13. Elementary Matrices and Determinants II
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1 3 Elementary Matrices and Determinants II In the last Section, we saw the definition of the determinant and derived an elementary matrix that exchanges two rows of a matrix Next, we need to find elementary matrices corresponding to the other two row operations, multiplying a row by a scalar, and adding a multiple of one row to another As a consequence, we will derive some important properties of the determinant Consider M = R R n, where R i are row vectors Let R i (λ) be the identity matrix, with the ith diagonal entry replaced by λ, not to be confused with the row vectors Then: R M = λr i = R i (λ)m R n What effect does multiplication by R i (λ) have on the determinant? det M = σ sgn(σ)m σ() λmi σ(i) mn σ(n) = λ σ sgn(σ)m σ() mi σ(i) mn σ(n) = λ det M Thus, multiplying a row by λ multiplies the determinant by λ Since R i (λ) is just the identity matrix with a single row multiplied by λ, then by the above rule, the determinant of R i (λ) is λ Thus: det R i (λ) = det λ = λ The final row operation is adding λr j to R i This is done with the matrix Sj i (λ), which is an identity matrix but with a λ in the i, j position
2 Sj(λ) i = λ Then multiplying Sj i (λ) by M gives the following: λ R i R i + λr j = R j R j What is the effect of multiplying by Sj i (λ) on the determinant? Let M = Sj i(λ)m, and let M be the matrix M but with R i replaced by R j det M = σ sgn(σ)m σ() (mi σ(i) + λmj σ(j) ) mn σ(n) = σ sgn(σ)m σ() mi σ(i) mn σ(n) + σ sgn(σ)m σ() λmj σ(j) mj σ(j) mn σ(n) = det M + λ det M Since M has two identical rows, its determinant is 0 Then det Sj i(λ)m = det M Notice that if M is the identity matrix, then we have det Sj i(λ) = det(sj i (λ)i) = det I = We now have an elementary matrices associated to each of the row operations 2
3 E i j = I with rows i, j swapped; det E i j = R i (λ) = I with λ in position i, i; det R i j (λ) = λ S i j (λ) = I with λ in position i, j; det Si j (λ) = We have also proved the following theorem along the way: Theorem If E is any of the elementary matrices Ej i, Ri (λ), Sj i (λ), then det(em) = det E det M We have seen that any matrix M can be put into reduced row echelon form via a sequence of row operations, and we have seen that any row operation can be emulated with left matrix multiplication by an elementary matrix Suppose that RREF(M) is the reduced row echelon form of M Then RREF(M) = E E 2 E k M where each E i is an elementary matrix What is the determinant of a square matrix in reduced row echelon form? If M is not invertible, then some row of RREF(M) contains only zeros Then we can multiply the zero row by any constant λ without changing M; by our previous observation, this scales the determinant of M by λ Thus, if M is not invertible, det RREF(M) = λ det RREF(M), and so det RREF(M) = 0 Otherwise, every row of RREF(M) has a pivot on the diagonal; since M is square, this means that RREF(M) is the identity matrix Then if M is invertible, det RREF(M) = Additionally, notice that det RREF(M) = det(e E 2 E k M) Then by the theorem above, det RREF(M) = det(e ) det(e k ) det M Since each E i has non-zero determinant, then det RREF(M) = 0 if and only if det M = 0 Then we have shown: Theorem For any square matrix M, det M 0 if and only if M is invertible Since we know the determinants of the elementary matrices, we can immediately obtain the following: Corollary Any elementary matrix Ej i, Ri (λ), Sj i (λ) is invertible, except for R i (0) In fact, the inverse of an elementary matrix is another elementary matrix 3
4 To obtain one last important result, suppose that M and N are square n n matrices, with reduced row echelon forms such that, for E i and F i elementary matrices, M = E E 2 E k RREF(M), N = F F 2 F k RREF(N) = N If RREF(M) is the identity matrix (ie, M is invertible), then: det(mn) = det(e E 2 E k RREF(M)F F 2 F k RREF(N)) = det(e E 2 E k IF F 2 F k RREF(N)) = det(e ) det(e k ) det(i) det(f ) det(f k ) det(rref(n) = det(m) det(n) Otherwise, M is not invertible, and det M = 0 = det ref(m) Then there exists a row of zeros in ref(m), so R n (λ)ref(m) = ref(m) Then: det(mn) = det(e E 2 E k ref(m)n) = det(e E 2 E k ref(m)n) = det(e ) det(e k ) det(ref(m)n) = det(e ) det(e k ) det(r n (λ)ref(m)n) = det(e ) det(e k )λ det(ref(m)n) = λ det(mn) Which implies that det(mn) = 0 = det M det N Thus we have shown that for any matrices M and N, References det(mn) = det M det N Hefferon, Chapter Four, Section I and I3 Wikipedia: Determinant Elementary Matrix Review Questions ( ) ( ) a b x y Let M = and N = Compute the following: c d z w 4
5 det M det N det(mn) det M det N det(m ) assuming ab cd 0 det(m T ) det(m + N) det M det N ( ) a b 2 Suppose M = is invertible Write M as a product of elementary row matrices times c d RREF(M) 3 Find the inverses of each of the elementary matrices, E i j, Ri (λ), S i j (λ) Make sure to show that the elementary matrix times its inverse is actually the identity 5
k=1 ( 1)k+j M kj detm kj. detm = ad bc. = 1 ( ) 2 ( )+3 ( ) = = 0
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