**MATH1025
AF**

**FW00F**

**Answers
to Tutorial 3 **

Section 1.4

15.
(a) If A and B are n by n matrices the entry in row i and column j of
the matrix product AB is a_{i1}b_{1j} + a_{i2}b_{2j}
+ ... + a_{in}b_{nj}. Thus if row i of A is all
zero this entry in the product is also zero.

This is true for any value of j which means that row i of AB is zero.

Now if B were the inverse of A, AB would be the unit matrix which has no zero rows. Hence we have a contradiction and A cannot have an inverse if it has a zero row.

(b)
Similarly, the entry in row i and column j of BA is b_{i }a_{1j}
+ b_{i2}a_{2j} + ... + b_{in}a_{nj}.
Thus if column j of A is zero, this entry is zero for any value of i.
Thus BA has a zero column and A cannot have an inverse.

17.
If A is invertible then A^{-1} exists. Multiply AB = 0 on
the left by this inverse to get A^{-1}AB = A^{-1}0
or B = 0.

18. Multiply the partitioned matrices together to get

Now
AA^{-1} = I so that if this product is to be the unit matrix,
then BA^{-1 }+ AC must be the zero matrix, i.e. AC = -BA^{-1}.
Multiply both sides by A^{-1 } we get A^{-1}AC = IC
= C = -A^{-1}BA^{-1}.

21.
Since (AB)^{T }= B^{T}A^{T} then ^{
}(BB^{T})^{T} = (B^{T})^{T}B^{T}
= BB^{T } and hence BB^{T} is symmetric.

Since
(A + B)^{T} = A^{T} + B^{T} then (B + B^{T})^{T}
= B^{T} + (B^{T})^{T }= B^{T} + B = B
+ B^{T} and hence B + B^{T} is symmetric.

Similarly,
(B - B^{T})^{T} = B^{T} - (B^{T})^{T
}= B^{T} - B = -(B - B^{T}) and hence B - B^{T}
is skew-symmetric.

Section 1.5

6. (a) The augmented matrix for the inverse is

R1 -> 1/3 R1; R2 -> R2 - R1; R3 -> R3 - 2 R1

R3-> 2/5 R3

R2 -> -3/4 R2; R3 -> R3 - 7/3 R2

R1-> R1-4/3R2

R1 -> R1 + 1/3 R3; R2 -> R2 + 5/2 R3

Thus the inverse is

(b) Augmented matrix is:

R1 -> -R1; R2 -> R2 -2 R1; R3 -> R3 - 4 R1

R2 -> 1/10 R2; R3 -> R3 + 10 R2

Since there is now a row over zeros in the left side of the third row, there is no solution and the matrix has no inverse.

(c) Augmented matrix is:

R3 -> R3 - R2

R3 -> R3 - R1

R3 -> (-1/2) R3

R1 -> R1 - R3; R2 -> R2 - R3

Thus the inverse is

8. (a) The inverse is

(b) Interchange the first and fourth rows and the second and third rows and then divide by the diagonal elements on the left to get the inverse

(c) The augmented matrix is

If we divide the first row by k and then subtract it from the second row; then divide the second row by k and subtract it from the third row; then divide the third row by k and subtract it from the fourth row; then finally divide the fourth row by k we get the inverse

Section 1.6

12. The augmented matrix is

R1 -> -R1; R2 -> R2 - R1; R3 -> R3 -6R1

R2 -> (1/13)R2; R3 -> R3 -28 R2

R3 -> (13/2)R3; R2 -> R2 + (1/13)R3; R1 -> R1 + R3

R1 -> R1 + 4 R2

Thus
the solution to the first system is x_{1} = -18, x_{2}
= -1, x_{3} = -14 and to the second system is

x_{1}
= -421/2, x_{2} = -25/2, x_{3} = -327/2.

Section 1.7

7. By Theorem 1.7.1 a triangular matrix is not invertible if at least one of its diagonal elements is zero. Thus for A and B both to be not invertible we must have a + b - 1 = 0 and 2a - 3b - 7 = 0 which yields the system

a + b = 1

2a -3b = 7

which has the solution a = 2 and b = -1.

