1 To solve the system x 1 + x 2 = 4 2x 1 9x 2 = 2 we find an (easier to solve) equivalent system as follows: Replace equation 2 with (2 times equation 1 + equation 2): x 1 + x 2 = 4 Solve equation 2 for x 2 : 5x 2 = 10 x 2 = 2 Substitute x 2 = 2 back into equation 1 and then solve for x 1 : implies that x 1 + (2) = 4 x 1 = 10 We conclude that the solution to the original system is x 1 = 10 x 2 = 2 3 The augmented matrix for the system x 1 3x 2 = 4 is 3x 1 + 9x 2 = 8 1 3 4 3 9 8 To find an equivalent matrix in row echelon form, we replace row 2 with (3 times row 1 + row 2) to obtain 1 3 4 0 0 20 The last row in the above matrix corresponds to the equation 0x 1 + 0x 2 = 20 which obviously has no solutions We conclude that the original system has no solutions 5 To find the point that the lines x 1 + 4x 2 = and x 1 x 2 = 1 have in common, we must solve the system x 1 + 4x 2 = x 1 x 2 = 1 Replacing equation 2 with (-1 times equation 1 + equation 2), we obtain the equivalent system x 1 + 4x 2 = 5x 2 = 8 1
This gives us x 2 = 8 5 and x 1 + 4 8 5 = or x 1 = 3 5 The point that lies on both lines is thus 3 5, 8 5 Multiply row 3 by 1/2 9 1 5 0 0 1 3 0 0 0 1 0 Replace row 2 with (-3 times row 3 + row 2) 1 5 0 0 1 0 0 0 0 1 0 Replace row 1 with (- times row 3 + row 1) Replace row 1 with (5 times row 2 + row 1) 1 5 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 The latter augmented matrix corresponds to the system x 1 = 0 x 2 = 0 x 3 = 0 We conclude that the original system has as its only solution x 1 = 0 x 2 = 0 x 3 = 0 11 2
1 1 0 0 5 0 1 2 0 0 0 1 3 2 0 0 0 1 4 Replace row 3 with (3 times row 4 + row 3) Replace row 2 with (2 times row 3 + row 2) 1 1 0 0 5 0 1 2 0 0 0 1 0 10 0 0 0 1 4 1 1 0 0 5 0 1 0 0 13 0 0 1 0 10 0 0 0 1 4 Replace row 1 with (row 2 + row 1) 1 0 0 0 8 0 1 0 0 13 0 0 1 0 10 0 0 0 1 4 The latter matrix is the augmented matrix of the system x 1 = 8 x 2 = 13 x 3 = 10 x 4 = 4 We conclude that the original system has as its only solution x 1 = 8 13 The system has augmented matrix x 2 = 13 x 3 = 10 x 4 = 4 x 2 + 5x 3 = 4 x 1 + 4x 2 + 3x 3 = 2 2x 1 + x 2 + x 3 = 1 3
0 1 5 4 1 4 3 2 2 1 1 Interchange rows 1 and 2 1 4 3 2 0 1 5 4 2 1 1 Replace row 3 with (-2 times row 1 + row 3) 1 4 3 2 0 1 5 4 0 1 5 3 Replace row 3 with (row 2 + row 3) 1 4 3 2 0 1 5 4 0 0 0 1 Since the last row in the above matrix corresponds to the equation 0x 1 + 0x 2 + 0x 3 = 1 which obviously has no solution, we conclude that the original system has no solution 15 The system x 1 + 2x 2 = 4 has augmented matrix which is equivalent to the matrix x 1 + 3x 2 + 3x 3 = 2 x 2 + x 3 = 0 1 2 0 4 1 3 3 2 0 1 1 0 1 0 0 2 0 1 0 1 0 0 1 1 We conclude that the original system has solution x 1 = 2 1 The system x 2 = 1 x 3 = 1 4
has augmented matrix 2x 1 3x 2 + 4x 3 = 5 x 2 2x 3 = 4 x 1 + 3x 2 x 3 = 2 2 3 4 5 0 1 2 4 1 3 1 2 By interchanging rows 1 and 2, we obtain the equivalent matrix 1 3 1 2 0 1 2 4 2 3 4 5 By replacing row 3 with (2 times row 1 + row 3), we obtain the equivalent matrix 1 3 1 2 0 1 2 4 0 3 2 9 By replacing row 3 with (-3 times row 2 + row 3), we obtain the equivalent matrix 1 3 1 2 0 1 2 4 0 0 8 3 This augmented matrix corresponds to a consistent system Thus the original system is consistent (and in fact has a unique solution) 19 The system 2x 2 + 2x 3 = 0 is equivalent to the system which is equivalent to the system 2x 1 + 3x 2 + 2x 3 + x 4 = 5 2x 2 + 2x 3 = 0 2x 1 + 3x 2 + 2x 3 + x 4 = 5 5
