University:
Massachusetts Institute of Technology
Course:
18.06SC | Linear Algebra
Academic year:

2023
Views:

369

Pages:

3
Author:

ilikemyicons0y5

Exercises on elimination with matrices Problem 2.1: In the two-by-two system of linear equations below, what multiple of the ﬁrst equation should be subtracted from the second equa tion when using the method of elimination? Convert this system of equa tions to matrix form, apply elimination (what are the pivots?), and use back substitution to ﬁnd a solution. Try to check your work before look ing up the answer. 2x + 3y = 5 6x + 15y = 12 Solution: One subtracts 3 times the ﬁrst equation from the second equa tion in order to eliminate the 6x. To convert to matrix form, use the general format Ax = b: � �� 2x + 3y = 5 2 3 x 5 −→ = . 6x + 15y = 12 6 15 y 12 We then apply elimination on matrix A. Using the ﬁrst pivot (the num ber 2 in the upper left corner of A), we subtract three times the ﬁrst row from the second row to get: � � � � 2 3 2 3 A= −→ U = 6 15 0 6 where U is an upper triangular matrix with pivots 2 and 6. Doing the same to the right side b = (5, 12) gives a new equation of the form Ux = c : � �� 2 3 x 5 = . 0 6 y −3 To solve our new equation, we use back substitution: 6y = −3 −→ y = − 12 and � 1 2x + 3y = 5 −→ 2x + 3 − 2 � = 5 −→ 2x = 5 + 3 13 = −→ x = 2 2 13 4 We know that our solution fulﬁlls the ﬁrst equation; let’s make sure that our values fulﬁll the second equation as a check on our work: � � � � 13 −1 78 − 30 6x + 15y = 6 + 15 = = 12 � 4 2 4 Problem 2.2: (2.3 #29. Introduction to Linear Algebra: Strang) Find the triangular matrix E that reduces “Pascal’s matrix” to a smaller Pascal: ⎡ 1 ⎢ 1 E⎢ ⎣ 1 1 0 1 2 3 0 0 1 3 ⎤ ⎡ 0 1 0 ⎢ 0 1 0 ⎥ ⎥=⎢ 0 ⎦ ⎣ 0 1 1 0 1 0 0 1 2 ⎤ 0 0 ⎥ ⎥. 0 ⎦ 1 Which matrix M (multiplying several E’s) reduces Pascal all the way to I? Solution: ⎡ 1 0 0 ⎢ −1 1 0 The matrix is E = ⎢ ⎣ 0 −1 1 0 0 −1 ⎤ 0 0 ⎥ ⎥ 0 ⎦ 1 One can eliminate the second column with the matrix ⎡ ⎤ 1 0 0 0 ⎢ 0 1 0 0 ⎥ ⎢ ⎥ ⎣ 0 −1 1 0 ⎦ 0 0 −1 1 and the third column with the matrix ⎡ 1 0 0 ⎢ 0 1 0 ⎢ ⎣ 0 0 1 0 0 −1 ⎤ 0 0 ⎥ ⎥ 0 ⎦ 1 Multiplying these together, we get ⎡ 1 ⎢ 0 M=⎢ ⎣ 0 0 0 0 1 0 0 1 0 −1 ⎤⎡ 0 1 0 0 ⎥ ⎢ 1 0 0 ⎥⎢ 0 ⎦ ⎣ 0 0 −1 1 0 −1 1 0 ⎤⎡ 0 1 0 0 ⎥ ⎢ 0 ⎥ ⎢ −1 1 0 ⎦ ⎣ 0 −1 0 1 0 −1 1 0 ⎤ ⎡ 0 1 0 0 ⎥ ⎢ 0 ⎥ ⎢ −1 1 0 =⎣ ⎦ 1 −2 0 1 3 −3 1 −1 Since M reduces the Pascal matrix to I, M must be the inverse matrix! ⎤ 0 0 ⎥ ⎥. 0 ⎦ 1