# MATH 1210 Assignment 4 Solutions 16R-T1

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1 MATH 1210 Assignment 4 Solutions 16R-T1 Attempt all questions and show all your work. Due November 13, Prove using mathematical induction that for any n 2, and collection of n m m matrices A 1, A 2,..., A n, det(a 1 A 2 A n ) = det(a 1 ) det(a 2 ) det(a n ). Fix m 1. For all n 2, let P n denote the statement that for any collection of n m m matrices A 1, A 2,..., A n, det(a 1 A 2 A n ) = det(a 1 ) det(a 2 ) det(a n ). Base Case. The statement P 2 says that for any collection of 2 m m matrices A, B, This is true by Theorem 7.6 in the text. det(ab) = det(a) det(b). Inductive Step. Fix k 2 and suppose that P k holds, that is, for any collection of k m m matrices A 1, A 2,..., A k, det(a 1 A 2 A k ) = det(a 1 ) det(a 2 ) det(a k ). It remains to show that P k+1 holds, that is, for any collection of k +1 m m matrices A 1, A 2,..., A k+1, det(a 1 A 2 A k+1 ) = det(a 1 ) det(a 2 ) det(a k+1 ). Let A 1, A 2,..., A k+1 be m m matrices. Then det(a 1 A 2 A k+1 ) = det( (A 1 A 2 A k ) A k+1 ) = det(a 1 A 2 A k ) det(a k+1 ) (by P 2 ) = det(a 1 ) det(a 2 ) det(a k ) det(a k+1 ) (by P k ). Therefore P k+1 holds. Thus by PMI, for all n 2, P n holds. 2. Prove using mathematical induction that for any n 1, the determinant of an uppertriangular n n matrix is the product of its diagonal entries. For any n 1, let P n denote the statement that the determinant of every upper-triangular n n matrix is the product of its diagonal entries. Base Case. The statement P 1 says that the determinant of every upper-triangular 1 1 matrix is the product of its diagonal entries. Every 1 1 matrix A = [a 1,1 ]

2 is upper-triangular, and A = a 1,1, which is the product of the diagonal entries. Therefore P 1 holds. Inductive Step. Fix k 1 and assume that P k holds, that is, the determinant of every upper-triangular k k matrix is the product of its diagonal entries. It remains to show that P k+1 holds, that is, the determinant of every upper-triangular k +1 k +1 matrix is the product of its diagonal entries. Let A = [a i,j ] k+1 k+1 be an upper-triangular matrix. First some notation: let A i,j denote the matrix formed from A by removing row i and column j. The last row of A is all zeros, except for the last entry, a k+1,k+1. Therefore expanding across the bottom row, we have: A = a k+1,j C k+1,j = a k+1,k+1 C k+1,k+1 = a k+1,k+1 ( 1) k+1+k+1 A k+1,k+1 = a k+1,k+1 ( 1) 2k+2 A k+1,k+1 = a k+1,k+1 A k+1,k+1. Note that A k+1,k+1 is a k k upper-triangular matrix. Therefore by P k, the determinant of A k+1,k+1 is the product of the diagonal entries, that is, Put these together and we get A k+1,k+1 = a 1,1 a 2,2 a k,k. A = a k+1,k+1 A k+1,k+1 = a k+1,k+1 a 1,1 a 2,2 a k,k. Therefore P k+1 holds. Thus by PMI, for all n 1, P n holds. 3. Is it true that for any two matrices A and B, det(a + B) = det(a) + det(b)? If so, prove it. If not, find a counter example. No. For instance, = 0, 0 0 = 0, but [ ] + [ 0 0 ] = 1 0 =

3 4. Solve the following system using Cramer s Rule: x 1 + 3x 3 = 1 x 2 + 2x 3 = 9 2x 1 + x 2 = 15 A = A 2 = x 1 x 2 x 3 =, A 1 =, A 3 = Then A = 4, A 1 = 20, A 2 = 20, and A 3 = 8. Therefore, by Cramer s rule: x 1 = A 1 A x 2 = A 2 A x 3 = A 3 A = 20 4 = 5 = 20 4 = 5 = 8 4 = Prove the following property: for all a, b, c R, a 0, b 0, c 0, ( b c = abc a + 1 b + 1 ). c b c R 2 R 2 R 1 = a b c R 3 R 3 (1 + c)r 1 = a b 0 1 (1 + a)(1 + c) c 0

