SSM: Linear Algebra Section 61 61 Chapter 6 1 2 1 Fails to be invertible; since det = 6 6 = 0 3 6 3 5 3 Invertible; since det = 33 35 = 2 7 11 5 Invertible; since det 2 5 7 0 11 7 = 2 11 5 + 0 + 0 0 0 0 = 110 0 0 5 7 This matrix is clearly not invertible, so the determinant must be zero 9 Invertible; since det 0 1 2 7 8 3 = 0 + 3 6 + 2 7 5 7 4 2 8 6 = 36 6 5 4 k 2 11 det 0 when 4k 6, or k 3 4 3 2 13 det k 3 5 0 2 6 = 8k, so k 0 will ensure that this matrix is invertible 0 0 4 15 det 0 k 1 2 3 4 = 6k 3 This matrix is invertible when k 1 2 5 6 7 17 det 1 1 1 1 k 1 = 2k 2 2 = 2(k 2 1) = 2(k 1)(k + 1) So k cannot be 1 or -1 1 k 2 1 19 det 1 1 k 1 k k = k 3 + 2k 2 k = k(k 1) 2 So k cannot be 0 or 1 k k k 21 det k 1 1 1 k 1 = k 3 3k + 2 = (k 1) 2 (k + 2) So k cannot be -2 or 1 1 1 k 1 λ 2 23 det(a λi 2 ) = det = (1 λ)(4 λ) = 0 if λ is 1 or 4 0 4 λ 149
Chapter 6 SSM: Linear Algebra 4 λ 2 25 det(a λi 2 ) = det = (4 λ)(6 λ) 8 = (λ 8)(λ 2) = 0 if λ is 2 or 4 6 λ 8 27 A λi 3 is a lower triangular matrix with the diagonal entries (2 λ), (3 λ) and (4 λ) Now, det(a λi 3 ) = (2 λ)(3 λ)(4 λ) = 0 if λ is 2, 3 or 4 29 det(a λi 3 ) = det 3 λ 5 6 0 4 λ 2 = (3 λ)(λ 8)(λ 3) = 0 if λ is 3 or 8 0 2 7 λ 31 This matrix is upper triangular, so the determinant is the product of the diagonal entries, which is 24 1 2 2 3 33 By Fact 618, the determinant is equal to det det = (7 16)(10 21) = 99 8 7 7 5 35 We will use Fact 615 and expand down the third column Our determinant equals: 2 det 2 3 2 6 0 3 7 0 4 6 3 Then we expand down the second column, so we have 2( 3) det = 6(24 21) = 7 4 18 5 4 37 Here we use Fact 618 to find det det 5 6 7 0 1 2 = (35 24)(5) = 55 6 7 0 0 1 39 We repeatedly expand down the first column, finding the determinant to be 0 1 5 4( 2) det = 120 3 0 0 0 1 2 1 3 5 7 41 Here we expand down the fourth column: 2 det, then down the second 2 0 4 6 0 0 3 4 column: 2(3) det 0 1 2 2 4 6 1 2 = 2(3)( 2) det = 12(4 6) = 24 3 4 0 3 4 a b a b 43 For a 2 2 matrix A =, we have det( A) = det = ad bc = det(a) c d c d Use Sarrus rule to see that det( A) = det(a) for a 3 3 matrix We may conjecture 150
SSM: Linear Algebra Section 61 that det( A) = det(a) for an n n matrix with even n, and det( A) = det(a) when n is odd; in both cases we can write det( A) = ( 1) n det(a) A proof of this conjecture can be based upon the following rule: If a square matrix B is obtained from matrix A by multiplying all the entries in the i th row of A by a scalar k, then det(b) = k det(a) We can show this by expansion along the i th row: n n det(b) = ( 1) i+j b ij det(b ij ) = ( 1) i+j ka ij det(a ij ) j=1 = k j=1 n ( 1) i+j a ij det(a ij ) = k det(a) j=1 We can obtain A by multiplying all the entries in the first row of A by k = 1, then all the entries in the second row, and so forth down to the n th row; each time the determinant gets multiplied by k = 1 Thus, det( A) = k n det(a) = ( 1) n det(a), as claimed a b a c 45 If A =, then det(a c d T ) = det = ad cb = det(a) It turns out that b d det(a T ) = det(a) 47 We have det(a) = (ah cf)k + bef + cdg aeg bdh Thus matrix A is invertible for all k if (and only if) the coefficient (ah cf) of k is 0, while the sum bef + cdg aeg bdh is nonzero A numerical