**MATH1025
AF**

**FW00F**

**Answers
to Tutorial 4**

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 - a^{2} R1

R3 -> R3 - (b+a) R2

so
that the determinant is (b-a)[(c^{2} - a^{2}) -
(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) = 3^{3} det(A) = -189

(b)
det(A^{-1}) = 1/det(A) = -1/7

(c)
det (2A^{-1}) = 2^{3} det(A^{-1}) = -8/7

(d)
det((2A)^{-1}) = 1/det (2A) = 1/(2^{3} 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) = k^{2} - 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 x_{1}
= x_{2} = t

solving the equations with the eigenvalue -1 gives -x_{1}
= x_{2} = t

(b) (i)

(ii) the eigenvalues are 6 and -1

(iii)
for the eigenvalue 6 the eigenvalue has components x_{1} = 3t
and x_{2} = 4t

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

for the eigenvalue -1 the eigenvalue has components x_{1}
= -t and x_{2} = t

(c) (i)

(ii) the eigenvalues are 2 and -2

(iii)
for the eigenvalue 2 the eigenvalue has components x_{1} = -t
and x_{2} = t

for the eigenvalue -2 the eigenvalue has components x_{1}
= -t and x_{2} = 5t

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

18.
Since det(A^{T}A) = det(A^{T}) det(A) = (det(A))^{2
}then

(i)
if A is invertible, det(A) is nonzero which means (det(A))^{2 }is
non zero and A^{T}A is invertible

(ii)
if A^{T}A 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