4 = = a b 0 a c ac c 0 a b a c ac c = ac b( a c ac) = ac + ba + bc + abc = abc ( a + 1 b + 1 ). c 6. (a) Let c R. Prove using mathematical induction that for any n 1 and any n n matrix A, ca = c n A. Fix c R. For all n 1, let P n denote the statement that for any n n matrix A, ca = c n A. Base Case. The statement P 1 says that for any 1 1 matrix A = [a 1,1 ], ca = c A. ca = [ca 1,1 ] = ca 1,1 = c [a 1,1 ] = c A. Therefore P 1 holds. Inductive Step. Fix k 1 and assume that P k holds, that is, for any k k matrix A, ca = c k A. It remains to show that P k+1 holds, that is, for any k + 1 k + 1 matrix A, ca = c k+1 A. First some notation: let A i,j denote the matrix formed from A by removing row i and column j. Then expanding across the first row we have: ca = ca 1,j C 1,j = ca 1,j ( 1) 1+j (ca) 1,j = ca 1,j ( 1) 1+j c k A 1,j = ca 1,j ( 1) 1+j c k A 1,j = c k+1 a 1,j ( 1) 1+j A 1,j = c k+1 A. By P k since (ca) 1,j is a k k matrix By P k since (ca) 1,j is a k k matrix Therefore P k+1 holds, and thus by PMI, for all n 1, P n holds. (b) A square matrix is called skew-symmetric if A T = A. Use part (a) and a property of determinants when taking transposes to show that every skew-symmetric

5 matrix has determinant 0. Let A be a skew-symmetric matrix. Then A T determinant of both sides, we get = A. Taking the A T = A A = ( 1)A = ( 1) 1001 A = A. Thus A = A, and so 2 A = 0, thus A = An elementary matrix is a matrix which is one elementary row operation away from the identity matrix. For instance, E 1 = 0, E 2 = 1 0 0, E 3 = are all elementary matrices. (a) Let k be any real number, k 0. Find an elementary matrix with determinant k. [ ] k 0 For any k 0, is an elementary matrix (for the row operation R 1 kr 1 ), and has determinant k. (b) BONUS: 3 MARKS. Let E be an n n elementary matrix formed by performing row operation r to the identity I n. Let A be any n n matrix. Then the matrix product EA will result in the result of performing r to A. Use this fact, and properties of determinants to formally prove the following theorem: If A is an n n matrix such that the row reduced row echelon form of A is I n, then det(a) 0. Assume A is an n n matrix with RREF I n. Then there exist row operations r 1, r 2,..., r k such that if we perform them to A (starting with r 1 and proceeding in order), we get I n. That is, there exist elementary matrices E 1,..., E k such that E k E 2 E 1 A = I n. Therefore det(e k E 2 E 1 A) = det(i n ) = 1. But we know from question 1 above that det(e k E 2 E 1 A) = det(e k ) det(e 1 ) det(a). Therefore det(a) cannot equal zero (since if it did, the left hand side would be zero, not 1). 8. Let u = [1, 1, 1], v = [ 1, 2, 5], w = [0, 1, 1]. Calculate each of the following: (a) (2u + v) (v 3w) (2u + v) (v 3w) = (2[1, 1, 1] + [ 1, 2, 5]) ([ 1, 2, 5] 3[0, 1, 1])

6 = ([2, 2, 2] + [ 1, 2, 5]) ([ 1, 2, 5] [0, 3, 3]) = [1, 4, 7] [ 1, 1, 2] = = 9. (b) u 2 v + ( 3)w u 2 v + ( 3)w = [1, 1, 1] 2 [ 1, 2, 5] + ( 3)[0, 1, 1] = [0, 3, 3] = = Prove the associative rule for addition of vectors in E 3 in the following two different ways: (u + v) + w = u + (v + w) (a) by writing each of u, v, w in terms of their coordinates and simplifying both sides algebraically in coordinate form Let u = (a, b, c), v = (d, e, f), w = (x, y, z). Then (u + v) + w = ((a, b, c) + (d, e, f)) + (x, y, z) = (a + d, b + e, c + f) + (x, y, z) = ((a + d) + x, (b + e) + y, (c + f) + z) = (a + (d + x), b + (e + y), c + (f + z)) = (a, b, c) + (d + x, e + y, f + z) = (a, b, c) + ((d, e, f) + (x, y, z)) = u + (v + w). (b) by a geometric argument using arrow representations for u, v, w

7 10. Find the points where the plane 3x 2y + 5z = 30 meets each of the x, y and z axes in E 3. Use these intercepts to provide a neat sketch of the plane. Here are the formulae for the axes: x-axis: y-axis: z-axis: x = (t, 0, 0), t R x = (0, t, 0), t R x = (0, 0, t), t R So to find the intersection of this plane and the x-axis, just plug in y = z = 0: 3x = 30 = x = 10 = (10, 0, 0) Similarly, 2y = 30 = y = 15 = (0, 15, 0) 5z = 30 = z = 6 = (0, 0, 6) Therefore we have the following plane:

8 11. (a) Find an equation for the line through points (1, 3) and (5, 4) in parametric form. The vector v = (5, 4) (1, 3) = (4, 1) is along the line. Therefore the line in point-parallel form is: which in parametric form becomes x = (1, 3) + t(4, 1), t R x = 1 + 4t, y = 3 + t, t R. (b) Find an equation for the line through points (1, 2, 3) and (5, 5, 0) in parametric form. The vector v = (5, 5, 0) (1, 2, 3) = (4, 3, 3) is along the line. Therefore the line in point-parallel form is: which in parametric form becomes x = (5, 5, 0) + t(4, 3, 3), t R x = 5 + 4t, y = 5 + 3t, z = 3t, t R.

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