Solution of simultaneous equation
Clement Radcliffe,Contributor

As we continue the review of matrices we will complete the presentation on solution of simultaneous equation.

Simultaneous equation

• The simultaneous equations are expressed in matrix form AX = B where A is the 2 x 2 coefficient matrix, X is the 2 x 1 matrix x and B the 2 x 1 matrix of the constant terms
• The 2 x 2 coefficient matrix A is converted to the unit matrix by pre-multiplying both sides by the inverse of A
  ... A -1 x A x X = A -1 B.
• By simplifying both sides, the equation of two 2 x 1 matrices remain.
• Equating terms will enable you to find the values of x and y, the solution of the original simultaneous equations.

Solution of homework

Solve the following simultaneous equations using matrices.

1. 2x + 5y = 6

3x + 4y = 8

Express in the matrix from C x X = D

\\ 2 5 x x = 6

3 4 y 8

The inverse of C = - 1 4 -5

7 - 3 2

Pre-multiply both sides by the inverse \\

- 1 4 -5 2 5 x = -1 4 -5 6

7 -3 2 3 4 y 7 -3 2 8

Simplify

á -1 -7 0 x = -1 -16

7 0 -7 y 7 -2

1 0 x =

0 1 y

\\ x = 16 and y = 2

7 7

2. Solve:

3x + 4y = 10

4x + 2y = 10

Express in matrix form

\\ 3 4 x = 10

4 2 y 10

The inverse of the coeffiecent matrix is -1 2 -4

10 - 4 3

\\ Pre-multiply by the inverse

-1 2 -4 3 4 x = - 1 2 -4 10

10 -4 3 4 2 y 10 -4 3 10

\\ - 1 - 10 0 x = -1 -20

10 0 - 10 y 10 -10

\\ x = 2

Y 1

\\ x = 2 and y = 1

I am sure that we can now proceed to the second application of matrices.

MATRIX TRANSFORMATION

Points to note

á In general, transformation maps a point (x, y) (object) on to its image (X1, Y1); that is (x, y) moves to (X1, Y1).

á In matrix transformation, the process involves pre-multiplying the column vector of the coordinates of the point by the 2 x 2 matrix. The product is the image.

For example, a b x = x1

c d y y1

Object image

[ALWAYS REMEMBER THIS FORMAT]

Example

The matrix A 3 -1 maps the point X (2, 1) onto its image X1. Determine the

0 2

coordinates of X1.

Solution

a b x = x1

c d y y1

3 -1 2 = 3 x 2 + -1 x 1 = 5

0 2 1 0 x 2 + 2 x 1 2

\\ Coordinates of X1 = (5, 2)

We will now look at another example.

Example

The equation of a straight line PQ is y = 3x + 1, and T is a transformation represented by the matrix

3 0

0 1

Determine

(i) the value of k if P is the point (2, k)

(ii) the coordinates of the image of P under the transformation T.

Solution

(i) Since y = 3x + 1 and P is the point (2, k)

Substituting x = 2, therefore y = 3 x 2 + 1 = 7. Since k is the y coordinate of the point P, then k = 7.

(ii) P (2, 7) under the transformation T is:

3 0 2 = 6

0 1 7 7

\\ The coordinates of the image of P is (6, 7)

For failing to write the answer in coordinate form (6, 7) you may be penalised.

As the 2 x 2 matrix maps the object on to its image, then given the coordinates of the object and its corresponding image. The 2 x 2 matrix can be found by using simultaneous equations.

Example

A transformation U maps the parallelogram OABC with vertices O (0,0) , A (1, 2), B (5, 1) and C (4, -1) on to 0 1 (0, 0), A1 (2, -1), B1 (1, -5) and C1 (-1, -4).

Find the matrix U which maps the object on to its image.

Solution

Let the unknown matrix be a b

c d

Since O is mapped on to O1 Similarly A is mapped on to A1

a b 0 = 0 É.. (1) a b 1 = 2....(2)

c d 0 0 c d 2 - 1

B is mapped on to B1 C is mapped on to C1

a b 5 = 1 É(3) a b 4 = - 1 ....(4)

c d 1 - 5 c d - 1 - 4

From (2) From (3) From (4)

a + 2b = 2 5a + b 1 4a - b = - 1

c + 2d - 1É. (5) 5c + d -5 ÉÉ (6) 4c - d - 4 ...(7)

From (6) and (7)

5a + b = 1

4a - b = -1 By adding both, 9a = 0, \\ a = 0,

Substituting b = 1

From (6) and (7)

5c + d = - 5

4c - d = -4 By adding both, 9c = -9, \\ c = -1,

Substituting d= 0

\\ the matrix U is 0 1

-1 0

It is advisable that you test your answer with A and A1 as follows.

0 1 1 = 2

-1 0 2 -1

Homework

Triangle PQR has vertices P (1, 4), Q (3, 1), and R (4, 2). The triangle is transformed by the matrix is 0 1

-1 0

Determine

(i) The coordinates of image of P1 Q1 R1.

(ii) The 2 x 2 matrix which transforms triangle P 1 Q1 R1 on to triangle PQR.

Next week will continue our review in this area.

Clement Radcliffe is the principal of Glenmuir High School in May Pen.