example is a = c = d = f = h = g = 1 and b = e = 2, but there are infinitely many other solutions as well 49 The kernel of T consists of all vectors x such that the matrix [ x v w] fails to be invertible This is the case if x is a linear combination of v and w, as discussed on Pages 247 and 248 Thus ker(t ) = span( v, w) The image of T isn t {0}, since T ( v w) 0, for example Being a subspace of R, the image must be all of R 51 We expand down the first column: 2 1 0 0 1 0 0 0 1 2 1 0 1 2 1 0 det(m n ) = 1 det 0 1 2 0 1 det 0 1 2 0 0 0 1 2 0 0 1 2 1 0 0 0 1 = det(a n 1 ) det 0 A = det(a n 1 ) 1 det(a n 2 ) Using Example 9, n 2 0 0 this equals n (n 1) = 1 151
Chapter 6 SSM: Linear Algebra 2 0 0 1 0 0 0 53 det(e n ) = det 1 A n 1 0 0 0 1 0 0 0 0 2 1 0 0 0 0 0 1 2 1 0 0 0 0 0 0 1 2 1 0 0 = 2 det(a n 1 ) 1 det 0 0 0 1 2 0 0 0 0 0 0 0 2 1 0 0 0 0 0 1 2 0 0 1 0 0 0 0 2 1 0 0 0 0 0 1 2 1 0 0 0 0 0 0 1 = 2 det(a n 1 ) 1 det 0 0 0 An 4 0 0 0 0 0 0 = 2 det(a n 1 ) det 0 0 1 2 1 0 det(a n 4 ) (by Fact 618) = 2 det(a n 1 ) 3 det(a n 4 ) 1 2 1 = 2n 3(n 3) = 2n 3n + 9 = 9 n So, det(e 9 ) = 0 1 1 0 0 0 0 1 1 1 0 0 0 0 1 1 1 0 0 55 a d n = det 0 0 1 1 1 0 0 0 0 1 1 0 0 0 0 0 0 1 152
SSM: Linear Algebra Section 62 1 1 0 0 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 0 0 1 1 1 0 = 1 det 0 1 1 1 0 0 0 1 1 1 det 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 = d n 1 1 det 0 M n 2 = d n 1 d n 2 0 1 1 b d 1 = det[1] = 1 d 2 = det = 0 d 1 1 3 = 0 1 = 1 d 4 = 1 0 = 1 d 5 = 1 ( 1) = 0 d 6 = 0 ( 1) = 1 d 7 = 1 0 = 1 d 8 = 1 1 = 0 Note that d 7 = d 1 = 1 and d 8 = d 2 = 0, so that the values are repeated, with a period of 6, meaning that d k = d k+6 for all k c Following part b, we have d 100 = d 100 16(6) = d 4 = 1 57 Repeatedly expanding down the first column, we see that the determinant will be 1 or -1, since it is the product of n terms that are all 1 or -1 59 Let M n be the number of multiplications required to compute the determinant of an n n matrix by Laplace expansion We will use induction on n to prove that M n > n!, for n 3 In the lowest applicable case, n = 3, we can check that M 3 = 9 and 3! = 6 Now let s do the induction step If A is an n n matrix, then det(a) = a 11 det(a 11 ) + + ( 1) n+1 a n1 det(a n1 ), by Definition 614 We need to compute n determinants of (n 1) (n 1) matrices, and then do n extra multiplications a i1 det(a i1 ), so that M n = nm n 1 + n If n > 3, then M n 1 > (n 1)!, by induction hypothesis, so that M n > n(n 1)! + n > n!, as claimed 153
Chapter 6 SSM: Linear Algebra 62 1 A = 1 1 1 1 3 3 I B = 1 1 1 0 2 2 Now det(a) = det(b) = 6, by Algorithm 623b 2 2 5 2I 0 0 3 1 3 2 4 1 3 2 4 1 6 4 8 3 A = I 0 3 2 4 B = Now det(a) = det(b) = 24, by 1 3 0 0 I 0 0 2 4 2 6 4 12 2I 0 0 0 4 Algorithm 623b 1 2 3 4 0 2 3 4 5 After three row swaps, we end up with B = Now, by Algorithm 623b, 0 0 3 4 0 0 0 4 det(a) = ( 1) 3 det(b) = 24 7 After two row swaps, we end up with an upper triangular matrix B with all 1 s along the diagonal Now det(a) = ( 1) 2 det(b) = 1, by Algorithm 623b 9 If we subtract the first row from every other row, then we have an upper triangular matrix B, with diagonal entries 1, 1, 2, 3 and 4 Then det(a) = det(b) = 24 by Algorithm 