# Formula for the inverse matrix. Cramer s rule. Review: 3 3 determinants can be computed expanding by any row or column

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1 Math 20F Linear Algebra Lecture 18 1 Determinants, n n Review: The 3 3 case Slide 1 Determinants n n (Expansions by rows and columns Relation with Gauss elimination matrices: Properties) Formula for the inverse matrix Cramer s rule (To solve nonhomogeneous systems of equations) Review: 3 3 determinants can be computed expanding by any row or column Slide 2 Claim 1 The determinant of a 3 3 matrix can be computed in terms of 2 2 determinants, expanding by any column or row, using the following sign convention for the addition, , Sign of coefficient aij is ( 1)i+j

2 Math 20F Linear Algebra Lecture 18 2 Review: Main properties of 3 3 determinants Let A = [a 1, a 2, a 3 ] be a 3 3 matrix Let c be a 3-vector Slide 3 det([a 1, a 1, a 2 ]) = 0 det([a 1, a 2, a 3 ]) = det([a 2, a 1, a 3 ]) det([a 1, a 2, a 3 ]) = det([a 1, a 3, a 2 ]) det([ca 1, a 2, a 2 ]) = c det([a 1, a 2, a 3 ]) det([a 1 + c, a 2, a 3 ]) = det([a 1, a 2, a 3 ]) + det([c, a 2, a 3 ]) Review: Important results concerning 3 3 determinants Slide 4 Theorem 1 Let A, B be 3 3 matrices Then, A is invertible det(a) 0 a 1, a 2, a 3 are ld det([a 1, a 2, a 3 ]) = 0 det(a) = det(a T ) det(ab) = det(a) det(b)

3 Math 20F Linear Algebra Lecture 18 3 Gauss elimination can be used to compute determinants! Theorem 2 Let A be a 3 3 matrix Slide 5 Let B be the result of adding to a row in A a multiple of another row in A Then, det(b) = det(a) Let B be the result of interchanging two rows in A Then, det(b) = det(a) Let B be the result of multiply a row in A by a number k Then, det(a) = (1/k) det(b) Notation needed for the n n case Slide a 11 a 1j a 1n A ij = a i1 a ij a in, a n1 a nj a nn eliminate the column j, and the row i, Sign of coefficient a ij is ( 1) i+j 7 5

4 Math 20F Linear Algebra Lecture 18 4 Determinant n n: expansion by the first row Definition 1 The determinant of an n n matrix A = [a ij ] is given by Slide 7 det(a) = det(a 11)a 11 det(a 12)a ( 1) 1+n det(a 1n)a 1n, = nx ( 1) 1+j det(a 1j) a 1j = j=1 nx a 1j C 1j, where C ij is called the cofactor of a matrix A and is the number given by C ij = ( 1) i+j det(a ij ) j=1 Determinants can be computed expanding along any row or any column Slide 8 Theorem 3 The determinant of an n n matrix A = [a ij ] can be computed by an expansion along any row or along any column That is, n det(a) = C ij a ij, for any i = 1,, n, = j=1 n C ij a ij, for any j = 1,, n i=1

5 Math 20F Linear Algebra Lecture 18 5 Use the row or column with the most number of zeros to compute the determinant Theorem 4 The determinant of a triangular matrix is the product of its diagonal elements Slide = (1) = = = (1) = = 15 Main properties of n n determinants Slide 10 Let A = [a 1,, a n ] be a n n matrix Let c be a n-vector det([a 1,, a i,, a i,, a n]) = 0 det([a 1,, a i,, a j,, a n]) = det([a 1,, a j,, a i,, a n]) det([a 1,, a j + c,, a n]) = det([a 1,, a j, a n]) + det([a 1,, c,, a n]) det([a 1,, ca j,, a n]) = c det([a 1,, a j,, a n])

