Transformation Using Matrix

Matrix is the powerful mathematical tool. It simplifies the mathematical problems in short cut and easy way. Most of the elementary geometrical transformation can be performed by the use of matrices. We have already discussed different types of transformations and the relation between the points and their corresponding image points. In this section, we show the important role of matrices in transforming the points.

The matrix which is used to transform a point is known as a transformation matrix.

Transformation by 2 x 1 matrix

Let us consider a column vector \(\begin{pmatrix}a\\b\\ \end{pmatrix}\). It can be written as 2 x 1 matrix \(\begin{bmatrix}a\\b\\ \end{bmatrix}\).

The matrix \(\begin{bmatrix}a\\b\\ \end{bmatrix}\) describes a translation of a unit in the x-direction and b units in the y-direction

The matrix \(\begin{bmatrix}3\\4\\ \end{bmatrix}\) describes the translation of 3 units in the x-direction and 4 units in the y-direction.

The image P' of any point P after translation by \(\begin{bmatrix}3\\4\\ \end{bmatrix}\) can be found by vector addition.

The image Q' of a point Q after translation by \(\begin{bmatrix}-3\\-4\\ \end{bmatrix}\) can be found as follows:

Let T = \(\begin{pmatrix}2\\1\\ \end{pmatrix}\) be the translation vector and p(x, y) will be the object point. Then the image P' (x', y') can be obtained as follows:

\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) = \(\begin{bmatrix}x\\y\\ \end{bmatrix}\) + \(\begin{bmatrix}2\\1\\ \end{bmatrix}\) = \(\begin{bmatrix}x + 2\\y + 1\\ \end{bmatrix}\)

∴ Image of P(x, y) is P' (x + 2, y + 1)

Hence, if\(\begin{pmatrix}a\\b\\ \end{pmatrix}\) be the translation vector, then P(x, y) will be translated into P(x + a, y + b).

Transformation using 2 x 2 matrices

A point (x, y) can be written as the column matrix\(\begin{bmatrix}x\\y\\ \end{bmatrix}\). If this column matrix is per-multiplied by any 2 x 2 matrix the result will be another column matrix such as

\(\begin{bmatrix}0&-1\\1&0\\ \end{bmatrix}\)\(\begin{bmatrix}x\\y\\ \end{bmatrix}\) =\(\begin{bmatrix}0.x - 1.y\\1.x + 0.y\\ \end{bmatrix}\) =\(\begin{bmatrix}-y\\x\\ \end{bmatrix}\)

Here, the matrix\(\begin{bmatrix}0&-1\\1&0\\ \end{bmatrix}\) maps the point (x, y) onto the point (-y, x). Thus any 2 x 2 matrix can be regarded as representing a geometrical transformation. It performs the transformation by pre-multiplying the column matrix which represent the object point.

Consider a 2 x 2 matrix\(\begin{bmatrix}a&b\\c&d\\ \end{bmatrix}\) and the point O (0, 0).

Now,\(\begin{bmatrix}a&b\\c&d\\ \end{bmatrix}\)\(\begin{bmatrix}0\\0\\ \end{bmatrix}\) =\(\begin{bmatrix}a.0 + b.0\\c.0 + d.0\\ \end{bmatrix}\) =\(\begin{bmatrix}0\\0\\ \end{bmatrix}\).

So, image of the origin (0, 0) under the transformation by the matrix\(\begin{bmatrix}a&b\\c&d\\ \end{bmatrix}\) is the origin itself. It shows that the origin is always invarient under a matrix transformation.

Again, consider a 2 x 2 matrix M =\(\begin{bmatrix}a&b\\c&d\\ \end{bmatrix}\) and a point P (x, y).

Then,

\(\begin{bmatrix}a&b\\c&d\\ \end{bmatrix}\)\(\begin{bmatrix}x\\y\\ \end{bmatrix}\) =\(\begin{bmatrix}ax + by\\cx + dy\\ \end{bmatrix}\).

Therefore, if\(\begin{bmatrix}x\\y\\ \end{bmatrix}\) denotes the position of a point (x, y), then 2 x 2 matrix \(\begin{bmatrix}a&b\\c&d\\ \end{bmatrix}\) transform this point onto (ax + by, cx + dy).

Reflection using 2 x 2 Matrices

(a) Reflection in X-axis

Let R_x denote the reflection in x-axis.