623b 11 By Fact 621a, the desired determinant is ( 9)(8) = 72 13 By Fact 621b, applied twice, since there are two row swaps, the desired determinant is ( 1)( 1)(8) = 8 15 By Fact 621c, the desired determinant is 8 17 The standard matrix of T is A = 2 3 0 0 2 6, so that det(t ) = det(a) = 8 0 0 2 19 The standard matrix of T is A = 1 0 0 0 1 0, so that det(t ) = det(a) = 1 0 0 1 1 0 0 0 0 1 0 0 21 The standard matrix of T is A =, so that det(t ) = det(a) = 1 0 0 1 0 0 0 0 1 154
SSM: Linear Algebra Section 62 23 Consider the matrix M of T with respect to a basis consisting of n(n + 1)/2 symmetric matrices and n(n 1)/2 skew-symmetric matrices (see Exercises 54 and 55 or Section 53) Matrix M will be diagonal, with n(n + 1)/2 entries 1 and n(n 1)/2 entries -1 on the diagonal Thus, det(t ) = det(m) = ( 1) n(n 1)/2 25 The standard matrix of T is A = 2 0 0 0 2 3, so that det(t ) = det(a) = 16 0 0 4 27 The matrix of T with respect to the basis cos(x), sin(x) is A = det(t ) = det(a) = a 2 + b 2 b a, so that a b 29 Expand down the first column, realizing that all but the first contribution are zero, since a 21 = 0 and A i1 has two equal rows for all i > 2 Therefore, det(p n ) = det(p n 1 ) Since det(p 1 ) = 1, we can conclude that det(p n ) = 1, for all n [ 1 1 31 a If n = 1, then A = a 0 a 1 ], so det(a) = a 1 a 0 (and the product formula holds) b Expanding the given determinant down the right-most column, we see that the coefficient k of t n is the n 1 Vandermonde determinant which we assume is (a i a j ) n 1 i>j Now f(a 0 ) = f(a 1 ) = = f(a n 1 ) = 0, since in each case the given matrix has two identical columns, hence its determinant equals zero Therefore f(t) = n 1 i>j (a i a j ) (t a 0 )(t a 1 ) (t a n 1 ) and det(a) = f(a n ) = (a i a j ), n i>j as required 155
Chapter 6 SSM: Linear Algebra 33 Think of the i th column of the given matrix as a i 1 a i a 2 i ai n 1 1 1 1 a 1 a 2 a n a 2 1 a 2 2 a 2 n, so by Fact 621a, the determi- nant can be written as (a 1 a 2 a n ) det The new determinant a1 n 1 a2 n 1 an n 1 is a Vandermonde determinant (see Exercise 31), and we get n a i (a i a j ) i=1 i>j 35 [ x1 x 2 ] must satisfy det 1 1 1 x 1 a 1 b 1 = 0, ie, must satisfy the linear equation x 2 a 2 b 2 (a 1 b 2 a 2 b 1 ) x 1 (b 2 a 2 ) + x 2 (b 1 a 1 ) = 0 x1 a1 x1 We can see that = and = x 2 a 2 x 2 has two identical columns in these cases [ b1 b 2 ] satisfy this equation, since the matrix 37 Applying Fact 624 to the equation AA 1 = I n we see that det(a) det(a 1 ) = 1 The only way the product of the two integers det(a) and det(a 1 ) can be 1 is that they are both 1 or both -1 Therefore, det(a) = 1 or det(a) = 1 39 det(a T A) = det(a T ) det(a) = [det(a)] 2 > 0 Fact 624 Fact 627 41 det(a) = det(a T ) = det( A) = ( 1) n (det A) = det(a), so that det(a) = 0 We have used Facts 627 and 621a ( ) v 43 det(a T T v v v w v 2 v w A) = det w T [ v w] = det = det v w w w v w w 2 = v 2 w 2 ( v w) 2 0 by the Cauchy-Schwarz inequality (Fact 5111) 156
