Definition The determinant of a 2 2 matrix A is denoted |A| and is given by Observe that the determinant of a 2 2 matrix is given by the different of the products of the two diagonals of the matrix. The notation det(A) is also used for the determinant of A. Definition The determinant of a 2 2 matrix A is denoted |A| and is given by Observe that the determinant of a 2 2 matrix is given by the different of the products of the two diagonals of the matrix. The notation det(A) is also used for the determinant of A.
Linear Algebra Chapter Determinants 3.1 Introduction to Determinants Definition The determinant of a × matrix A is denoted |A| and is given by a11 a12 a21 a22 = a11a22 − a12 a21 Observe that the determinant of a × matrix is given by the different of the products of the two diagonals of the matrix The notation det(A) is also used for the determinant of A Example A = 4 − 1 det( A) = = (2 ×1) − (4 × (−3)) = + 12 = 14 −3 Ch03_2 Definition Let A be a square matrix The minor of the element aij is denoted Mij and is the determinant of the matrix that remains after deleting row i and column j of A The cofactor of aij is denoted Cij and is given by Cij = (–1)i+j Mij Note that Cij = Mij or −Mij Ch03_3 Example Determine the minors and cofactors of the elements a11 and a32 of the following matrix A 3 1 A = − 2 0 − 1 Solution Minor of a11 : M 11 = − = − = (−1× 1) − (2 × (−2)) = −2 −2 Cofactor of a11 : C11 = (−1)1+1 M 11 = (−1) (3) = 3 Minor of a32 : M 32 = − = = (1× 2) − (3 × 4) = −10 −2 Cofactor of a32 : C32 = (−1)3+2 M 32 = (−1) (−10) = 10 Ch03_4 Definition The determinant of a square matrix is the sum of the products of the elements of the first row and their cofactors If A is × 3, A = a11C11 + a12C12 + a13C13 If A is × 4, A = a11C11 + a12C12 + a13C13 + a14C14 If A is n × n, A = a11C11 + a12C12 + a13C13 + + a1nC1n These equations are called cofactor expansions of |A| Ch03_5 Example Evaluate the determinant of the following matrix A − 1 A = 3 1 1 4 Solution A = a11C11 + a12C12 + a13C13 = 1(−1) + 2(−1)3 + (−1)(−1) 4 = [(0 × 1) − (1× 2)] − 2[(3 ×1) − (1× 4)] − [(3 × 2) − (0 × 4)] = −2 + − = −6 Ch03_6 Theorem 3.1 The determinant of a square matrix is the sum of the products of the elements of any row or column and their cofactors A = ai1Ci1 + 2Ci + + ainCin ith row expansion: jth column expansion: A = a1 j C1 j + a2 j C2 j + + anj Cnj Example Find the determinant of the following matrix using the second row − 1 A = 3 1 1 4 Solution A = a21C21 + a22C22 + a23C23 = −3 − + − − 1 2 4 = −3[(2 ×1) − (−1× 2)] + 0[(1×1) − (−1× 4)] − 1[(1× 2) − (2 × 4)] = −12 + + = −6 Ch03_7 Example Evaluate the determinant of the following × matrix 4 2 2 0 − 5 7 − − 3 0 Solution A = a13C13 + a23C23 + a33C33 + a43C43 = 0(C13 ) + 0(C23 ) + 3(C33 ) + 0(C43 ) = −1 −3 = 6(3 − 2) = = 3(2) − 1 −3 Ch03_8 Example Solve the following equation for the variable x x x +1 = −1 x − Solution Expand the determinant to get the equation x( x − 2) − ( x + 1)(−1) = Proceed to simplify this equation and solve for x x2 − 2x + x +1 = x2 − x − = ( x + 2)( x − 3) = x = −2 or There are two solutions to this equation, x = – or Ch03_9 Computing Determinants of × and × Matrices a11 A= a21 a12 a22 ⇒ A = a11a22 − a12 a21 a11 A = a21 a 31 a12 a22 a32 a13 a11 a23 ⇒ a21 a a33 31 a12 a22 a32 a13 