MATRIX A1 = A2 = NULL SPACE 4 −1 −8 2 1 5 2 2 1 0 A3 = −1 1 1 0 A4 = A5 = 4 3 1 0 1 1 A9 = A10 = 0 0 α 1 2 −1 3 2 4 −2 6 3 6 −3 9 1 2 −1 3 α α −2 1 0 0 2 1 0 0 +β +β 1 α −1 −1 0 0 0 1 −2 0 −1 −2 4 1 2 1 −2 −1 −1 1 −2 3 −1 2 3 1 0 α 2 + β 1 −2 1 1 0 α 0 +β 1 1 −2 R2 −2 α −1 1 1 0 α −1 + β 1 1 0 −1 2 1 A8 = 1 5 4 3 1 −2 −1 2 1 A7 = 1 −2 −1 1 −2 −1 R2 1 −2 α 0 +β 1 1 0 1 −2 0 0 1 α −2 2 1 0 2 A6 = 2 1 5 −2 1 −3 α 2 4 −2 3 6 −3 1 4 α RANGE 1 0 1 0 R3 1 0 0 1 α +γ −3 0 0 1 1 −2 1 1 +β α 1 2 3 1 0 1 −1 3 True–False, circle correct answer, or give a brief answer. T–F (1) There exists a solution to A1 x = 3 2 . T–F . (2) There is a unique solution to the system of linear equations T–F (3) The general solution to A2 x = T–F T–F 6 4 is α 1 1 . x + 5y = 0 2x + 2y = 0 x + 2z =0 (4) There is a unique solution to the system of linear equations 2x + y + 5z = 0 . −2x + y − 3z = 0 1 (5) There is a unique solution to A8 x = 2 . 3 (6) The range of A7 is a point, line, or plane. (7) The null space of A5 is a point, line, or plane. T–F T–F T–F T–F (8) The equation A3 x = b has a solution for every b ∈ R3 . 1 (9) There exists a solution to A3 x = 0 . 1 1 (10) There exists a unique solution to A3 x = 0 . 1 (11) There exists a unique solution to A3 x = b for every b ∈ R3 . T–F (12) There exists a solution to A5 x = 4 6 . 4 6 T–F (13) There exists a unique solution to A5 x = T–F (14) The equation A4 x = b has a solution for every b ∈ R2 . T–F (15) The equation A4 x = b has a unique solution for every b ∈ R2 . T–F (16) The general solution to A4 x = 8 2 . 1 1 is 1 + α −2 . 1 2 (17) Give two independent (i.e. non-parallel) solutions to (18) Find the general solution to A9 x = 1 −1 0 4 −x + 2y + z =0 x − 2y − z = 0 . x − 2y − z = 0 . x+2 z =0 2x + y + 5z = 2 (19) Find the general solution to −2x + y − 3 z = 2 T–F (20) Circle the matrices which have inverses: A1 , A2 , A6 , A7 , A8 , A9 , A10 . 1 −2 is a point, line, plane, or a 3-space. (21) The solution space for A9 x = 1 1 2 4 (22) The solution space for A10 x = is a point, line, plane, or a 3-space. 6 2 (23) There exists a solution to A10 x = b for every b ∈ R4 . T–F . (24) The system of bi ’s. x1 − 2x2 − x4 −2x1 + 4x2 + x3 + 2x4 x1 − 2x2 − x3 − x4 x1 − 2x2 + 3x3 − x4 = b1 = b2 has a solution for all choices = b3 = b4 (25) Find a vector perpendicular to the null space of A5 . (26) Explain why (27) Explain why 2 1 1 4 cannot be in the null space of A9 . 22 −64 42 −38 is in the range of A9 .