Exercises on solving Ax = 0: pivot variables, special solutions Problem 7.1: a) Find the row reduced form of: ⎡ ⎤ 1 5 7 9 4 1 7 ⎦ A=⎣ 0 2 −2 11 −3 b) What is the rank of this matrix? c) Find any special solutions to the equation Ax = 0. Solution: a) To transform A into its reduced row form, we perform a series of row operations. Different operations are possible (same answer!). First, we multiply the ﬁrst row by 2 and subtract it from the third row: ⎡ ⎤ ⎡ ⎤ 1 5 7 9 1 5 7 9 ⎣ 0 1 7 ⎦. 4 1 7 ⎦ −→ ⎣ 0 4 2 −2 11 −3 0 −12 −3 −21 We then multiply the second row by 14 to make the second pivot 1: ⎡ ⎤ ⎡ ⎤ 1 5 7 9 1 5 7 9 ⎣ 0 1 1/4 7/4 ⎦ . 4 1 7 ⎦ −→ ⎣ 0 0 −12 −3 −21 0 −12 −3 −21 Multiply the second row by 12 and add it to the third row: ⎡ ⎤ ⎡ ⎤ 7 9 1 5 7 9 1 5 ⎣ 0 1 1/4 7/4 ⎦ −→ ⎣ 0 1 1/4 7/4 ⎦ . 0 −12 −3 −21 0 0 0 0 Finally, multiply the second row by 5 and subtract it from the ﬁrst row: ⎡ ⎤ ⎡ ⎤ 1 0 23/4 1/4 1 5 7 9 ⎣ 0 1 1/4 7/4 ⎦ −→ ⎣ 0 1 1/4 7/4 ⎦ 0 0 0 0 0 0 0 0 1 b) The matrix is of rank 2 because it has 2 pivots. c) The special solutions to Ax = 0 are: ⎡ ⎤ – 23/4 ⎢ −1/4 ⎥ ⎢ ⎥ and ⎣ 1 ⎦ 0 ⎡ ⎤ −1/4 ⎢ −7/4 ⎥ ⎥ ⎢ ⎣ 0 ⎦ 1 (3.3 #17.b Introduction to Linear Algebra: �Strang)� Find A1 1 1 . and A2 so that rank(A1 B) = 1 and rank(A2 B) = 0 for B = 1 1 Problem 7.2: Take A1 = I2 and A2 �= 02 . � 1 −1 . A less trivial example is A2 = 1 −1 Solution: 2