MATH136 Lecture Notes - Lecture 29: Elementary Matrix, Mathematical Induction, Natural Number
Friday, July 7
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Lecture 29 : Determinants of elementary matrices. (Refers to 5.3)
Concepts:
1. Recognize the effects the 3 elementary row (column) operations have on the value
of the determinant of a matrix.
2. Determinants of elementary matrices
3. Recognize that det(A) = det(AT).
4. Recognize conditions on A that force det(A) = 0.
5. 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 = c det A” is easily obtained by computing
det B by expanding B along row Ri. Proof is given:
(Note that, for each j = 1 to n, crj for the matrix B is equal to crj for the matrix A.)
- 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. We will only prove the case cRi + Rj.
Claim “A → cRi + Rj→ B implies det B = det A”. Proof of claim by induction.
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, kR2 + R1) and second row is (c, d).
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Friday, july 7 lecture 29 : determinants of elementary matrices. (refers to 5. 3) 29. 1 theorem suppose a and b are both n n matrices. The proof of a cri b implies det b = c det a is easily obtained by computing det b by expanding b along row ri. Proof is given: (note that, for each j = 1 to n, crj for the matrix b is equal to crj for the matrix a. ) 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. We will only prove the case cri + rj. Claim 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).