Math 220 Linear Algebra (Spring 2018)
Homework 1.1 and 1.2 WITH SOLUTIONS
Due Thursday January 25
These will be graded in detail and will count as two (TA graded) homeworks. Be sure to start each of these problems on a new sheet of paper, summarize the problem, and explain what you are doing so that a classmate who is struggling can follow what you are doing by just reading your work (do not refer to any external references, including the text).
1. (9 points) Determine whether each of the following equations are linear or nonlinear. If it is linear, put it into the canonical form,
a1x1 + a2x2 + ... + anxn = b
and if it is nonlinear, explain why it is nonlinear. cos(ln 1) (a) e x1 + x2 = −0.9x3 + .1x2 Solution: The equation is linear—note that ecos(ln(1)) = ecos(0) = e1 = e is just a constant—and the canonical form is
ex1 + 0.9x2 + 0.9x3 = 0.
(b) x1 + 2x2 − 3x2x3 = −21
Solution: This equation is not linear because of the x2x3 term in the left-hand side. 2 (c)( x1 − x2) = 15x3 Solution: This equation is not linear either, though it might not be readily visible 2 2 before expanding the square in the left-hand side. Doing so yields x1 −2x1x2 +x2 = 15x3, which is evidently nonlinear because of the x1x2 term.
2. (15 points) Solve the system of equations, unless the system of equations is inconsistent, and then write inconsistent. Use elementary row operations and be sure to get the augmented matrix in at least row echelon form. (No points if the augmented matrix is not, at some point, in row echelon form). (a)
x1 + 2x2 + 2x3 = 4
x1 + 3x2 + 3x3 = 5
2x1 + 6x2 + 5x3 = 6
Solution: We set up the augmented matrix 1 2 2 4 1 3 3 5 . 2 6 5 6 We add −1 times the first row to the second row, and −2 times the first row to the second row, yielding 1 2 2 4 0 1 1 1 . 0 2 1 −2 Next add −2 times the second row to the third row, giving us row echelon form: 1 2 2 4 0 1 1 1 . 0 0 −1 −4
From here on we can either do backwards substitution to solve for x3, x2, and x1, or we can keep on performing elementary row operations to reach reduced row echelon form. We will do the latter in order to practice matrix manipulation. We multiply the last row by −1, making the pivot element 1 there, 1 2 2 4 0 1 1 1 0 0 1 4 and then add −1 times the last row to the second row, and −2 times the last row to the first row: 1 2 0 −4 0 1 0 −3 . 0 0 1 4 We are almost done, it only remains to get right of the 2 in the first row, which we accomplish by adding −2 times the second row to the first row, leaving us with 1 0 0 2 0 1 0 −3 . 0 0 1 4 Since this is now in reduced row echelon form we can read off the unique solution x1 = 2, x2 = −3, and x3 = 4, and we see also that this system is consistent. (b)
x1 + 2x2 + x4 = 7
x1 + x2 + x3 − x4 = 3
3x1 + x2 + 5x3 − 7x4 = 1
Solution: We set up the augmented matrix
1 2 0 1 7 1 1 1 −1 3 . 3 1 5 −7 1
Adding −1 times the first row and −3 times the first row to the second and third row respectively we get 1 2 0 1 7 0 −1 1 −2 −4 , 0 −5 5 −10 −20 and then add −5 times row two to row three, whence
1 2 0 1 7 0 −1 1 −2 −4 . 0 0 0 0 0
Multiplying the second row by −1 and subtracting 2 times this new row from row 1 we end up with 1 0 2 −3 −1 0 1 −1 2 4 . 0 0 0 0 0 Reading off this matrix we now see that the system is consistent with infinitely many solutions, namely x1 = −1 − 2x3 + 3x4, x2 = 4 + x3 − 2x4, x is free, 3 x4 is free.
(c)
x1 + 2x2 + 2x3 = −2
3x1 − 2x2 + 3x3 = −4
4x1 + 5x3 = 0 Solution: As per usual we set up an augmented coefficient matrix:
1 2 2 −2 3 −2 3 −4 . 4 0 5 0
In an effort to get a pivot in the first column we add −3 times row 1 to row 2, and −4 times row 1 to row 3: 1 2 2 −2 0 −8 −3 2 . 0 −8 −3 8 To get a pivot in the second column we now add −1 times row two to row three:
1 2 2 −2 0 −8 −3 2 . 0 0 0 6
The augmented matrix is now in row echelon form, but note that the last column has a pivot. This means that the system is not consistent. Another way of seeing the same is to spell out what the last row is really saying: 0x1 +0x2 +0x3 = 6, or in other words 0 = 6, which is impossible. All this to say, the system is inconsistent.
3. (9 points) (a) Explain the difference between a coefficient matrix and an augmented matrix. Solution: A coefficient matrix is a matrix made up of the coefficients from a system of linear equations. An augmented matrix is similar in that it, too, is a coefficient matrix, but in addition it is augmented with a column consisting of the values on the right-hand side of the equations of the linear system. (b) If you know that a system of linear equations has at least two solutions, how many solutions must it actually have? Explain. Solution: It must have infinitely many solutions. There are three options for a system of linear equations: it has no solutions at all (called inconsistent), exactly one solution, or infinitely many solutions. Hence if it has more than one solution, it has infinitely many. (c) How many solutions can a consistent system of linear equations have? Discuss the different possibilities. Solution: A system being consistent means it has solutions at all. There are now two possibilities: either there is exactly one unique solution, or there are infinitely many solutions.
4. (12 points) (a) Give an example of two row equivalent matrices in echelon form. Note that although the reduced echelon form is unique, there will be several different row equivalent matrices in echelon form. Solution: For example
1 0 0 1 0 0 0 1 0 and 0 1 1 . 0 0 1 0 0 1
There are infinitely many solutions to this problem; the particular one shown above was obtained by starting with a matrix (in this case the identity matrix, which happens to be in reduced row echelon form) and adding its third row to its second row. Since this is an elementary row operation they remain row equivalent, and since we haven’t introduced any nonzero elements under the pivot elements it remains in row echelon form. (b) Suppose a 4×6 coefficient matrix has 4 pivot columns. Is the corresponding system of equations consistent? Justify your answer. Solution: The system is consistent with infinitely many solutions since there’s a pivot in every row but not every column. The columns without pivot elements correspond to free variables. (c) if a 7 × 5 augmented matrix has a pivot in every column, what can you say about the solutions to the corresponding system of equations? Justify your answer. Solution: Because the matrix at hand is augmented and has a pivot in every column, the rightmost column in particular must be a pivot column. This means that the system is not consistent, since it means that the row corresponding to the last pivot element looks like 0 0 0 0 b where b 6= 0, which means that 0 = b, which is impossible. Hence the system is inconsistent. (d) If a consistent system of equations has more unknowns than equations, what can be said about the number of solutions? Solution: A system with more unknowns than equations has either no solutions or infinitely many solutions. Since our system is specified to be consistent (meaning it has solutions), it must be the second option.