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Lecture

# Lecture 30.pdf

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School
Department
Mathematics
Course
MATH 136
Professor
Robert Sproule
Semester
Winter

Description
Friday, March 21 − Lecture 30 : Determinants and invertibility. Concepts: 1. Recognize that det(AB) = det(A)det(B). 2. Recognize that det(cA) = c det(A). 3. Recognize that a matrix-1 is invertible iff det(A) ≠ 0. 4. Recognize that det(A ) = 1 / det(A). 30.1 Theorem − If A and B are two square matrices of same dimension then Proof : We first see that if A is an elementary matrix E and B is any matrix, then det(AB) = det Adet B: So the statement det(AB) = det Adet B holds true if A is an elementary matrix. We prove det(AB) = det Adet B by considering two cases. Case 1: Suppose A is not invertible. Claim 1: det(A) = 0: If A is not invertible then ARREF must have a row of zeroes. Then det(A) = det(E E …E A ) = det(E ) det(E )… det(E ) det(A ) 1 2 k RREF 1 2 k RREF Now A RREF has a row of zeroes so det(A RREF) = 0 = det(A) as claimed. Claim 2: AB is not invertible Suppose AB is invertible. As previously shown, B must then be invertible. Since A is non-invertible Ax = 0 has a non-zero solution, say u. Then Bx = u has a unique non-trivial solution, say v. Consider (AB)x = 0. Then 0 = Au = A(Bv) = (AB)v. Then (AB)x = 0 has a non-trivial solution v. So AB cannot be invertible. Then det(AB) = 0. (This follows from claim 1). So det(AB) = 0det(B) = det(A)det(B). Case 2: Suppose A is invertible. Then A is a product of elementary matrices E E … E . 1 2 k n 30.2 Theorem − If A is an n × n matrix and c is a scalar, then the det(cA) = c det(A). Proof is straightforward and so is given in class. 30.3 Theorem − The matrix A is invertible if and only if detA ≠ 0. Proof: ( ⇒ ) Suppose A is invertible. We are required to show that detA ≠ 0. - Then A is a product of elementary matrices,A = E E E ..1 2 3 n - Then det(A) = det(E E 1 2 3E ) =ndet(E )det1E )det2E )...d3t(E ). n - The determinant of an elementary matrix is either −1, a non-zero scalar or 1, depending on whether it is an elementary matrix of type I, II or III, respectively. So the determinant of an elementary matrix is never zero. - Since det(A) is the product of non-zero numbers, det(A) ≠ 0. (⇐) Suppose now that det(A) ≠ 0. We are required to show that A is invertible. - Note that A is a product of elementary matrices with A RREF. Then A RREF is the product of elementary matrices with A. Since the determinant of an elementary matrix is never zero an it is given that det(A) ≠ 0, then det(A RREF) is not zero. - Thus A RREF does not have a row of zeros. - Thus A is the identity matrix. RREF - Thus A is invertible. −1 30.4 Theorem − If the matrix A is invertible then det(A ) = 1 / det(A). Proof : The proof is straightforward and so is left as an exercise. T 30.5 For any matrix A, det(A) = det(A ) Proof : Case 1 : A is not invertible: - If A is not invertible then det(A) = 0. T - If A is not invertible A is not invertible (Since A B = I implies B A = I implies A is invertible – CTD!) T - Then det(A ) = 0. T - Thus, if A is not invertible, det(A) = 0 = det(A ). Case 2 : A is invertible: Case 2.1: A is an elementary matrix − If A is an elementary matrix of type I or II, we see that Ais symmetric and so A = T A . T − Hence in this ca
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