5x 2 = 10. x 1 + 7(2) = 4. x 1 3x 2 = 4. 3x 1 + 9x 2 = 8
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1 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 = To find an equivalent matrix in row echelon form, we replace row 2 with (3 times row 1 + row 2) to obtain 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
2 This gives us x 2 = 8 5 and x = or x 1 = 3 5 The point that lies on both lines is thus 3 5, 8 5 Multiply row 3 by 1/ Replace row 2 with (-3 times row 3 + row 2) Replace row 1 with (- times row 3 + row 1) Replace row 1 with (5 times row 2 + row 1) 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 =
3 Replace row 3 with (3 times row 4 + row 3) Replace row 2 with (2 times row 3 + row 2) Replace row 1 with (row 2 + row 1) 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
4 Interchange rows 1 and Replace row 3 with (-2 times row 1 + row 3) Replace row 3 with (row 2 + row 3) 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 = We conclude that the original system has solution x 1 = 2 1 The system x 2 = 1 x 3 = 1 4
5 has augmented matrix 2x 1 3x 2 + 4x 3 = 5 x 2 2x 3 = 4 x 1 + 3x 2 x 3 = By interchanging rows 1 and 2, we obtain the equivalent matrix By replacing row 3 with (2 times row 1 + row 3), we obtain the equivalent matrix By replacing row 3 with (-3 times row 2 + row 3), we obtain the equivalent matrix 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
6 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 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 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 Thus, in order for the system to be consistent, we must have h 1/2 6
7 1 4 g h k 1 4 g h g + k 1 4 g h g + 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 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 x x 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
8 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 The six linear equations involving the unknown temperatures T 1,,T 6 are T 1 = (20 + T 2 + T )/4 T 2 = (20 + T 3 + T 5 + T 1 )/4 T 3 = ( T 6 + T 2 )/4 T 4 = (T 1 + T )/4 T 5 = (T 2 + T T 4 )/4 T 6 = (T 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 =
9 The temperature distribution on the metal plate is thus T 1 = T 2 = T 3 = T T T
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