SSM: Linear Algebra Section 62 45 f(x) is a linear function, so f (x) is the coefficient of x (the slope) Expanding down the 1 2 3 4 0 2 3 4 first column, we see that the coefficient of x is det = 24, so f 0 0 3 4 (x) = 0 0 0 4 24 x b 47 Yes! For example, T = dx + by is given by the matrix [d b], so that T is linear in y d the first column 49 For example, we can start with an upper triangular matrix B with det(b) = 13, such as B = 1 1 1 0 1 1 Adding the first row of B to both the second and the third to make 0 0 13 all entries nonzero, we end up with A = 1 1 1 1 2 2 Note that det(a) = det(b) = 13 1 1 14 51 Notice that it takes n row swaps (swap row i with n + i for each i between 1 and n) to turn A into I 2n So, det(a) = ( 1) n det(i 2n ) = ( 1) n 53 a See Exercise 37 a b d b b If A =, then A c d 1 = 1 det(a) has integer entries c a 55 We start out with a preliminary remark: If a square matrix A has two equal rows, then D(A) = 0 Indeed, if we swap the two equal rows and call the resulting matrix B, then B = A, so that D(A) = D(B) = D(A), by property b, and D(A) = 0 as claimed Next we need to understand how the elementary row operations affect D Properties a and b tell us about how row multiplications and row swaps, but we still need to think about row additions We will show that if B is obtained from A by adding k times the i th row to the j th, then D(B) = D(A) Let s label the row vectors of A by v 1, v n By linearity of D in the j th row (property c) we have 157
Chapter 6 SSM: Linear Algebra D(B) = D v i v j + k v i = D(A) + kd v i v i = D(A) Note that in the last step we have used the preliminary remark Now, using the terminology introduced on Page 262, we can write D(A) = ( 1) s k 1 k 2 k r D(rref A) Next we observe that D(rref A) = det(rref A) for all square matrices A Indeed, if A is invertible, then rref(a) = I n, and D(I n ) = 1 = det(i n ) by property c of function D If A fails to be invertible, then D(rref A) = 0 = det(rref A) by linearity in the last row It follows that D(A) = ( 1) s k 1 k 2 k r D(rref A) = ( 1) s k 1 k 2 k r det(rref A) = det(a) for all square matrices, as claimed 57 Note that matrix A 1 is invertible, since det(a 1 ) 0 Now y y T = [A x 1 A 2 ] = A x 1 y + A 2 x = 0 when A 1 y = A 2 x, or, y = A 1 1 A 2 x This shows that for every x there is a unique y (that is, y is a function of x); furthermore, this function is linear, with matrix M = A 1 1 A 2 0 0 1 0 59 det = 0 = det, but these matrices fail to be similar 0 0 0 0 [ ] In 0 A B A B 61 We follow the hint: = C A C D CA + AC CB + AD ( ) A B In 0 A B = So, det = det(a) det(ad CB) 0 AD CB C A C D () A B Thus, det(i n ) det(a) det = det(a) det(ad CB), which leads to C D () A B det = det(ad CB), since det(a) 0 C D 63 1 By Fact 633, the area equals det 3 8 = 50 = 50 7 2 158
SSM: Linear Algebra Section 63 Figure 61: for Problem 633 6 1 3 Area of triangle = 1 2 det = 13 (See Figure 61) 2 4 5 The volume of the tetrahedron T 0 defined by e 1, e 2, e 3 is 1 3 (base)(height) = 1 6 Here we are using the formula for the volume of a pyramid (See Figure 62) Figure 62: for Problem 635 The tetrahedron T defined by v 1, v 2, v 3 can be obtained by applying the linear transformation with matrix [ v 1 v 2 v 3 ] to T 0 Now we have vol(t ) = det[ v 1 v 2 v 3 ] vol(t 0 ) = 1 6 det[ v 1 v 2 v 3 ] = 1 6 V ( v 1, v 2, v 3 ) Fact 638 and Page 282 Fact 635 7 Area = 1 10 2 2 det + 11 13 1 8 10 2 det = 110 (See Figure 63) 2 11 9 By Fact 633, det[ v 1 v 2 ] = area of the parallelogram defined by v 1 and v 2 But v 1 is the base of that parallelogram and v 2 sin θ is its height, hence det[ v 1 v 2 ] = v 1 v 2 sin θ 159
