**MATH1025
AF**

**FW00F**

**Answers
to Tutorial 2 **

Section 1.2

6. Using Gauss-Jordan elimination:

(a) Augmented matrix is

R2 -> R2 + R1 (row 2 is replaced by row 2 plus row 1)

R3 -> R3 - 3 R1 (row 3 is replaced by row 3 - 3 times row 1)

R2 -> -R2; R3 -> R3 + 10 R2

R3 -> R3/(-52)

R2 -> R2 + 5 R3; R1 -> R1 -2 R3

R1 -> R1 - R2

Thus
the solution can be read off as x_{1} = 3, x_{2} = 1,
x_{3} = 2.

(c) Augmented matrix is

R2 -> R2 - 2 R1 ; R3 -> R3 + R1 ; R4 -> R4 - 3 R1

R2 -> (1/3)R2; R3 -> R3 - R2; R4 -> R4 - 3 R2

R1 -> R1 + R2

Thus if we let z = s and w = t the solution can be read off as x = -1 + t , y = 2s.

7. Using Gaussian elimination:

(b) Augmented matrix is

R1 -> (1/2)R1; R2 -> R2 + 2 R1; R3 -> R3 - 8R1

R2 -> (1/7)R2; R3 -> R3 + 7R2

If
we let x_{3} = t then x_{2} = 1/7 -4t/7 and x_{1}
= -1/7 -3t/7

7(d) Augmented matrix is

R1 <-> R2

R1 -> (1/3)R1; R3 -> R3 -6 R1

R2 -> R2/(-2); R3 -> R3 + 6 R2

Since the last row represents the equation 0 = 6 this is an inconsistent system of equations and there is no solution.

13. Homogeneous systems

(a) Using Gauss-Jordan elimination. Augmented matrix is

R1 -> (1/2)R1; R2 -> R2 - R1

R2-> (2/3)R2; R3 -> R3 - R2

R3 -> (1/2)R3; R2 -> R2 + R3; R1 -> R1 - (3/2)R3

R1 -> R1 - (1/2)R2

From
this we see that the solution is the trivial solution x_{1} =
x_{2} = x_{3} = 0.

(b) Gaussian elimination. Augmented matrix

R1 -> (1/3)R1; R2 -> R2 - 5 R1

R2 -> (-3/8) R2

If
we let x_{4} = t and x_{3} = s then x_{2} =
-s/4 -t and x_{1} = -s/4. (The answer in the text is
obtained by letting x_{3} = 4s which gives a simpler form).

17. Augmented matrix is

Using Gaussian elimination

R2 -> R2 - 3 R1; R3 -> R3 - 4 R1

R2 -> (-1/7) R2; R3 -> R3 +7 R2

For no solution we must have an inconsistent system of equations, i.e.

For exactly one solution we must have

For infinitely many solutions we must have a zero row, i.e.

Section 1.3

See the text for the answers to questions 3 and 5.

12.(a) Let A be a matrix with dimensions m by n and B be a matrix with dimension p by q. If AB is defined we must have n = p (see equation (3) in section 1.3). Similarly, if BA is defined, q = m. Thus AB has dimensions m by q, i.e. m by m which is square. Similarly, BA has dimensions p by n, i.e. n by n which is square.

(b) From part (a), if BA is defined, q = m and BA has dimensions p by n. Thus if A(BA) is defined we must have n = p. Thus B has dimensions n by m.

15.(a) Using the notation of this problem we have

A_{11}
=
;
A_{12} =
;
A_{21} =
;
A_{22} =

B_{11}
=
;
B_{21} =
;
B_{12} =
;
B_{22} =

Then
A_{11}B_{11} + A_{12}B_{21} =

A_{11}B_{12}
+ A_{12}B_{22 }=

A_{21}B_{11}
+ A_{22}B_{21 }=

A_{21}B_{12}
+ A_{22}B_{22 }= 14 + 27 = 41

Thus AB =

(b) For this partitioning we have

A_{11}
=
;
A_{12} = 5; A_{21} =
;
A_{22} =

B_{11}
=
;
B_{12} =
;
B_{21} =
;
B_{22} = -3

Then
A_{11}B_{11} + A_{12}B_{21} =

A_{11}B_{12}
+ A_{12}B_{22 }= 5 - 15 = -10

A_{21}B_{11}
+ A_{22}B_{21 }=

A_{21}B_{12}
+ A_{22}B_{22 }=

Thus AB =