Wednesday, March 19 − Lecture 29 : Determinants of elementary matrices.
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:
- 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 ×