A. MillerM340 Exam 1February 27, 1991There is a total of 100 points. Put your answers on separatesheets of paper. Keep this exam, hand in the answers and all worksheets.1. (16 points) Let A ∈ Fm×n, B ∈ Fn×p, and C ∈ Fp×r. Prove thatA(BC) = (AB)C.2. (16 points) Find the inverse of the following matrix:1 2 01 3 1−1 −1 23. (16 points) Find a 4 × 4 elementary matrix E such that for any 4 × nmatrix A, EA is the matrix obtained from A by replacing the third row ofA by the third row of A plus 5 times the fourth row of A. What is E−1?4. (18 p oints) Find a matrix R in rref such that for some invertible matrixM, MA = R. Find two solutions to the equation AX = Z where Z is the3 × 1 zero matrix.A =1 2 −2 −3 2−1 −2 3 4 03 6 −3 −6 −35. (18 points) Let A ∈ Fn×nand suppose that A is not invertible. Provethat there is a Y ∈ Fn×1for which there does not exist X ∈ Fn×1. such thatAX = Y .6. (16 points) Let T : V → W is a mapping and X, Y, Z ⊆ V . Define T (Z).Prove that T (X ∩ Y ) ⊆ T (X) ∩ T (Y ). Give an example of T, X, Y such thatT (X ∩ Y ) 6= T (X) ∩ T (Y ).A. MillerM340 Exam 1 answersFebruary 27, 19911. see example 2.20 page 20.2.7 −4 2−3 2 −12 −1 13. E =1 0 0 00 1 0 00 0 1 50 0 0 1E−1=1 0 0 00 1 0 00 0 1 −50 0 0 14. R =1 2 0 −1 00 0 1 1 00 0 0 0 1X1=00000X2=−210005. see Thm 4.11 page 70-71.6. Define T (Z) = {T (z) : z ∈ Z}, or (in English) T (Z) is the set of allT (z) for z an element of Z.THM. T (X ∩ Y ) ⊆ T (X) ∩ T (Y ).proof. Let z ∈ T (X ∩ Y ). Then for some x ∈ X ∩ Y we must have thatT (x) = z. Since x ∈ X ∩ Y we have that both x ∈ X and x ∈ Y . So thatz = T (x) ∈ T (X) and z = T (x) ∈ T (Y ) and therefore z ∈ T (X) ∩ T (Y ).Example. Let T : R → R be defined by T (x) = x2. Let X = {0, 1} andY = {0, −1}. Then T (X ∩ Y ) = {0} 6= {0, 1} = T (X) ∩ T (Y
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