which is equivalent to the system which is equivalent to the system which is equivalent to the system 2x 2 + 2x 3 = 0 3x 2 + 2x 3 3x 4 = 1 x 2 + x 3 = 0 3x 2 + 2x 3 3x 4 = 1 x 2 + x 3 = 0 x 3 3x 4 = 1 x 2 + x 3 = 0 0 = 5 which is obviously not consistent Thus, the original system is not consistent 21 Since 1 3 h 2 6 5 1 3 h 0 0 2h 5 then either of the above matrices is the matrix of a consistent system if and only if 2h 5 = 0 Thus, in order for the system to be consistent, we must have h = 5/2 23 Since, 1 h 2 4 2 10 1 h 2 0 4h + 2 2, 25 then either of the above matrices is the matrix of a consistent system if and only if 4h + 2 0 Thus, in order for the system to be consistent, we must have h 1/2 6
1 4 g 0 3 5 h 2 5 9 k 1 4 g 0 3 5 h 0 3 5 2g + k 1 4 g 0 3 5 h 0 0 0 2g + k + h Thus, the given matrix is the augmented matrix of a consistent system if and only if 2g + k + h = 0 2 In order for the three lines to have a common point of intersection, the system 2x 1 + 3x 2 = 1 6x 1 + 5x 2 = 0 2x 1 5x 2 = must be consistent Since the augmented matrix for this system is and this matrix is equivalent to 2 3 1 6 5 0 2 5 1 0 0 0 1 0 0 0 1 which is the augmented matrix of an inconsistent system (because the last line corresponds to the equation 0x 1 + 0x 2 = 1), we conclude that the three lines do not have a common point of intersection Here are the graphs of the three lines:, 4 x2 2-4 -2 0 2 x1 4-2 -4 29 To transform the first matrix into the second, multiply row 2 by 1/2 To transform the second matrix into the first, multiply row 2 by 2 31 To transform the first matrix into the second, replace row 3 with (-2 times row 2 + row 3) To transform the second matrix into the first, replace row 3 with (2 times
33 34 row 2 + row 3) a True b False A 5x6 matrix has 5 rows and 6 columns c True d True a True b False Two matrices are row equivalent if each matrix can be transformed into the other one via a sequence of elementary row operations c False An inconsistent system has no solutions d True 35 36 The six linear equations involving the unknown temperatures T 1,,T 6 are T 1 = (20 + T 2 + T 4 + 10)/4 T 2 = (20 + T 3 + T 5 + T 1 )/4 T 3 = (20 + 40 + T 6 + T 2 )/4 T 4 = (T 1 + T 5 + 20 + 10)/4 T 5 = (T 2 + T 6 + 20 + T 4 )/4 T 6 = (T 3 + 40 + 20 + T 5 )/4 which can be written in standard form as 4T 1 T 2 T 4 = 30 The augmented matrix for this system is which is equivalent to T 1 + 4T 2 T 3 T 5 = 20 T 2 + 4T 3 T 6 = 60 T 1 + 4T 4 T 5 = 30 T 2 T 4 + 4T 5 T 6 = 20 T 3 T 5 + 4T 6 = 60 4 1 0 1 0 0 30 1 4 1 0 1 0 20 0 1 4 0 0 1 60 1 0 0 4 1 0 30 0 1 0 1 4 1 20 0 0 1 0 1 4 60 8
1 0 0 0 0 0 120 0 1 0 0 0 0 150 0 0 1 0 0 0 190 0 0 0 1 0 0 120 0 0 0 0 1 0 150 0 0 0 0 0 1 190 The temperature distribution on the metal plate is thus T 1 = 120 114 T 2 = 150 2143 T 3 = 190 214 T 4 114 T 5 2143 T 6 214 9