Then, R_x: P (x, y)→ P'(x, y)

If P'(x', y') represents the image of P(x, y) under R_x, then

x' = x = 1.x + 0.y

y' = -y = 0.x - 1.y

This system of linear equations may be written in matrix form as

\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) =\(\begin{bmatrix}1.x + 0.y\\0.x - 1.y\\ \end{bmatrix}\) =\(\begin{bmatrix}1&0\\0&-1\\ \end{bmatrix}\)\(\begin{bmatrix}x\\y\\ \end{bmatrix}\)

Hence, the matrix\(\begin{bmatrix}1&0\\0&-1\\ \end{bmatrix}\) represent the reflection in X-axis.

So, the image of the point (4, 5) under the reflection in X-axis is obtained as follows:

\(\begin{bmatrix}1&0\\0&-1\\ \end{bmatrix}\)\(\begin{bmatrix}4\\5\\ \end{bmatrix}\) =\(\begin{bmatrix}1.4 + 0.5\\0.4 - 1.5\\ \end{bmatrix}\) =\(\begin{bmatrix}4\\-5\\ \end{bmatrix}\)

∴ Image of (4, 5) is (4, -5).

(b) Reflection in y-axis

Let R_y denote reflection in Y-axis.

Then, R_y: P (x, y)→ P'(-x, y).

If P'(x', y') is the image of P (x, y), then

x' = -x = -1.x + 0.y y' = y = 0.x + 1.y

This system of linear equations may be written in matrix form as

\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) =\(\begin{bmatrix}-1.x +0.y\\0.x + 1.y\\ \end{bmatrix}\) =\(\begin{bmatrix}-1&0\\0&1\\ \end{bmatrix}\)\(\begin{bmatrix}x\\y\\ \end{bmatrix}\)Hence, the matrix\(\begin{bmatrix}-1&0\\0&1\\ \end{bmatrix}\) represents the reflection in Y-axis.

(c) Reflection in the line y = x

Let R be the reflection in the line y = x.

Then, R: P (x, y)→ P'(y, x)

If P'(x', y') is the image of P(x, y), then

x' = y = 0.x + 1.y

y' = x = 1.x + 0.y

This system of linear equations may be written in matrix form as

\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) =\(\begin{bmatrix}0.x + 1.y\\1.x + 0.y\\ \end{bmatrix}\) =\(\begin{bmatrix}0&1\\1&0\\ \end{bmatrix}\)\(\begin{bmatrix}x\\y\\ \end{bmatrix}\)

Hence, the matrix\(\begin{bmatrix}0&1\\1&0\\ \end{bmatrix}\) represents the reflection in the line y = x.

(c) Reflection in the line y = x

Let R be the reflection in the line y = -x,

Then, R: P(x, y)→ P'(-y, -x).

If P'(x', y') is the image of P(x, y), then

x' = -y = 0.x - 1.y

y' = -x = -1.x + 0.y

In the matrix form, this system can be written as

\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) =\(\begin{bmatrix}0.x - 1.y\\-1.x + 0.y\\ \end{bmatrix}\) =\(\begin{bmatrix}0&-1\\-1&0\\ \end{bmatrix}\)\(\begin{bmatrix}x\\y\\ \end{bmatrix}\).

Hence, the matrix\(\begin{bmatrix}0&-1\\-1&0\\ \end{bmatrix}\) represents the reflection in the line y = -x.

Rotation using 2 x 2 Matrices

(a) Positive Quarter turn about the origin

Let Q⁺ be the rotation through 90⁰ about the origin.

Then, Q⁺: P (x, y)→ P'(-y, x)

If P'(x', y') represents the image of P (x, y) after the rotation through 90⁰, then

x' = -y = 0.x - 1.y

y' = x = 1.x + 0.y

This system of linear equations can be written in matrix form as

\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) =\(\begin{bmatrix}0.x - 1.y\\1.x + 0.y\\ \end{bmatrix}\) =\(\begin{bmatrix}0&-1\\1&0\\ \end{bmatrix}\)\(\begin{bmatrix}x\\y\\ \end{bmatrix}\)

So, the matrix\(\begin{bmatrix}0&-1\\1&0\\ \end{bmatrix}\) represent the rotation through 90⁰ about the origin.

The images of a point after rotation through 90⁰ about the origin and after rotation through -270⁰ about the origin are always same.

So, the matrix\(\begin{bmatrix}0&-1\\1&0\\ \end{bmatrix}\) also represent the rotation through -270⁰ about the origin.

(b) Negative Quarter turn about the origin

Let Q be the rotation through -90⁰ about the origin.

Then Q: P (x, y)→ P'(y, -x).