a11 a23 a21 a33 a31 a12 a22 a32 ⇒ A = a11a22 a33 + a12 a23 a31 + a13 a21a32 (diagonal products from left to right) − a13 a22 a31 − a11a23 a32 − a12 a21a33 (diagonal products from right to left) Ch03_10 Theorem 3.6 Let A be a square matrix with |A| ≠ A is invertible with −1 A = adj( A) A Proof Consider the matrix product A⋅adj(A) The (i, j)th element of this product is (i, j ) th element = (row i of A) × (column j of adj( A)) C j1 C j = [ ai1 ain ] C jn = ai1C j1 + 2C j + + ain C jn Ch03_29 Proof of Theorem 3.6 If i = j, ai1C j1 + 2C j + + ain C jn = A If i ≠ j, let A ⇒ Rj is replaced by Ri B Matrices A and B have the same cofactors Cj1, Cj2, …, Cjn So ai1C j1 + 2C j + + ain C jn = B = A if i = j Therefore (i j ) th element = 0 if i ≠ j 1 Since |A| ≠ 0, A adj( A) = I n A 1 Similarly, adj( A) A = I n A row i = row j in B ∴ A⋅ adj(A) = |A|In Thus A = adj( A) A −1 Ch03_30 Theorem 3.7 A square matrix A is invertible if and only if |A| ≠ Proof (⇒) Assume that A is invertible ⇒ AA–1 = In ⇒ |AA–1| = |In| ⇒ |A||A–1| = ⇒ |A| ≠ (⇐) Theorem 3.6 tells us that if |A| ≠ 0, then A is invertible A–1 exists if and only if |A| ≠ Ch03_31 Example Use a determinant to find out which of the following matrices are invertible − 3 − 1 − A= B= C = 12 − D = − 1 2 3 2 2 1 1 − 0 Solution |A| = ≠ |B| = |C| = |D| = ≠ A is invertible B is singular The inverse does not exist C is singular The inverse does not exist D is invertible Ch03_32 Example Use the formula for the inverse of a matrix to compute the inverse 3 of the matrix A = − − 2 5 − Solution |A| = 25, so the inverse of A exists.We found adj(A) in Example 14 − − 12 adj( A) = 1 8 − 14 12 − − 14 − − 12 25 25 25 1 −1 A = adj( A) = 1 = A 25 25 25 25 8 8 − − 25 25 25 Ch03_33 Exercise 3.3 page 178-179: 4, Exercise Show that if A = A-1, then |A| = ±1 Show that if At = A-1, then |A| = ±1 Ch03_34 Theorem 3.8 Let AX = B be a system of n linear equations in n variables (1) If |A| ≠ 0, there is a unique solution (2) If |A| = 0, there may be many or no solutions Proof (1) If |A| ≠ ⇒ A–1 exists (Thm 3.7) ⇒ there is then a unique solution given by X = A–1B (Thm 2.9) (2) If |A| = ⇒ since A ≈…≈ C implies that if |A|≠0 then |C|≠0 (Thm 3.2) ⇒ the reduced echelon form of A is not In ⇒ The solution to the system AX = B is not unique ⇒ many or no solutions Ch03_35 Example Determine whether or not the following system of equations has an unique solution x1 + x2 − x3 = x1 + x2 + x3 = −5 Solution x1 + x2 + x3 = Since 3 −2 =0 Thus the system does not have an unique solution Ch03_36 Theorem 3.9 Cramer’s Rule Let AX = B be a system of n linear equations in n variables such that |A| ≠ The system has a unique solution given by A1 A2 An x1 = , x2 = , , xn = A A A Where Ai is the matrix obtained by replacing column i of A with B Proof |A| ≠ ⇒ the solution to AX = B is unique and is given by X = A−1 B = adj( A) B A Ch03_37 Proof of Cramer’s Rule xi, the ith element of X, is given by xi = [row i of adj( A)] × B A b1 = [ C1i C2i Cni ] b2 A b n = (b1C1i + b2C2i + + bnCni ) A Ai Thus xi = A the cofactor expansion of |Ai| in terms of the ith column Ch03_38 Example Solving the following system of equations using Cramer’s rule x1 + x2 + x3 = −2 x1 + x2 + x3 = −5 x1 + x2 + x3 = Solution The matrix of coefficients A and column matrix of