Determinants Chapter 3 of Lay


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1 Determinants Chapter of Lay Dr. Doreen De Leon Math 152, Fall Introduction to Determinants Section.1 of Lay Given a square matrix A = [a ij, the determinant of A is denoted by det A or a 11 a 1j a 1n... a i1 a ij a in.... a n1 a nj a nn Recall, [ a b  If A =, then c d det A = a b c d = ad bc. a 11 a 12 a 1  If A = a 21 a 22 a 2, then we can find det A by a 1 a 2 a a 11 a 12 a 1 a 11 a 12 a 21 a 22 a 2 a 21 a 22 a 1 a 2 a a 1 a 2 to obtain det A = a 11 a 22 a + a 12 a 2 a 1 + a 1 a 21 a 2 a 1 a 22 a 1 a 2 a 2 a 11 a a 21 a 12. For larger matrices, there is no similar closed form expression. We can define the determinant of a matrix recursively. In general, an n n determinant is defined by determinants of (n 1) (n 1) submatrices. 1
2 Denote the submatrix of A formed by deleting the i th row and j th column of A by A ij. 5 Example: A = Find A 12 and A Solution: 7 0 A 12 = 2 1 1, A = Definition. For n 2, the determinant of an n n matrix A = [a ij is the sum of n terms given by det A = a 11 det A 11 a 12 det A ( 1) 1+n a 1n det A 1n n = ( 1) 1+j a 1j det A 1j. i=1 5 Example: Find det A for A = Solution: Using the formula given in the definition, we have = ( 1)1+1 (1) ( 1)1+2 (2) ( 1)1+ (5) = 2( 7) + 5() = 26. Definition. Given an n n matrix A, the (i, j) cofactor of A, denoted C ij, is given by Then, using this, we define C ij = ( 1) i+j det A ij. det A = a 11 C 11 + a 12 C a 1n C 1n. (1) Equation (1) is a cofactor expansion across the first row. Theorem 1. The determinant of an n n matrix A can be computed by a cofactor expansion across any row i, det A = a i1 C i1 + a i2 C i2 + + a in C in, or down any column j, det A = a 1j C 1j + a 2j C 2j + + a nj C nj. 2
3 Example: Compute det A, where A = , 5 0 using a cofactor expansion. Solution: We choose to do a cofactor expansion along row 2. det A = a 21 C 21 + a 22 C 22 + a 2 C 2 + a 2 C = 0 C C 22 + ( 1) C = We will use a cofactor expansion along row to compute this determinant. det A = (a 1 C 1 + a 2 C 2 + a C ) ( ) = 5( 1) C 22 + ( 1) = [5( 10 ( 12)) + ( 6 ( )) = 6. Theorem 2. If A is a triangular matrix, then det A is the product of the entries on the main diagonal. Example: Evaluate Solution: = 1(1)(6)(2)(7)(5) =
4 2 Properties of Determinants Section.2 of Lay Theorem (Row Operations). Let A be a square matrix. (a) If a multiple of one row of A is added to another row to produce a matrix B, then det B = det A. (b) If two rows of A are interchanged to produce B, then det B = det A. (c) If one row of A is multiplied by k to produce B, then det B = k det A. Example: Let A = We form matrix B by r 2 2r = B Then, det A = 0 C C 2 + 5C = 5( 1) = 5( ) = 20. det B = 1( )(5) = 20. So, we see that det B = 20 = det A Example: Let A = 0 5. We form matrix B as follows: r 1 2r = B Then, det A = 1 ()(6) = 9 2 and det B = 1()(6) = 18. So, we see that det B = 2 det A. Example: Let A = We form matrix B by r 2 r
5 Then, So, we see that det B = det A. det A = 1(1)(5) = 5. det B = 0 C C C 2 = 5( 1) = 5. Example: Compute det A by row reducing to echelon form, where A = Solution: = r 2 r 2 +2r 1 r r r 1 r r r 1 = r 7 r = 27(1)(1)(1)() = = r r +r 2 r r +r 2 = r r 18r Lemma 1. If E is an elementary matrix, then det(ea) = det(e) det(a). Theorem. A square matrix A is invertible if and only if det A 0. Proof. A can be reduced to echelon form U with a finite number of row operations, so U = E k E k 1 E 1 A, where E i represents an elementary matrix. Then, det(u) = det(e k E k 1 E 1 A) = det(e k ) det(e k 1 ) det(e 1 ) det(a). det(e i ) 0 for all i. Therefore, det(a) = 0 if and only if det(u) = 0. 5