If P'(x', y') represents the image of P(x, y) after the rotation through -90⁰, then

x' = y = 0.x + 1.y

y' = -x = -1.x + 0.y

This system of linear equation can be written in the matrix form as

\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) =\(\begin{bmatrix}0.x + 1.y\\-1.x + 0.y\\ \end{bmatrix}\) =\(\begin{bmatrix}0&1\\-1&0\\ \end{bmatrix}\) \(\begin{bmatrix}x\\y\\ \end{bmatrix}\)

So, the matrix \(\begin{bmatrix}0&1\\-1&0\\ \end{bmatrix}\) represents the rotation through -90⁰ about the origin. It also represents the rotation through 270⁰ about the origin.

(c) Half turn about origin

Let H be the half turn i.e. rotation through 180⁰ about the origin.

Then, H: P(x, y) → P'(-x, -y).

If P'(x', y') represents the image of P(x, y) under the rotation through 180⁰, then

x' = -x = -1.x + 0.y

y' = -y = 0.x - 1.y

This system of linear equations can be written as

\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) =\(\begin{bmatrix}-1.x + 0.y\\0.x - 1.y\\ \end{bmatrix}\) = \(\begin{bmatrix}-1&0\\0&-1\\ \end{bmatrix}\) \(\begin{bmatrix}x\\y\\ \end{bmatrix}\)

So, the matrix \(\begin{bmatrix}-1&0\\0&-1\\ \end{bmatrix}\) represents rotation through 180⁰ about the origin. It also represents the rotation through -180⁰ about the origin.

Enlargement using 2 x 2 Matrices

(a) Enlargement with centre (0, 0) and scale factor k.

Let E be the enlargement with centre at the origin and scale factor k.

Then, E: P(x, y) → P'(kx, ky)

If P'(x', y') represents the image of P(x, y) under E, then

x' = kx = k.x + 0.y

y' = ky = 0.x + k.y

This system of linear equations can be written in the matrix form as

\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) =\(\begin{bmatrix}k.x + 0.y\\0.x + k.y\\ \end{bmatrix}\) =\(\begin{bmatrix}k&0\\0&k\\ \end{bmatrix}\)\(\begin{bmatrix}x\\y\\ \end{bmatrix}\)

So, the matrix\(\begin{bmatrix}k&0\\0&k\\ \end{bmatrix}\) represents the enlargement with the centre at the origin and the scale factor k.

(b) Enlargement with centre (a,b) and scale factor k.

Let P (a, b) be the centre of the enlargement. Then, the position vector of P is

\(\overrightarrow{O}{P}\) =\(\begin{pmatrix}a\\b\\ \end{pmatrix}\)

Let A(x, y) be any other point on the plane. Then, the position vector of A is

\(\overrightarrow{O}{A'}\) =\(\begin{pmatrix}x\\y\\ \end{pmatrix}\)

Now, \(\overrightarrow{P}{A}\) = \(\overrightarrow{O}{A}\) - \(\overrightarrow{O}{P}\) =\(\begin{bmatrix}x\\y\\ \end{bmatrix}\) -\(\begin{bmatrix}a\\b\\ \end{bmatrix}\) =\(\begin{bmatrix}x&-a\\y&-b\\ \end{bmatrix}\)

Let A' (x', y') be the image of A (x, y) under the enlargement with scale factor k. Then, position vector of A' is

\(\overrightarrow{O}{A'}\) =\(\begin{pmatrix}x'\\y'\\ \end{pmatrix}\)

\(\overrightarrow{P}{A'}\) - \(\overrightarrow{O}{P'}\) =\(\begin{pmatrix}x'\\y'\\ \end{pmatrix}\) -\(\begin{pmatrix}a\\b\\ \end{pmatrix}\) =\(\begin{pmatrix}x'&-a\\y'&-b\\ \end{pmatrix}\)

Since the scale factor is k, then

\(\overrightarrow{P}{A'}\) = k.\(\overrightarrow{P}{A}\) 

\(\begin{pmatrix}x'&-a\\y'&-b\\ \end{pmatrix}\) = k\(\begin{pmatrix}x&-a\\y&-b\\ \end{pmatrix}\) =\(\begin{pmatrix}kx&-ka\\ky&-kb\\ \end{pmatrix}\)