6 Math 20F Linear Algebra Lecture 18 6 Important results concerning n n determinants Theorem 5 Let A, B be n n matrices Then, Slide 11 A is invertible det(a) 0 a 1,, a n are ld det([a 1,, a n ]) = 0 det(a) = det(a T ) det(ab) = det(a) det(b) The properties of the determinant on the column vectors of A and the property det(a) = det(a T ) imply the following results on the rows of A Theorem 6 (Determinants and elementary row operations) Let A be a n n matrix Let B be the result of adding to a row in A a multiple of another row in A Then, det(b) = det(a) Let B be the result of interchanging two rows in A Then, det(b) = det(a) Let B be the result of multiply a row in A by a number k Then, det(b) = k det(a) Determinant and Gauss elimination operations Theorem 7 If E represents an elementary row operation and A is an n n matrix, then det(ea) = det(e) det(a) The proof is to compute the determinant of every elementary row operation matrix, E, and then use the previous theorem Theorem 8 (Determinant of a product) If A, B are arbitrary n n matrices, then det(ab) = det(a) det(b) Determinant of a product of matrices Proof: If A is not invertible, then AB is not invertible, then the theorem holds, because 0 = det(ab) = det(a) det(b) = 0 Suppose that A is invertible Then there exist elementary row operations E k,, E 1 such that A = E k E 1

7 Math 20F Linear Algebra Lecture 18 7 Then, det(ab) = det(e k E 1 B), = det(e k ) det(e k 1 E 1 B), = det(e k ) det(e 1 ) det(b), = det(e k E 1 ) det(b), = det(a) det(b) Formula for the inverse matrix Slide 12 Theorem 9 Let A = [a ij ] be an n n matrix, C ij = ( 1) i+j det(a ij ) be the ijth cofactor, and = det(a) Then the inverse matrix A 1 is given by C 11 C 21 C n1 A 1 = 1 C 12 C 22 C n2 C 1n C 2n C nn Formula for the inverse matrix Proof: It is a straightforward computation Let us denote B the matrix with components (B) ij = C ji / Then, a 11 a 12 a 1n C 11 C 21 C n1 a 21 a 22 a 2n 1 C 12 C 22 C n2 AB = a n1 a n2 a nn C 1n C 2n C nn Compute each component of the product AB (AB) 11 = 1 (C 11a 11 + C 12 a C 1n a 1n ) = 1, because the factor in the numerator in the right hand side is precisely det(a) = The second component is given by (AB) 12 = 1 (C 11a 21 + C 12 a C 1n a 2n )

8 Math 20F Linear Algebra Lecture 18 8 The factor between brackets in the right hand side is an expansion by the first row of the determinant of a matrix whose first row is a 21, a 22, a 2n That is, (AB) 12 = 1 a 21 a 22 a 2n a 21 a 22 a 2n a 31 a 32 a 3n a n1 a n2 a nn An analogous calculation shows that (AB) ij is given by = 0 (AB) ij = 1 (C j1a i1 + C j2 a i2 + + C jn a in ), The factor between brackets in the right hand side is an expansion by the j row of the determinant of a matrix whose j row is is the i row of A, That is, a i1, a i2, a in a 11 a 12 a 1n (AB) ij = 1 a i1 a i2 a in a n1 a n2 a nn in the j-row Therefore, when i j the factor between brackets is the determinant of a matrix with two identical rows, so (AB) ij = 0 for i j If i = j, the the that factor is precisely det(a), then (AB) ii = 1 Summarizing, (AB) ij = a 11 a 12 a 1n 1 a i1 a i2 a in in the j-row a n1 a n2 a nn = I ij Repeat this calculation for BA

9 Math 20F Linear Algebra Lecture 18 9 Cramer s rule is a formula to solve nonhomogeneous linear equations Slide 13 Theorem 10 Let A be an invertible n n matrix, so the system of linear equations Ax = b has a unique solution for every vector b Then the components x i of the solution x are given by x i = 1 det(a i(b)) where we introduced the matrix A i(b) = [a 1,, b,, a n], with b placed in the i-column Proof: On the one hand, A invertible means that the solution can be written as x = A 1 b From the formula of the inverse matrix one obtains x i = 1 (C 1i b 1 + C 2i b C ni b n ), Slide 14 where b i are the components of b On the other hand, if one expands the det(a i (b)) by the i row one gets det(a i (b)) = (C 1i b 1 + C 2i b C ni b n )

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