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3.1 Determinants The Laplace Expansion What’s the use? • All square matrices have a real number determinant • We can use the determinant to see if the matrix is invertible. • We can use the determinant to find the inverse. • We can also use determinants to solve systems of equations (using Cramer’s Rule) Definition • Given a (1 x 1) matrix A = [a]: • • • • det[a] = a We know A is invertible as long as a ≠ 0. So for a (1x1) A is invertible iff a≠0. A-1 = [1/a] Definition for (2x2) • Recall what the inverse of a (2x2) matrix was: a b A c d 1 d b A ad bc c a 1 • The inverse clearly exists as long as (ad - bc) ≠ 0. • We call det A = ad - bc • Again, A is invertible iff det A ≠ 0 Minors and Cofactors • Mij(A) = (i,j) minor of an n x n matrix, A: • is the determinant of the (n-1)x(n-1) matrix formed from A by deleting row i and column j. • Cij(A) = (i,j) cofactor of A • = (-1)i+jMij(A) • C = ± M depending on row and column …continued • Since Cij(A) = (-1)i+jMij(A), note the simple method we can use for finding the sign to apply to M to get C based on the position of cofactor in A: ... ... ... ... ... ... ... ... ... Example: • Find the minors and cofactors of positions (2,1),(2,2) and (2,3). 2 3 1 A 0 1 2 3 1 4 Determinant of an (nxn), n≥2 • Laplace Expansion: • We can use the Laplace Expansion along any row or column to find the determinant of (n x n) mtx, A: – Expansion along rowi: det(A) = ai1Ci1(A) + ai2Ci2(A) + … + ainCin(A) – Expansion along column j: det(A) = a1jC1j(A) + a2jC2j(A) + … + anjCnj(A) Example • Find the determinant of our earlier matrix: 2 3 1 A 0 1 2 3 1 4 Example • Find the determinant of the following matrix: 0 3 0 2 4 1 B 1 1 3 0 3 2 0 0 1 1 Helpful Hints to reduce work • Note that if we have a matrix with a row or column of zeros, the determinant is 0. • Recall also that we can apply elementary row operations to matrix A to create a row of zeros in some matrices. • We can create some zeros in a row by applying elementary row operations in all matrices. • The question is, what effect will applying the row operations have on the determinant. Theorem 2 • Let A be an (n x n) matrix. 1 If A has a row or column of zeros, det A = 0. 2 If two distinct rows (or columns) of A are interchanged, the determinant of the resulting matrix is (- det A). 3 If a row (or column) of A is multiplied by a constant, u, the determinant of the resulting matrix is u(det A). 4 If two distinct rows (or columns) of A are identical, det A = 0. 5 If a mult of a row of A is added to a diff’t row, the determinant of the resulting mtx is det A (cols too). Proof • Property 2: Proof by induction, • Prove for n=2: a11 A a21 a12 ,det A a11 a22 a21 a12 a22 a21 B a11 a22 , det B a21 a12 a11 a22 det A a12 Prop 2 (cont) • For A = (nxn),n>2, we swap rows c and d to get B. – In finding det A and det B, we expand on a row other than c or d, we get the same coefficients w/ same signs, and the minors are the same except rows c and d are swapped. – We continue this process, always getting the same coefficients w/ the same signs until we get down to (2x2)’s – These (2x2)’s will be c d – We have shown that for (2x2)’s, det B = -det A, so for each of our terms (which have same coeff’s so far), the det B = detA, so all of our terms will simply have opp signs. Proof of Property 4 • If two distinct rows (or cols) of A are same, det A=0 – – – – First swap the two rows that are the same to get B B = A so det A = det B Also, by prop. 2, det B = - det A So det A = - det A which can only happen if det A = 0. • Proof of property 5 • Proof: B is obtained from A by adding k times row c to row d. – So row d of B is now: [ad1 + kac1, ad2 + kac2,…,adn + kacn) – The cofactors of this row do not depend on their values, only on their position and on the rest of the matrix, so they remain the same. So Cdj (B) = Cdj (A) n – If we expand along row d of B: det B (a ka )C (B) dj cj dj j 1 n n j 1 j 1 adj Cdj (A) k acj Cdj (A) det A k det C Continued... • This matrix C is just obtained from A by replacing row d by row c since the coefficients are from row c and we expand along d so that row c remains in the minor matrix. • Therefore, matrix C has two identical rows, and by property 4, det C=0 • Therefore, det B=det A• Finding determinants is easier • Now we can simplify the matrices of which we need to find the determinant: • row or column reduce to create more zeros • swapping rows/cols changes sign • adding multiples of a row/col to another doesn’t change • factor out a common factor from a row or column • pulls out as a multiplier on the determinant Examples 2 3 1 0 1 1 2 3 4 a if ...det d g b e h c f 2 i 2g 2h find...det d a e b g h 2i f c i Examples • Find all x such that det A = 0 (making A not invertible). 1 x x A x 1 x x x 1 Examples • show that: 1 a1 2 a1 1 a2 2 a2 1 a3 a3 a2 a3 a1 a2 a1 2 a3 • Matrix of this type is called a Vandermonde matrix. • Can extend formula to (n x n) case Theorem 3 • If A is (n x n), then det(uA) = un det A for any u (try it…) • Proof: R1 uR1 R2 uR2 A ,,,,uA ... ... R uR n n R1 R1 R1 R uR2 2 R2 n 2 det (uA) u det u ... u ... ... ... uR uR R n n n Triangular matrices • Lower Triangular: all entries above main diagonal are zero • Upper Triangular: all entries below main diagonal are zero • Note what happens with the determinant: a 0 0 0 d g 0 h i 0 0 j b c e f • Theorem 4: the determinant of a triangular matrix is the product of the entries on the main diagonal. This makes life easy! • Can we carry all matrices to triangular form using row operations? Of course! • So to find the determinant, reduce to triangular form noting the row operations you have used along the way. • Remember: • when you swap rows, switch sign of determinant • when you mult a row times a value, the det of what remains must be multiplied by recipricol • when you add a mult of one row to another, nothing changes. Continued.. • Note that if when you reduce, you end up with a row of zeros, the determinant will be zero (as it should be since the matrix will not be invertible. Theorem 5 • Given the following block matrices w/ A and B square: A det 0 X det A det B B A 0 det A det B det Y B Proof of Theorem 5(ind) • show true for A (1 x 1) A det 0 a11 X det B 0 X a11 M11 a11 det B det A det B B Proof of Theorem 5(ind) • assume true for A (k x k) A det 0 X det A det B B For A (k x k) • Prove true for A ((k+1) x (k+1)) A X T 0 B det T a11 M11 a21 M21 ... ak 1 Mk 1 Proof of Theorem 5(ind) • Mi1 is the det of submatrix (Si1(T)) formed by deleting the ith row and 1st col of T Si1 (A) Xi Since only getting rid of row of A Si1 (T ) B and same row of X, and col 1 ofA 0 and col 1 of 0. • Si1 (A) will be (k x k). • Mi1(T) =det(Si1(T)) = det(Si1(A))detB (by induction) =Mi1(A)detB Proof of Theorem 5(ind) det T a11 M11 a21 M21 ... ak 1 Mk 1 a11 M11 (A) det B a21 M21 (A)det B ... ak 1 Mk 1 (A)det B det B(a11 M11 (A) a21 M21 (A) ... ak 1 Mk 1 ( A)) det B(det A) •