∴ x' - a = kx - ka and y' - b = ky - kb

or, x' = kx - ka + a = k(x-a) + a

y' = ky - kb + b = k(y-b) + b

This system of linear equations can be represented in matrix form as

\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) =\(\begin{bmatrix}k(x - a) + a\\k(y - b) + b\\ \end{bmatrix}\) =\(\begin{bmatrix}k&0\\0&k\\ \end{bmatrix}\)\(\begin{bmatrix}x - a\\y - b\\ \end{bmatrix}\) +\(\begin{bmatrix}a\\b\\ \end{bmatrix}\)

Hence, if the centre of enlargement is (a, b) and the scale factor is k, then the image (x', y') of the point (x, y) can be obtained as follows:

\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) =\(\begin{bmatrix}k&0\\0&k\\ \end{bmatrix}\)\(\begin{bmatrix}x - a\\y - b\\ \end{bmatrix}\) +\(\begin{bmatrix}a\\b\\ \end{bmatrix}\) =\(\begin{bmatrix}kx - ka + a\\ky - kb + b\\ \end{bmatrix}\)

Hence, the image of the point (x, y) is (kx - ka + a, ky - kb + b).

Translations using 2 x 2 matrices

Let T =\(\begin{pmatrix}a\\b\\ \end{pmatrix}\) be the translation vector or the translation matrix. Then the image P' (x', y') of the point P (x, y) under the 2 x 1 matrix \(\begin{bmatrix}a\\b\\ \end{bmatrix}\) can be follows:\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) =\(\begin{bmatrix}x\\y\\ \end{bmatrix}\) +\(\begin{bmatrix}a\\b\\ \end{bmatrix}\) =\(\begin{bmatrix}x + a\\y + b\\ \end{bmatrix}\)

It also can be done using the 2 x 2 unit matrix\(\begin{bmatrix}1&0\\0&1\\ \end{bmatrix}\) as follows:

\(\begin{bmatrix}x'\\y'\\ \end{bmatrix}\) =\(\begin{bmatrix}1&0\\0&1\\ \end{bmatrix}\)\(\begin{bmatrix}x\\y\\ \end{bmatrix}\) +\(\begin{bmatrix}a\\b\\ \end{bmatrix}\) =\(\begin{bmatrix}x\\y\\ \end{bmatrix}\) +\(\begin{bmatrix}a\\b\\ \end{bmatrix}\) =\(\begin{bmatrix}x + a\\y + b\\ \end{bmatrix}\)

Geometric Transformations using Matrices 

Consider the vertices of a square O(0, 0), A(1, 0), B(1, 1) and C(0, 1). These coordinates can be displayed in the matrix form as:

\(\begin{bmatrix}O&A&B&C\\0&1&1&0\\0&0&1&1\\ \end{bmatrix}\)

The square matrix OABC with vertices O(0, 0), A(1, 0), B(1, 1) and C(0, 1) is called a unit square.

Again, consider other points O(0, 0), A(2, 0), B(2, 1) and C(0, 1). These coordinates can be displayed in the matrix form as

\(\begin{bmatrix}O&A&B&C\\0&2&2&0\\0&0&1&1\\ \end{bmatrix}\)

The quafrilateral OABC with the vertices O(0, 0), A(2, 0), B(2, 1) and C(0, 1) is a rectangle.

In order to transform any object or figure by using a 2 x 2 matrix, the matrix formed by the coordinates of the vertices of the object is pre-multiplied by the transformation matrix.

For example, the coordinates of the vertices A(2, 2), B(2, 6), C(5, 2) and D(5, 6) of rectangle ABCD are written as

\(\begin{bmatrix}A&B&C&D\\2&2&5&5\\2&6&2&6\\ \end{bmatrix}\)

It is transformed by a matrix M =\(\begin{bmatrix}2&0\\0&1\\ \end{bmatrix}\) as follows:

\(\begin{bmatrix}2&0\\0&1\\ \end{bmatrix}\)\(\begin{bmatrix}A&B&C&D\\2&2&5&5\\2&6&2&6\\ \end{bmatrix}\)=\(\begin{bmatrix}A'&B'&C'&D'\\2.2 + 0.2&2.2 + 0.6&2.5 + 0.2&2.5 + 0.6\\0.2 + 1.2&0.2 + 1.6&0.5 + 1.2&0.5 + 1.6\\ \end{bmatrix}\)=\(\begin{bmatrix}A'&B'&C'&D'\\4&4&10&10\\2&6&2&6\\ \end{bmatrix}\)

Hence, the coordinates of the vertices of the image are A'(4, 2), B'(4,6), C'(10, 2) and D'(10,6). Object figure ABCD and Image figure A'B'C'D' are shown in the graph.

Things to remember

• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.

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