6 If A is invertible, then U = I, so det(u) = 1 = det(a) 0. If det(a) = 0, then det(u) = 0 = U contains a row consisting entirely of zeros (since det U = u 11 u 22 u nn ). Therefore, A is not invertible. Column Operations We can perform column operations on a matrix in the same way we perform row operations. Column operations have the same effect on determinants as row operations. NOTE: Do NOT perform column operations when solving systems of equations. Theorem 5. If A is a square matrix, then det A T = det A. Theorem 6. If A and B are n n matrices, then det AB = (det A)(det B). Proof. If A or B is not invertible, then neither is AB. So, det AB = 0 = (det A)(det B). If A is invertible, then A = E k E k 1 E 1 I = E k E k 1 E 1 (since A is rowequivalent to I n ). So, det AB = det(e k E k 1 E 1 B) = det(e k ) det(e k 1 ) det(e 1 ) det(b) = det(e k E k 1 E 1 ) det(b) = det(a) det(b). Corollary 1. Let A be an n n matrix. Then det(a k ) = [det A k. 2 7 Example: Let A = Find det(a ) Solution: Since det(a ) = [det(a), det A = 2(9)(2) = 6. det(a ) = [6 =
7 Example: Show that if A is invertible, then Solution: Since A is invertible, det(a 1 ) = 1 det A. AA 1 = I = det(aa 1 ) = det I = (det A)(det A 1 ) = 1 = det(a 1 ) = 1 det A. Cramer s Rule Section. of Lay We first need the following notation. For an n n matrix A and any b R n, let A i (b) be the matrix obtained from A by replacing column i of A by the vector b; so, A i (b) = [ a 1 a i 1 b a i+1 a n. Theorem 7 (Cramer s Rule). Let A be an invertible n n matrix. For any b R n, the unique solution x of Ax = b has entries given by x i = det(a i(b)), for i = 1, 2,..., n. det A Proof. Let A = [ a 1 a 2 a n and I = [ e1 e 2 e n. If Ax = b, then AI i (x) = A [ e 1 e i 1 x e i+1 e n Then, (det A)(det(I i (x)) = det(a i (b), and = [ Ae 1 Ae i 1 Ax Ae i+1 Ae n = [ Ae 1 Ae i 1 b Ae i+1 Ae n = [ a 1 a i 1 b a i+1 a n = Ai (b). x i det(a) = det(a i (b)) = x i = det(a i(b)) det A. Note that det A 0 since A is invertible. Example: Use Cramer s rule to solve 2x 1 + x 2 = 7 x 1 + x = 8 x 2 + 2x = 7.
8 Solution: For this problem, A = 0 1 and det A =. 0 Applying Cramer s rule, we have x 1 = = = x 2 = = = x = = = 1. So, x = (, 1, 1). Use Cramer s Rule for Engineering Applications Systems of first order differential equations solved using Laplace transforms can lead to systems of equations like 6sx 1 + x 2 = 5 9x 1 + 2sx 2 = 2. We need to know (a) for what values of s the is solution unique, and (b) the solution. First, we know from Cramer s rule that if the coefficient matrix A is invertible, the solution is unique. Since [ 6s A =, 9 2s 8
9 we have 6s 9 2s = 12s2 6 = 12(s 2 ) = 12(s + )(s ). Therefore, the system has a unique solution if s ±. For such an s, we have 5 2 2s x 1 = A = 10s 8 12(s 2 ). 6s x 2 = A 12s 5 = 12(s 2 ). A Formula for A 1 Cramer s rule leads to a general formula for the inverse of an n n matrix as follows. The j th column of A 1 is a vector x that satisfies Ax = e j. The i th entry of x is the (i, j) th entry of A 1. By Cramer s rule, the We can show that where So, (i, j) th entry of A 1 = x i = det(a i(e j )) det A. det(a i (e j )) = ( 1) i+j det A ji = C ji, A ji = (n 1) (n 1) matrix formed by deleting row j and column i of A. A 1 = 1 det A C 11 C 21 C n1 C 12 C 22 C n2.... C 1n C 2n C nn The matrix [C ij is the adjoint of A, adj A. This leads us to the following theorem. 9
10 Theorem 8. Let A be an invertible n n matrix. Then 0 0 Example: Find A 1 if A = A 1 = 1 adj A. det A Solution: First, note that det A = (1)(2) = 6. Then, C 11 = ( 1) = 2 C 12 = ( 1) = 2 C 1 = ( 1) = ( 2) = 1 C 21 = ( 1) = 0 C 22 = ( 1) = 6 C 2 = ( 1) = 9 C 1 = ( 1) = 0 C 2 ( 1) = 0 C = ( 1) = So, A 1 = =
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