MAT1332 Assignment #5 solutions

Transcription

1 1 MAT133 Assignment #5 solutions Question 1 Determine the solution of the following systems : a) x + y + z = x + 3y + z = 5 x + 9y + 7z = 1 The augmented matrix associated to this system is Replace the second row R with R R1, and replace the third row R3 with R3 R1 (R1 means times row 1) : Replace the third row R3 with R3 5R : The bottom row, when expressed as equations, says 0 = 15. Since that never holds, the original system of equations has no solution. b) x + y + z = 6 x + 3y + z = 18 x + y + 7z = 1 The augmented matrix associated to this system is For convenience, we switch the first two rows R1 R. This way the leading entry in the first row is a 1 instead of a. (It is also correct to avoid this step.)

2 Replace R with R R1 and R3 with R3 R1, to clear our the rest of the third column : The leading entry in the second row is in the second column. We next clear out the entries below it, by replacing R3 with R3 R : Our matrix is now in row-echelon form. We now continue through to reduced row-echelon form. We first make all leading entries equal to one, by scaling each row. We leave R1 alone, multiply R by 1/4, and multiply R3 by 1/6, to get : /4 30/4 1 Next we clear the entries above the right-most leading entry, by replacing R with R 3R3, and replacing R1 with R1 R /4. 1 Lastly we clear the remaining entry that is above a leading entry, by replacing R1 with R1 3R (note that 16 3 (7/4) = 17/4) : /4 7/4. 1 With the system in reduced row-echelon form, we can just read off the values. The first variable, here x, equals 17/4, the second, y, is 7/4, and the third, z is 1. This is the only solution. Question Consider the system. x 1 + x + x 3 + x 4 = 1 3x 1 + x + x 3 + 3x 4 = x 1 + x + 4x 3 + 9x 4 = 3 a) Write the augmented matrix M associated with this system. M =

3 b) Reduce M to upper triangular (i.e. row-echelon) form. For convenience, switch the first and third lines, R1 R (see comment in Question 1(b)) : Apply the operation R 3R1 (that is, replace R with this), and then the operation R3 R1 : Again for convenience, switch the second and third line : R R3, then multiply the (new) second row by 1 : R The leading entry in the second row is in the second column. We next clear out the entries below it, by replacing R3 with R3 + R : ( ) This matrix is in row echelon form. Note : There are many augmented matrices that would count as a row-echelon form of the original augmented matrix. In contrast, each matrix has only one reduced row-echelon form. c) Solve the system. Solution #1 : One way to solve the system is to further reduce the matrix to reduced row-echelon form. From the row-echelon form above, first make all leading entries equal to one, by replacing R3 with 1/4R3 : / 3/4 Next we clear the entries above the right-most leading entry, by replacing R with R 7R3, and replacing R1 with R1 4R / 1/4. 5/ 3/4. 3

4 4 Lastly we clear the remaining entry that is above a leading entry, by replacing R1 with R1 R : / 1/4 1/ 1/4. ( ) 5/ 3/4 This is the reduced row-echelon form of our system. We see that the fourth column has no leading entry, so we set x 4 = t, with t arbitrary. The equations encoded in the augmented matrix (**) tell us that and so the solution is x 1 t/ =1/4 x t/ = 1/4 x 3 + 5/t =3/4, x 1 = t/ + 1/4, x = t/ 1/4, x 3 = 5/t + 3/4, x 4 = t, t arbitrary. Solution # : Another solution works directly from the row-echelon form (*). Again, since the fourth column of (*) has no leading entry, we set x 4 = t, with t arbitrary. The last row of (*) corresponds to the equation x 3 + 5/t = 3/4. Solving for x 3 gives x 3 = 5/t + 3/4. The next-to-last row of (*) corresponds to the equation x + 7x x 4 = 5. Using the already solved formulas for x 4 and x 3, we can solve for x and get x = 5 7x 3 17x 4 = 5 7( 5/t + 3/4) 17t = t/ 1/4. The top row of (*) corresponds to the equation x 1 + x + 4x 3 + 9x 4 = 3. Using the already solved formulas for x 4, x 3, and x we can solve for x 1 and get (skipping the steps) x 1 = t/ + 1/4. The final answer is the same as in solution #1. Question 3 Solve the following for x (as complex numbers) : a) x 10x + 4 = 0 By the quadratic formula, x = or 5 1 Notice that the roots here are real. That s OK since all real numbers are also complex numbers. b) x cos(a)x + 1 = 0, where a is a constant. By the quadratic formula, the solutions are x = cos(a) ± ( cos(a)) 4 = cos(a) ± cos (a) 1

5 5 so We can simplify this, though : since cos (a) + sin (a) = 1, we have that cos (a) 1 = sin (a), ± cos (a) 1 = ± sin (a) = ±i sin(a), and so the roots can be more simply written as x = cos(a)±i sin(a). An even more concise way to write the roots is : x = e ia or e ia Question 4 Consider z 1 = e iπ/3 and z = e iπ/4. a) Express z 1 and z is the form a + ib where a and b are real. b) Find z 1 /z (in the form a + ib). z 1 z = i + i = z 1 = cos(π/3) + i sin(π/3) = i z = cos(π/4) + i sin(π/4) = + i ( ( c) Find z 1 z (in the form a + ib). z 1 z = i ) ( i ) + i ) ( i ) = ( ) ( ) i i = i i 4 4 Question 5 Consider A = a) Calculate A A = 1 0 b) Calculate A 3.

6 6 A 3 = c) Determine the form of A n for n 1. A n = 1 0 n