constants B are 1 − 2 A = 2 1 and B = − 5 3 6 It is found that |A| = –3 ≠ Thus Cramer’s rule be applied We get − 1 − 1 − 2 A1 = − 5 1 A2 = 2 − 1 A3 = 2 − 5 3 6 3 Ch03_39 Giving A1 = −3, A2 = 6, A3 = −9 Cramer’s rule now gives A1 − A2 A3 − x1 = = = 1, x2 = = = −2, x3 = = =3 A −3 A −3 A −3 The unique solution is x1 = 1, x2 = −2, x3 = Ch03_40 Example Determine values of λ for which the following system of equations has nontrivial solutions.Find the solutions for each value of λ (λ + 2) x + (λ + 4) x = Solution 2 x1 + (λ + 1) x2 = homogeneous system ⇒ x1 = 0, x2 = is the trivial solution ⇒ nontrivial solutions exist ⇒ many solutions ⇒ λ+2 λ+4 =0 λ λ +1 ( λ + )( λ + ) − ( λ + ) = λ ⇒ ⇒ + λ − = ⇒ (λ − 2)(λ + 3) = ⇒ λ = – or λ = Ch03_41 λ = – results in the system − x1 + x2 = x1 − x2 = This system has many solutions, x1 = r, x2 = r λ = results in the system x1 + x2 = x1 + 3x2 = This system has many solutions, x1 = – 3r/2, x2 = r Ch03_42 Homework Exercise 3.3 pages 179-180: 8, 12, 14, 15, 17 Ch03_43 [...]... 2 3 5 2 = 3 R2 + R1 4 R3 + (−2)R1 1 R4 + (−6)R1 1 −1 0 2 0 0 2 5 =0 0 0 3 0 0 0 5 − 11 diagonal element is zero and all elements below this diagonal element are zero Ch03_26 3.4 Determinants, Matrix Inverse, and Systems of Linear Equations Definition Let A be an n × n matrix and Cij be the cofactor of aij The matrix whose (i, j)th element is Cij is called the matrix of cofactor of A The transpose of... determinant − 1 − 6 3 2 9 −3 Solution 3 4 −2 −1 − 6 3 2 9 −3 = C2 + 2 C3 3 0 −2 3 −2 −1 0 3 = (−3) = −21 −1 3 2 3 −3 Ch03_13 Example 2 4 3 1 2 5, |A| = 12 is known If A = 0 − 2 − 4 10 Evaluate the determinants of the following matrices 4 3 1 12 3 1 1 4 3 (a ) B1 = 0 6 5 (b) B2 = − 2 − 4 10 (c) B3 = 0 2 5 − 2 − 12 10 0 0 4 16 2 5 Solution (a) A ≈ B1 Thus |B1| =... (assuming A–1 exists) Proof (a) (d) A ≈ cR1, cR 2 , , cRn cA ⇒ cA = c n A A ⋅ A−1 = A ⋅ A−1 = I = 1 ⇒ A−1 = 1 A Ch03_17 Example 4 If A is a 2 × 2 matrix with |A| = 4, use Theorem 3.4 to compute the following determinants (a) |3A| (b) |A2| (c) |5AtA–1|, assuming A–1 exists Solution (a) |3A| = (32)|A| = 9 × 4 = 36 (b) |A2| = |AA| =|A| |A|= 4 × 4 = 16 1 t –1 2 t –1 t –1 = 25 A = 25 (c) |5A A | = (5 )|A A | =... singular implies that either A or B is singular The inverse is not true Ch03_19 Homework Exercise 3.2 pp 170-171: 4, 5, 9, 10, 11, 13, 19 Exercise 11 Prove the following identity without evaluating the determinants a+ b c+ d e+ f Solut io n a+b c+d p q u v p u q v a c e b d f r = p q r+ p q r w u v w u v w e+ f q r p r p q r = ( a + b) − (c + d ) + (e + f ) v w u w u v w Ch03_20 3.3 Numerical Evaluation...Homework Exercise 3.1 pages161-162: 3, 6, 9, 11, 13, 14 Ch03_11 3.2 Properties of Determinants Theorem 3.2 Let A be an n × n matrix and c be a nonzero scalar (a) If A ≈ B then |B| = c|A| cRk (b) If A ≈ B then |B| = –|A| Ri↔ Rj (c) If A ≈ B then |B| = |A| Ri+ cRj Proof (a) |A| = ak1Ck1... 1 6 8 − 1 − 25 25 25 Ch03_33 Exercise 3.3 page 178-179: 4, 7 Exercise Show that if A = A-1, then |A| = ±1 Show that if At = A-1, then |A| = ±1 Ch03_34 Theorem 3.8 Let AX = B be a system of n linear equations in n variables (1) If |A| ≠ 0, there is a unique solution (2) If |A| = 0, there may be many or no solutions Proof (1) If |A| ≠ 0 ⇒ A–1 exists (Thm 3.7) ⇒ there is then a unique solution