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Lecture 29.pdf

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MATH 136
Robert Sproule

Wednesday, March 19 − Lecture 29 : Determinants of elementary matrices. Concepts: 1. Recognize the effects the 3 elementary row (column) operations have on the value of the determinant of a matrix. 2. Recognize that det(A) = det(A ). T 3. Recognize conditions on A that force det(A) = 0. 4. Find determinants by using the properties of determinants. 29.1 Theorem − Suppose A and B are both n × n matrices. Then: Proof : - The proof of “A → cRi→ B implies det B = cdet A” is easily obtained by computing det B by expanding B along row R . i - The proofs of “A → Pij B implies det B = −det A” and “A → cRi+ Rj B implies det B = det A” can both be done by induction. To illustrate how this is done we prove “A → cRi+ Rj→ B implies det B = det A”. Base case: Let A be the 2 × 2 matrix whose first row is (a, b) and second row is (c, * d). Let Let A be the 2 × 2 matrix whose first row is (kc + a, kb+ d) (obtained by the ERO, kR 2 + R )1and second row is (c, d). * The same result is obtained if A is obtained by applying kR + R to A. So 1he 2 statement holds true for n = 2. Induction hypothesis: Let n be a natural number greater than two. Suppose the statement “A → cRi+ Rj→ B implies det B = det A” holds true for all matrices A m × m where m ≤ n – 1. Claim: “A → cRi+ Rj B implies det B = det A” holds true if A is n × n. Let A = [a ] be an n × n matrix (n ≥ 3) and A be obtained by applying cR + R to A. ij * * i j We compute det A by expanding along row R of A . where t ≠ it t ≠j. That is, * For each j = 1 to n, let M be thtjmatrix such that M is obtained by tjplying cR + R to i j M .tjere cR + R iefersjto rows belonging to M , but indexed as tjws of A where i, j ≠ t. * Now, for each j, M is an ij −1) × (n −1) matrix. By the induction hypothesis, det(M ) = ij det(M ).ijo we have Conclusion: By the principle of mathematical induction, “A → cRi+ Rj→ B implies det B = det A” holds true for A n ×
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