1.

If A = B =C = D =

Find:

(a)2A + B

(b)3A + 2B - 4C

(c)A.B

(d)B.A

(e)A.B.C

(f)

(g)

Answer Guidelines

(a)2A + B= 2* + = + =

=

(b)3A + 2B - 4C = 3* +2* - 4 =

=

(c)A.B =`. = =

(d)B.A = . = =

Note that AB≠BA

(e)A.B.C = A.B *C = .

=

=

(f)= = (2*4) - (-5*3) = + 23

(g) = = = -198

2.

A catering firm produces three main menus, M1, M2, and M3. Company C1 orders 30, 17 and 23 units of M1, M2, and M3 respectively for one week and Company C2 orders 40, 15 and 27 units of M1, M2, and M3 respectively for the same week. Matrix B represents the amounts of different inputs (I1 to I5) that are required to produce the different menus, M1, M2, and M3. Matrix C describes the cost (in p) of each of these inputs.

B = C =

(a)Write the orders for the firm as a matrix. Label this matrix A.

(b)Find and interpret:

  1. A.B
  2. B.C
  3. A.B.C
  4. IT.A where I =
  5. If the price of the menus are 75p, 50p and 100p for M1, M2, and M3 respectively, use matrix algebra to find this catering firm’s profit on these two orders.

Answer Guidelines

(a)A =

This could be written as A = and would be an equally correct answer to part (a) but this matrix would need to be transposed in order to carry out the operations in (b).

(b)

  1. A is a 2x3 matrix and B is a 3x5 matrix so multiplication is possible as the number of columns in B is equal to the number of rows in B. It will result in a 2x5 matrix. AB = . This shows the ‘input requirements’ for each company’s order.
  2. B is a 3x5 matrix and C is a 5x1 matrix so multiplication is possible as the number of columns in B is equal to the number of rows in C. It will result in a 3x1 matrix. B.C = . This is the unit cost of making the three different menus M1, M2, and M3 respectively.
  3. A.B.C = AB.C = A.BC. AB is a 2x5 matrix and C is a 5x1 matrix so multiplication is possible as the number of columns in AB is equal to the number of rows in C and will result in a 2x1 matrix. Equally correct is to pre-multiply BC by A. A is a 2x3 matrix and BC is a 3x1 matrix so multiplication is possible as the number of columns in A is equal to the number of rows in BC and will result in a 2x1 matrix. A.B.C = . This is the cost of producing the ordered number of menus by company C1 and C2 respectively.
  4. I = IT =

IT is a 1x2 matrix and A is a 2x3 matrix so multiplication is possible as the number of columns in IT is equal to the number of rows in A and will result in a 1x3 matrix. IT.A = . This is the total quantities of each menu M1, M2, and M3 respectively.

  1. Price matrix is P = (Note: this could equally correctly be written as but it would need to be transposed before multiplying)

Profit = Revenues – Costs where:

Revenues = price * quantities which is A.P

A is a 2x3 matrix and P is a 3x1 matrix so multiplication is possible as the number of columns in A is equal to the number of rows in P and will result in a 2x1 matrix. Let the resulting matrix be labelled R.

R = AP =

Costs = A.B.C =

Profit = R – Costs = =

3.

Write the following simultaneous set of equations into matrix form using the method of Gauss Jordan pivoting to solve.

5p1 / - 2p2 / =15
p1 / + 8 p2 / =16

Answer Guidelines

  1. Putting the equations into matrix form: =
  2. For the Gauss Jordan pivot

Step 1 / Initialise by creating partitioned matrix /
Step 2
Iteration 1 / Select first pivot =P = a11 = 5;
Apply rule 1 /
Apply rule 2 /
Apply rule 3
Ni =O – R * where:
O is the corresponding element in the previous matrix
R is the value in this column in the pivot row in the previous matrix
C is the value in this row in the pivot column of the previous matrix / =
=
Iteration 2 / Select second pivot = P = a22= 8.4
Apply rule 1 /
Apply rule 2 /
Apply rule 3 /
=
=
Step 3 / Stop, solution obtained / p1 = 3.62 p2 =1.55

4.

Find the inverse of matrix A, A-1, and show that A. A-1 = I where I is the identity matrix.

A =

Answer Guidelines

  1. First, check whether inverse exists:

a)the matrix is square – matrix is square; and

b)by ensuring determinant of matrix is non-zero.

Value of determinant ≠ 0 therefore inverse exists.

  1. Find the matrix of co-factors:
  1. Find the transpose of the matrix of co-factors:
  2. A-1 = matrix of cofactors transposed

=

  1. A. A-1 = .

5.

A coal mine owner sells in three related markets: to coal fired power stations, to industry and to domestic properties. From the mines, three grades of coal can be produced – best, nuts and rough. In order to have an equilibrium in the three related markets, the following system of simultaneous equations must hold:

11pr / - pn / - pb / = 31
-2 pr / + 12 pn / + 4 pb / = 52
- pr / - 2 pn / + 7 pb / = 24

where pb, pn, and pr are the prices of best, nuts and rough respectively.

Using the method of matrix inverse,find the prices of rough, nuts and best coal which simultaneously solve this set of equations.

Answer Guidelines

  1. First, put the equation into matrix form:

Axb

  1. First, check whether inverse exists:
  2. the matrix is square – matrix is square; and
  3. by ensuring determinant of matrix is non-zero: = 802 ≠ 0 therefore the inverse exists
  4. Find the matrix of co-factors:
  5. Find the transpose of the matrix of co-factors:
  6. The inverse of A, A-1 is * transpose of co-factor matrix =
  7. To solve the system of equations:

= A-1b = * =

Hence the price of rough is £4 per ton, the price of nuts is £7 per ton and the price of best is £6