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 x1 = 3, x2 = 1, x3 = 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 x3 = t then x2 = 1/7 -4t/7 and x1 = -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 x1 = x2 = x3 = 0.


(b) Gaussian elimination. Augmented matrix

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

R2 -> (-3/8) R2

If we let x4 = t and x3 = s then x2 = -s/4 -t and x1 = -s/4. (The answer in the text is obtained by letting x3 = 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

A11 = ; A12 = ; A21 = ; A22 =


B11 = ; B21 = ; B12 = ; B22 =


Then A11B11 + A12B21 =


A11B12 + A12B22 =


A21B11 + A22B21 =


A21B12 + A22B22 = 14 + 27 = 41


Thus AB =



(b) For this partitioning we have

A11 = ; A12 = 5; A21 = ; A22 =


B11 = ; B12 = ; B21 = ; B22 = -3


Then A11B11 + A12B21 =


A11B12 + A12B22 = 5 - 15 = -10


A21B11 + A22B21 =


A21B12 + A22B22 =


Thus AB =