MATRICES: SYSTEMS OF EQUATIONS 5 minute review. Remind students about the different types of systems of equations: homogeneous versus non-homogeneous, and singular versus non-singular. Recall how many solutions there are in each case: non-singular = exactly one (which is 0 if homogeneous), singular + non-homogeneous = zero (inconsistent) or infinitely many (use parameters to describe), singular + homogeneous = infinitely many (since 0 is a solution). Class warm-up. Discuss what kinds of systems each of the nature of their solution sets. Find the solutions. x1 − x2 + x3 2x − 5y = 1 (b) 3x1 + x2 − 4x3 (a) x + 3y = 6 5x1 − x2 − 2x3 these are, and therefore = 0 = 0 = 0 Problems. Choose from the below. 1. Standard systems of equations. Solve the following systems of equations. (a) 2x1 + 7x3 = 0 x1 − x2 + 3x3 = 0 4x1 + 5x2 + 2x3 = 0 2x1 + 3x3 = 1 (b) x1 − x2 + 5x3 = 1 2x2 − 4x3 = −2 (c) 2x1 + 7x3 = 29 x1 − x2 + 3x3 = 0 4x1 + 5x2 + 2x3 = −29 2x1 + 3x3 = 0 (d) x1 − x2 + 5x3 = 0 2x2 − 4x3 = 0 x1 + 2x2 + 6x3 + 5x4 x2 − 2x3 + 4x4 (e) x3 + 3x4 3x1 + x4 = = = = 1 0 0 2 8x1 − x2 + 3x3 5x1 + 2x2 + x3 (f) x1 + x2 + 7x3 2x1 − 4x2 − 5x3 = = = = 7 0 0 7 2. Singular systems. Find the value(s) of α for which the following systems of equations have infinitely many solutions and then find those solutions. −x1 + 4x2 + αx3 = −2 2x1 − x2 = −1 (a) 5x1 + x2 − x3 = −5 (c) 6x1 + 3x2 − 7x3 = 1 x1 + 4x2 + x3 = −1 (b) 9x1 + 15x2 − 4x3 = α (α + 1)x1 + 6x2 + 5x3 = 6 (α − 2)x2 + x3 = −3 (α − 3)x3 = 0 3. Linear combinations revisited. (a) Let A and B be any 2 × 2 matrices. Investigate whether you can find scalars α, β, γ, δ such that αA + βAT + γB + δB T = I2 . 1 2 MATRICES: SYSTEMS OF EQUATIONS For the warm-up, (a) is non-singular, solution [3 [ 43 λ 74 λ λ]T . 1]T ; (b) is singular, solutions are Selected answers and hints. 1. (a) [0 (b) [1 0]T 0 −1 T 3 ] −5 3 (c) [−24 9 (d) [0 0]T 0 67 (e) [ 100 1 10 (f) [ 13 21 −34 21 11]T 3 100 −1 T 100 ] 1 T 7] 2. (a) α = −1, solutions are x1 = λ, x2 = 2λ + 1, x3 = 7λ + 6. (b) α = −2, solutions are x1 = λ, x2 = −13λ−6 , x3 31 = 21λ−7 31 (c) Possible values of α are −1, 2, 3. There are no solutions when α = 2. • α = −1 solutions are [λ 1 0]T . • α = 3 solutions are [λ 21 − 4λ − 24 + 4λ]T . a b e f 3. Let A = and B = . Writing out the four equations that c d g h result gives a a e e α 1 b c f g β 0 c b g f γ = 0 , d d h h δ 1 which is a singular system. This means there is either no solution or infinitely many solutions depending on the matrices A and B. For more details, start a thread on the discussion board.
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