3.2soln

# 3.2soln - Solutions to problems from section 3.2 6 Find the...

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Unformatted text preview: Solutions to problems from section 3.2 6. Find the determinant by row reduction to echelon form. 1 5 −3 3 −3 3 2 13 −7 Solution: 1 5 −3 3 −3 3 2 13 −7 N R2 =R2 −3R1 NR3 =R3 −2R1 = 1 5 −3 0 −18 12 0 3 −1 N R3 =R3 + 1 R2 6 = 1 5 −3 0 −18 12 0 0 1 = 1(−18)(1) = −18. 8. Find the determinant by row reduction to echelon form. 1 3 3 −4 0 1 2 −5 2 5 4 −3 −3 −7 −5 2 Solution: 1 3 3 −4 0 1 2 −5 2 5 4 −3 −3 −7 −5 2 N R3 =R3 −2R1 NR4 =R4 +3R1 = 1 3 3 −4 0 1 2 −5 0 −1 −2 5 0 2 1 −6 N R3 = R3 + R2 NR4 =R4 −2R2 = 1 0 0 0 3 3 −4 1 2 −5 0 0 0 0 −3 4 3 3 −4 1 2 −5 R3 ↔R4 = = −(1)(1)(−3)(0) = 0. 0 −3 4 0 0 0 5 0 −1 22. Use determinants to ﬁnd out if the matrix 1 −3 −2 is invertible. 0 5 3 Solution: 5 0 −1 1 −3 −2 0 5 3 −3 −2 5 3 1 −3 0 5 1 0 − 0 0 =5 −1 = 5(−9 + 10) − 1(5 − 0) = 0. Thus the matrix in question is not invertible. 3 5 26. Use determinants to decide if the set of vectors −6 4 linearly independent. Solution: 3 2 −2 0 5 −6 −1 0 −6 0 30 4 7 03 3 2 −2 =3 5 −6 −1 −6 0 3 = 3 −6 2 −2 −6 −1 +3 3 2 5 −6 2 −2 0 −6 −1 0 , 0 , 3 , 0 7 0 3 , is = 3(−6(−2 − 12) + 3(−18 − 10)) = 0. Thus the set of vectors given in the exercise are linearly dependent. 28. Assume A and B are n × n matrices. True or false. Justify your answer. a. If two row interchanges are made in succession, then the new determinant equals the old determinant. Solution: TRUE. If we start with a square matrix A and interchange two rows to get a matrix B and then interchange two rows to get a matrix C then we have det B = − det A and det C = − det B so that det C = − det B = −(− det A) = det A. b. The determinant of A is the product of the diagonal entries in A. 11 = 1 − 1 = 0 whereas the product of the diagonal Solution: FALSE. For example 11 11 entries in is 1. 11 c. If det A is zero, then two rows or two columns are the same, or a row or column is zero. 12 Solution: FALSE. For example, let A = . Then det A = 0 but A but no two 24 rows or two columns of A are the same, and no row or column of A is zero. d. det AT = (−1) det A. Solution: FALSE. det AT = det A, so any matrix with nonzero determinant will be a counterexample to the statement. 32. Find a formula for det(rA) where A is an n × n matrix. Solution: I claim det(rA) = r n · det A. Here’s a proof: det(rA) = = = = det(r (In A)) det((rIn )A) det(rIn ) det A r n det A (since A = In A) (property of matrix multiplication.) (multiplicative property of determinants) (Since rIn is a n × n diagonal matrix with r ’s on the main diagonal) [Alternatively, one could say that multiplying a row of A by r changes the determinant by a factor of r . Since rA is obtained from A by multiplying each of the n rows by r , the determinant of rA will be obtained by multiplying the determinant of A by n factors of r , i.e. det(rA) = r n det A.] 34. Let A and P be square matrices, with P invertible. Show that det(P AP −1 ) = det A. Solution: det(P AP −1 ) = = = = = = det(P ) det(A) det(P −1 ) det(P ) det(P −1 ) det(A) det(P P −1 ) det(A) det(I ) det(A) 1 · det(A) det A (multiplicative property of determinants) (commutativity of real numbers) (multiplicative property of determinants) (properties of inverses) (because I is a diagonal matrix with all 1’s on the diagonal) 40. Let A and B be 4 × 4 matrices, with det A = −1 and det B = 2. Compute: a. det AB Solution: Use the multiplicative property of determinants to get det AB = (det A)(det B ) = −1(2) = −2. b. det B 5 . Solution: Use the multiplicative property of determinants to get det(B 5 ) = (det B )5 = 25 = 32. c. det(2A) Solution: Use the result of exercise 32 to get det(2A) = 24 det A = 16(−1) = −16. d. det AT A Solution: Use the multiplicative property of determinants as well as theorem 5 on page 196 to get det AT A = (det AT )(det A) = (det A)(det A) = (−1)(−1) = 1. e. det B −1 AB . Solution: [First notice that since det B = 0 we know that B is invertible so the question makes sense.] Use the result of exercise 34 with P = B −1 to get det B −1 AB = det A = −1. ...
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