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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