Chapter 6 SSM: Linear Algebra Figure 63: for Problem 637 11 The matrix of the transformation T with respect to the basis v 1, v 2 is B = that det(a) = det(b) = 12, by Fact 625 13 By Fact 637, the desired 2-volume is 1 1 det 1 1 1 1 1 2 4 10 = det = 20 1 2 3 4 1 3 10 30 1 4 3 0, so 0 4 15 If v 1, v 2,, v m are linearly dependent and if A = [ v 1 v m ], then det(a T A) = 0 since A T A and A have equal and nonzero kernels (by Fact 542), hence A T A fails to be invertible On the other hand, since the v i are linearly dependent, at least one of them will be redundant For such a redundant v i, we will have v i = v i and v i = 0, so that V ( v 1,, v m ) = 0, by Definition 636 This discussion shows that V ( v 1,, v m ) = 0 = det(a T A) if the vectors v 1,, v m are linearly dependent 160
SSM: Linear Algebra Section 63 17 a Let w = v 1 v 2 v 3 Note that w is orthogonal to v 1, v 2 and v 3, by Exercise 6244c Then V ( v 1, v 2, v 3, w) = V ( v 1, v 2, v 3 ) w = V ( v 1, v 2, v 3 ) w b By Exercise 6244e, Definition 636 V ( v 1, v 2, v 3, v 1 v 2 v 3 ) = det [ v 1 v 2 v 3 v 1 v 2 v 3 ] = det [ v 1 v 2 v 3 v 1 v 2 v 3 ] = v 1 v 2 v 3 2 c By parts a and b, V ( v 1, v 2, v 3 ) = v 1 v 2 v 3 If the vectors v 1, v 2, v 3 are linearly dependent, then both sides of the equation are 0, by Exercise 15 and Exercise 6244a 19 det[ v 1 v 2 v 3 ] = v 1 ( v 2 v 3 ) = v 1 v 2 v 3 cos θ where θ is the angle between v 1 and v 2 v 3 so det[ v 1 v 2 v 3 ] > 0 if and only if cos θ > 0, ie, if and only if θ is acute (0 θ π 2 ) (See Figure 64) Figure 64: for Problem 6319 21 a Reverses Consider v 2 and v 3 in the plane (not parallel), and let v 1 = v 2 v 3 ; then v 1, v 2, v 3 is a positively oriented basis, but T ( v 1 ) = v 1, T ( v 2 ) = v 2, T ( v 3 ) = v 3 is negatively oriented b Preserves 161
Chapter 6 SSM: Linear Algebra Consider v 2 and v 3 orthogonal to the line (not parallel), and let v 1 = v 2 v 3 ; then v 1, v 2, v 3 is a positively oriented basis, and T ( v 1 ) = v 1, T ( v 2 ) = v 2, T ( v 3 ) = v 3 is positively oriented as well c Reverses The standard basis e 1, e 2, e 3 is positively oriented, but T ( e 1 ) = e 1, T ( e 2 ) = e 2, T ( e 3 ) = e 3 is negatively oriented 23 Here A = 5 3, det(a) = 17, 6 7 b = det x 1 = 1, so by Fact 639 0 1 3 0 7 = 7 det 17 17, x 2 = 5 1 6 0 = 6 17 17 25 By Fact 6310, the ij th entry of adj(a) is given by ( 1) i+j det(a ji ), so since A = 1 0 1 0 1 0 1 0 for i = 1, j = 1, we get ( 1) 2 det = 1, and for i = 1, j = 2 we 0 1 2 0 [ 1 ] 0 1 get ( 1) 3 det = 0, and so forth 0 1 Completing this process gives adj(a) = 1 0 1 0 1 0, hence by Fact 6310, 2 0 1 A 1 = 1 det(a) adj(a) = 1 1 0 1 1 0 1 0 = 1 0 1 0 1 0 2 0 1 2 0 1 [ a b 27 By Fact 639, using A = b a x = det ( ) 1 b 1 0 a a 2 +b = a 2 a 2 +b, y = det 2 ], det(a) = a 2 + b 2, b = [ 1 0 ], we get ( ) a 1 1 b 0 a 2 +b = b 2 a 2 +b, 2 so x is positive, y is negative (since a, b > 0), and x decreases as b increases 29 By Fact 639, dx 1 = 2 det6 4 0 R 1 (1 α) 0 1 α (1 α) 2 R 2 de 2 R 2 (1 α)2 3 7 5 α D = R1R2(1 α)2 de 2 R 2(1 α) 2 de 2 D 162
