MATH1025 AF

FW00F

Section 2.2

2. If a matrix is triangular, its determinant is the product of the diagonal elements. Thus for part (a) the determinant is -30 and for part (b) it is -2. In part (c) rows 1 and 3 are identical so the determinant is zero. In part (d) row 2 is twice row 1 so the determinant is zero.

4. Use Gaussian elimination to evaluate the determinant. Thus

R2 <-> R3

R1 -> 1/3 R1

R3 -> R3 + 2 R1

Since this is now in triangular form the determinant is (-3) 5 (-2) = 30.

7. R1 -> 1/3 R1; R2 -> R2 + 2 R1

R2 <-> R3; R3 -> R3 - 3 R2

Since this is now in triangular form the determinant is (-3)(-11) = 33.

8.R2 -> R2 - 5R1; R3 -> R3 + R1; R4 -> R4 - 2R1

R3 -> (-1/3) R3; R4 -> R4 -108 R3

R4 -> R4 - 12 R1

Since this is now in triangular form the determinant is (-3)(-13) = 39.

12. (a) -6 (to get the new determinant from the original interchanage rows 1 and 2 and rows 2 and 3 so determinant is multiplied by (-1)2).

(b) 72 (multiply row 1 by 3, row 2 by (-1) and row 3 by 4 so determinant is multiplied by 3(-1)4

(c) -6 (add row 3 to row 1 so no change)

(d) 18 (subtract 4 times row 2 from row 3 (no change) and multiply row 1 by (-3) so determiinant is multiplied by (-3)).

13. Use gaussian elimination to get

R2 -> R2 - a R1; R3 -> R3 - a2 R1

R3 -> R3 - (b+a) R2

so that the determinant is (b-a)[(c2 - a2) - (c-a)(b+a)] = (b-a)[(c-a)(c+a) - (c-a)(b+a)]

= (b-a)(c-a)(c-b)

Section 2.3

5. (a) det(3A) = 33 det(A) = -189

(b) det(A-1) = 1/det(A) = -1/7

(c) det (2A-1) = 23 det(A-1) = -8/7

(d) det((2A)-1) = 1/det (2A) = 1/(23 det(A)) = -1/56

(e) This determinant is got from det(A) by intercahnging rows 2 and 3 and then taking the transpose. Thus the determinant is (-1) det(A) = 7. Taking the transpose does nt change the value of the determinant.

12. A is not invertible if det(A) = 0.

(a) det(A) = (k-3)(k-2) - (-2)(-2) = k2 - 5k + 2 which equals zero if

(b) Using the special rule for 3 by 3 determinants we get det(A) = 2 + 12k + 36 - 4k - 18 - 12 = 8k + 8 which is zero if k = -1.

14 (a) (b)

(c)

15. (a) (i) Characteristic polynomial is the determinant of the coefficient matrix, i.e.

(ii) the eigenvalues are the roots of the characteristic polynomial, i.e 3 and -1

(iii) solving the original set of equations with the eigenvalue 3 gives x1 = x2 = t

solving the equations with the eigenvalue -1 gives -x1 = x2 = t

(b) (i)

(ii) the eigenvalues are 6 and -1

(iii) for the eigenvalue 6 the eigenvalue has components x1 = 3t and x2 = 4t

(NB this is equivalent to the answer in the text)

for the eigenvalue -1 the eigenvalue has components x1 = -t and x2 = t

(c) (i)

(ii) the eigenvalues are 2 and -2

(iii) for the eigenvalue 2 the eigenvalue has components x1 = -t and x2 = t

for the eigenvalue -2 the eigenvalue has components x1 = -t and x2 = 5t

(NB this is equivalent to the answer in the text)

18. Since det(ATA) = det(AT) det(A) = (det(A))2 then

(i) if A is invertible, det(A) is nonzero which means (det(A))2 is non zero and ATA is invertible

(ii) if ATA is invertible, (det(A))2 is nonzero which means det(A) is non zero and A is invertible

Section 2.4

31 (a) det(A) = But

and R2 -> R2 + 2 R1; R3 -> R3 - 3 R1

R2 -> 1/12 R2; R3 -> R3 + 4 R2

= 12(-9) = -108

Thus det(A) = 10(-108) = -1080