A. Miller M340 Exam 1 Feb 971. Let A =h1 −1 0iand B =h0 1 −2i. Compute ABT, ATB, and(AB)T, or say that they have incompatible dimensions.2. LetA =1 1 1 1 10 1 1 1 10 0 1 1 10 0 0 1 10 0 0 0 1Calculate A−1if it exists, otherwise show that A is noninvertible.3. LetA =1 0 0 10 x 1 00 2x x 0−1 0 0 1For what values of x, if any, is A noninvertible?4.(a) State the definition: A is an invertible matrix iff ???.(b) Let A, B, and C be n × n and let I be the n × n identity matrix.Suppose AB = I and CA = I. Prove that B = C and therefore A isinvertible.5. An n×n matrix B is called a diagonal matrix iff all entries off the diagonalare zero, i.e., entryi,j(B) = 0 for any i 6= j. Let D be the n × n diagonalmatrix w ith entryi,i(D) = i for i = 1, 2, . . . , n. So D looks like this:D =1 0 0 . . . 00 2 0 . . . 00 0 3 0............0 0 0 . . . nProve that if an n×n matrix A commutes with D, then A must be a diagonalmatrix.1Answers1. (AB)Tundefined.2.A−1=1 −1 0 0 00 1 −1 0 00 0 1 −1 00 0 0 1 −10 0 0 0 13. The determinant of A is 2x2− 4x and this is zero iff x = 0 or x = 2.4. (a) A is invertible iff there exists a matrix B such that AB = I andBA = I.Remarks: It is not part of the definition that B be unique. This is thepoint of proving part (b). You can make the definition more precise byincluding that B is n × n and I is the n × n identity matrix, if A is ann × n matrix. This emphasises the fact that nonsquare matrices are notinvertible. It cannot be the definition of invertible that the determinate isnonzero, since we have been proving theorems about invertible matrices forthree weeks before the word determinant came up. You have confused thedefinition with a theorem about determinates.4. (b)proof: By the associative law of multiplication of matrices(CA)B = C(AB)Since CA = I we have(CA)B = IBSince I is the identity matrix we haveIB = BSimilarly, we have C(AB) = CI = C and so putting these togetherB = IB = (CA)B = C(AB) = CI = Cand thus B = C as was to be proved. Hence BA = AB = I and therefore Ais invertible. QEDRemark: You cannot assume that A is invertible, particularly since thatis part of what you are asked to prove.25. Proof: By the definition of matrix multiplication:entryi,j(AD) =nXk=1entryi,k(A)entryk,j(D)But by definition of D, entryk,j(D) = 0 if k 6= j, sonXk=1entryi,k(A)entryk,j(D) = entryi,j(A)entryj,j(D)Also, entryj,j(D) = j and so entryi,j(AD) = j entryi,j(A). Similarly,entryi,j(DA) =nXk=1entryi,k(D)entryk,j(A) = i entryi,j(A)Since AD = DA we have that entryi,j(AD) = entryi,j(DA) soj entryi,j(A) = i entryi,j(A)and if i 6= j, then this implies entryi,j(A) = 0 and so A is diagonal. QEDIf you don’t understand this proof tryA =a b cd e fg h jand D =1 0 00 2 00 0 3Multiply out AD and DA. What must be true if AD = DA?“If you can’t solve the problem find a simplier version of theproblem, then a solve