SSM: Linear Algebra Section 63 dy 1 = 2 det6 4 2 det6 4 R 1 0 (1 α) α 0 (1 α) 2 R 2 R 2 de 2 (1 α)2 D R 1 R 1 0 α 1 α 0 R 2 R 2 R 2 de 2 α 3 7 5 3 7 5 = R2de2(R1(1 α)2 +α(1 α)) D > 0 dp = D = R1R2de2 D > 0 31 Using the procedure outlined in Exercise 25, we find adj(a) = 6 0 1 3 5 2 4 5 1 24 0 0 0 0 12 0 0 33 Using the procedure outlined in Exercise 25, we find that adj(a) = 0 0 8 0 0 0 0 6 Note that the matrix adj(a) is diagonal, and the i th diagonal entry of adj(a) is the product of all a jj where j i 35 det(adj(a)) = det(det(a)a 1 ) Taking the product det(a)a 1 amounts to multiplying each row of A 1 by det(a), so that det(adj(a)) = (det A) n det(a 1 ) = (det A) n 1 det(a) = (det A) n 1 37 adj(a 1 ) = det(a 1 )(A 1 ) 1 = (det A) 1 (A 1 ) 1 = (adja) 1 39 Yes, let S be an invertible matrix such that AS = SB, or SB 1 = A 1 S Multiplying both sides by det(a) = det(b), we find that S(det(B)B 1 ) = (det(a)a 1 )S, or, S(adjB) = (adja)s, as claimed 41 If A has a nonzero minor det(a ij ), then the n 1 columns of the invertible matrix A ij will be independent, so that the n 1 columns of A, minus the j th, will be independent as well Thus, the rank of A (the dimension of the image) is at least n 1 Conversely, if rank(a) n 1, then we can find n 1 independent columns of A The n (n 1) matrix consisting of those n 1 columns will have rank n 1, so that there will be exactly one redundant row (compare with Exercises 3349 through 51) Omitting this redundant row produces an invertible (n 1) (n 1) submatrix of A, giving us a nonzero minor of A 43 A direct computation shows that A(adjA) = (adja)a = (det A)(I n ) for all square matrices Thus we have A(adjA) = (adja)a = 0 for noninvertible matrices, as claimed Let s write B = adj(a), and let s verify the equation AB = (det A)(I n ) for the diagonal entries; the verification for the off-diagonal entries is analogous The i th diagonal entry 163
Chapter 6 SSM: Linear Algebra of AB is ith [i th row of A] column = a i1 b 1i + + a in b ni = of B n a ij b ji j=1 Since B is the adjunct of A, b ji = ( 1) j+i det(a ij ) So, our summation equals n a ij ( 1) i+j det(a ij ) j=1 which is our formula for Laplace expansion across the i th row, and equals det(a), proving our claim for the diagonal entries a b a c d b 45 Let A = We want A c d T = adj(a), or = So, a = d and b d c a a b b = c Thus, the equation A T = adj(a) holds for all matrices of the form b a 47 Note that 1 2 det x1 x 2 is the area of the triangle OP y 1 y 1 P 2, where O denotes the origin 2 This is likewise true[ for one-half ] the second matrix However, because of the reversal 1 in orientation, 2 det x3 x 4 is negative the area of the triangle OP y 3 y 3 P 4 ; likewise for 4 the last matrix (See the discussion on Page 277) Finally, note that the area of the quadrilateral P 1 P 2 P 3 P 4 is equal to: the area of triangle OP 1 P 2 + area of triangle OP 2 P 3 area of triangle OP 3 P 4 area of triangle OP 4 P 1 49 We will use the terminology introduced in the solution [ of Exercise 48 throughout Note 2 2/ 3 that the transformation L 1, with matrix A 1 = ], maps the circle C (with 0 3 radius 1/ 3) into the ellipse E Now consider