18.
If A^{T}A = A then taking the transpose of each side we get
(A^{T}A)^{T} = A^{T}(A^{T})^{T}
= A^{T}A = A^{T}. Thus A = A^{T} and A is
symmetric. Also from the original equation we get A^{2} = A.

22.
(a)Since (A^{-1})^{T} = (A^{T})^{-1}
by Theorem 1.4.10

= (-A)^{-1} since A is skew-symmetric

= -A^{-1} by Theorem 1.4.8

then
A^{-1} is skew-symmetric if A is invertible and
skew-symmetric.

(b) Let B=A^{T}. Then
B^{T} =A. But A^{T} = -A so B = -B^{T} and
therefore A^{T} is skew-symmetric if A is.

(A + B)^{T} = A^{T}
+ B^{T} = -A - B = -(A + B) so A + B is skew-symmetric if A
and B are.

(A - B)^{T} = A^{T}
- B^{T} = -A - (-B) = -(A - B) so A - B is skew-symmetric if
A and B are.

(kA)^{T} = kA^{T}
= k(-A) - -(kA) so kA is skew-symmetric if A is.

(c) Since (A + A^{T})^{T}
is symmetric for any square matrix A and A - A^{T} is
skew-symmetric for any square matrix A (question 21 of section 1.4)
we can write

then any square matrix can be written as the sum of a symmetric and a skew-symmetric matrix.

24.
(a) Solving L**y** = **b** where L =and
b =

we
get y_{1} = 1, y_{2} = (-2 + 2y_{1})/3 = 0
and y_{3} = -2y_{1} - 4y_{2} = -2. Now solve
U**x** = **y** where

U
=we
get x_{3} = y_{3}/4 = -1/2, x_{2} = y_{2 }-
2x_{3} = 1 and x_{1} = (y_{1} + x_{2}
- 3x_{3})/2 = 7/4.

(b)
In a similar manner we get y_{1} = 4/2 = 2, y_{2} =
-5 - 4y_{1} = -13 and y_{3} = (2 + 3y_{1} +
2y_{2})/3 = -6. Then x_{3} = y_{3}/2 = -3,
x_{2} = (y_{2} - x_{3})/4 = -5/2 and x_{1}
= (y_{1} + 5x_{2} - 2x_{3})/3 = -3/2.

Section 11.6

2.
(a) **x**^{(1)} = P**x**^{(0) }=,
**x**^{(2)}
= P**x**^{(1)} =,
**x**^{(3)}
= P**x**^{(2)} =.

(b)
P is regular since all of its entries are positive. Let the
steady-state vector be **q**. Then P**q** = **q** or (P -
I)**q** = **0**. The coefficient matrix is

R1 -> 1/(-0.8)R1; R2 -> R2 - (0.6)R1; R3 -> R3 - (0.2)R1

R2 -> 1/(-0.525)R2; R3 -> R3 - (0.525)R2; R1 -> R1 + (0.125)R2

Thus
the solution is q_{1 }= 22t, q_{2} = 29t and q_{3}
= 21t. But q_{1} + q_{2} + q_{3} = 1 or
72t = 1. Thus

q_{1
}= 22/72, q_{2} = 29/72 and q_{3} = 21/72.

8.
The transition matrix has elements p_{ij} which represents
the proportion of people moving from region j to region i each year.
Thus P =.
We solve for the steady state vector as in question 2. Thus P - I
=

R1 -> 1/(-0.1)R1; R2 -> R2 - (0.05)R1; R3 -> R3 - (0.05)R1

R2 -> 1/(-0.175)R2; R3 -> R3 - (0.175)R2; R1 -> R1 + (1.5)R2

Thus q_{1 }= 13t, q_{2}
= 4t and q_{3} = 7t. But q_{1} + q_{2} +
q_{3} = 1 or 24t = 1.

Thus q_{1 }= 13/24, q_{2}
= 4/24 and q_{3} = 7/24 and after a long time 13/24 of the
population lives in region 1, 1/6 in region 2 and 7/24 in region 3.