If Ax = b has a solution, then it has __solutions.arrow_forward6) Find the initial point of the vector that is equivalent to u= (1. 2) and whose terminal point is B(2. 0).arrow_forward
- Linear Algebra: A Modern IntroductionAlgebraISBN:9781285463247Author:David PoolePublisher:Cengage Learning