a radial vector v = cos θ 1 3 of C, sin θ and find the maximal value M and the minimal value m of A 1 v 2 = 4 3 + 1 9 sin2 θ 8 3 25 (sin θ)(cos θ) = 3 18 1 4 18 (cos 2θ) 3 (sin 2θ) (we are taking the square to facilitate 3 the computations) Then M and m will be the lengths of the semi-axes of E The function above is sinusoidal with average value 25 18 and amplitude 1 18 + 16 2 27 = 193 18 Thus M = 25+ 193 18 and m = 25 193 18, so that the length of the semi-major axis of E is M = 25+ 193 18 147, and for the semi-minor axis we get 164
SSM: Linear Algebra True or False m = 25 193 18 079 True or False 1 T, by Fact 616 (a diagonal matrix is triangular as well) 3 T, by Definition 611 5 F; Let A = B = I 5, for example 7 F; In fact, det(a) = 0, since A fails to be invertible 9 T, by Fact 621a, applied to the columns 11 T, by Fact 625 13 T, by Fact 624 15 T; The determinant is 0 for k = 1 or k = 2, so that the matrix is invertible for all positive k 17 T; It would be tedious to compute the exact value of the determinant, but we can show that the determinant is nonzero without finding its exact value One method is to show that the determinant is an odd integer Expanding down the third column, we see that det(a) = 3 det 5 4 8 100 9 7 + even terms 6 5 100 5 4 = 3( 7 det + even terms ) + even terms 6 5 = 21 + even terms = odd, as claimed Alternatively, we can use the permutation formula for the determinant (Fact 6210), and argue that one of the 24 terms is very large (namely 100 4 = 10 8 ) compared to the other terms [those are less than (100 2 )(10 2 ) = 10 6 each in absolute value], so that the determinant turns out to be positive (In fact, this determinant is 97,763,383) 19 T; Matrix A is invertible 21 T, by Fact 634, since v i v i = 1 for all column vectors v i 165
Chapter 6 SSM: Linear Algebra 3 2 23 F; Let A =, for example 5 3 1 1 1 1 1 1 1 1 25 T; Let A = The column vectors of A are orthogonal and they all 1 1 1 1 1 1 1 1 have length 2 27 F; In fact, det(a) = det[ u v w] = det[ v u w] = v ( u w) We have used Fact 621b and Definition 611 29 F; Note that det 1 1 0 0 1 1 = 2 1 0 1 31 F; Let A = 2I 2, for example 33 F; Let A = I 2 and B = I 2, for example 35 T; Let s do Laplace expansion along the first row, for example (see Fact 615) n Then det(a) = ( 1) 1+j a 1j det(a 1j ) 0 Thus det(a 1j ) 0 for at least one j, so that A 1j is invertible j=1 37 T, since adj(a) = (det A)(A 1 ), by Fact 6310 39 F; Note that det(s 1 AS) = det(a) but det(2a) = 2 3 (det A) = 8(det A) 41 T; We can use induction on n to show that det(a) is an odd integer If we expand det(a) down the first column, then the first term, a 11 det(a 11 ), will be odd, since a 11 is odd, and det(a 11 ) is odd by the induction hypothesis However, all the other terms a i1 det(a i1 ), where i > 1, will be even, since a i1 is even in this case Thus, det(a) is odd as claimed, and A is invertible a b 43 T; Let A = c d [ 0 1 B = 0 0 [ 0 0 If a 0, let B = 0 1 1 0 0 0 ], and if d 0, let B = ] ; if b 0, let B = 0 0 ; if c 0, let 1 0 45 T; A(adjA) = A(det(A)A 1 ) = det(a)i n = det(a)a 1 A = adj(a)a 47 F; A = 1 0 0 0 0 1 and B = 1 0 0 0 1 0 are both orthogonal and det(a) = 0 1 0 0 0 1 det(b) = 1 However, AB BA 166