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Long Questions

A line segment AB joining the points A(4, 1) and B(7, 5) is transformed to the line segment A'B' joining the points A'(-4, 1) and B'(-7, 5). Find the 2 × 2 matrix that represents this transformation.

Let the 2 × 2 transformation matrix (T.M.) be [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}
Object = [4715]\begin{bmatrix} 4 & 7 \\ 1 & 5 \end{bmatrix}
Image = [−4−715]\begin{bmatrix} -4 & -7 \\ 1 & 5 \end{bmatrix}

Now, we know, Image = T.M. × Object or, [−4−715]\begin{bmatrix} -4 & -7 \\ 1 & 5 \end{bmatrix} = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} [4715]\begin{bmatrix} 4 & 7 \\ 1 & 5 \end{bmatrix}
or, [−4−715]\begin{bmatrix} -4 & -7 \\ 1 & 5 \end{bmatrix} = [4a+b7a+5b4c+d7c+5d]\begin{bmatrix} 4a + b & 7a + 5b \\ 4c + d & 7c + 5d \end{bmatrix}

Comparing the corresponding value of equal matrix,

4a + b = -4 or, b = -4 - 4a --- (i) 7a + 5b = -7 or, 7a + 5(-4 - 4a) = -7
or, 7a - 20 - 20a = -7
so, a = -1
also, b = 0 (from 'i') 4c + d = 1 or, d = 1 - 4c --- (ii) 7c + 5d = 5 or, 7c + 5(1 - 4c) = 5
or, 7c + 5 - 20c = 5
or, -13c = 0
so, c = 0
also, d = 1 (from 'ii')

∴ T.M. = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} = [−1001]\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}

Find the 2 × 2 transformation matrix which transforms the unit square into a parallelogram [04620132]\begin{bmatrix} 0 & 4 & 6 & 2 \\ 0 & 1 & 3 & 2 \end{bmatrix}

Let the 2 × 2 transformation matrix (T.M.) be [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}
Object = The unit square, [01100011]\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}
Image = [04620132]\begin{bmatrix} 0 & 4 & 6 & 2 \\ 0 & 1 & 3 & 2 \end{bmatrix}

Now, we know, Image = T.M. × Object or, [04620132]\begin{bmatrix} 0 & 4 & 6 & 2 \\ 0 & 1 & 3 & 2 \end{bmatrix} = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} [01100011]\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}
or, [04620132]\begin{bmatrix} 0 & 4 & 6 & 2 \\ 0 & 1 & 3 & 2 \end{bmatrix} = [0+0a+0a+b0+b0+0c+0c+d0+d]\begin{bmatrix} 0 + 0 & a + 0 & a + b & 0 + b \\ 0 + 0 & c + 0 & c + d & 0 + d \end{bmatrix}

Comparing the corresponding value of equal matrix,
a = 4, b = 2
c = 1, d = 2

∴ T.M. = [abcd]\begin{bmatrix} a & b \ c & d \end{bmatrix} = [4212]\begin{bmatrix} 4 & 2 \ 1 & 2 \end{bmatrix}

Find the 2 × 2 matrix, which transforms [02320351]\begin{bmatrix} 0 & 2 & 3 & 2 \\ 0 & 3 & 5 & 1 \end{bmatrix} into [03510232]\begin{bmatrix} 0 & 3 & 5 & 1 \\ 0 & 2 & 3 & 2 \end{bmatrix}

Let the 2 × 2 transformation matrix (T.M.) be [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}
Object = [02320351]\begin{bmatrix} 0 & 2 & 3 & 2 \\ 0 & 3 & 5 & 1 \end{bmatrix}
Image = [03510232]\begin{bmatrix} 0 & 3 & 5 & 1 \\ 0 & 2 & 3 & 2 \end{bmatrix}

Now, we know, Image = T.M. × Object or, [03510232]\begin{bmatrix} 0 & 3 & 5 & 1 \\ 0 & 2 & 3 & 2 \end{bmatrix} = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} [02320351]\begin{bmatrix} 0 & 2 & 3 & 2 \\ 0 & 3 & 5 & 1 \end{bmatrix}
or, [03510232]\begin{bmatrix} 0 & 3 & 5 & 1 \\ 0 & 2 & 3 & 2 \end{bmatrix} = [0+02a+3b3a+5b2a+b0+02c+3d3c+5d2c+d]\begin{bmatrix} 0 + 0 & 2a + 3b & 3a + 5b & 2a + b \\ 0 + 0 & 2c + 3d & 3c + 5d & 2c + d \end{bmatrix}

Comparing the corresponding value of equal matrix,
2a + 3b = 3 --- (i)
2a + b = 1 --- (ii)
2c + 3d = 2 --- (iii)
2c + d = 2 --- (iv)

Subtracting (i) by (ii),
2a+3b=32a+b=12b=2\begin{array}{cc} 2a + 3b = 3 \\ 2a + b = 1 \\ \hline 2b = 2 \end{array}
So, b = 1
Then, a = 0 (from 'i')

Again, Subtracting (iii) by (iv),
2c+3d=22c+d=22d=0\begin{array}{cc} 2c + 3d = 2 \\ 2c + d = 2 \\ \hline 2d = 0 \end{array}
So, d = 0
Then, c = 1 (from 'iii')

∴ T.M. = [abcd]\begin{bmatrix} a & b \ c & d \end{bmatrix} = [0110]\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}

Find the 2 × 2 matrix which transform the unit square [01100011]\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix} to a parallelogram [03410132]\begin{bmatrix} 0 & 3 & 4 & 1 \\ 0 & 1 & 3 & 2 \end{bmatrix}

Let the 2 × 2 transformation matrix (T.M.) be [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}
Object = [01100011]\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}
Image = [03410132]\begin{bmatrix} 0 & 3 & 4 & 1 \\ 0 & 1 & 3 & 2 \end{bmatrix}

Now, we know, Image = T.M. × Object or, [03410132]\begin{bmatrix} 0 & 3 & 4 & 1 \\ 0 & 1 & 3 & 2 \end{bmatrix} = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} [01100011]\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}
or, [03410132]\begin{bmatrix} 0 & 3 & 4 & 1 \\ 0 & 1 & 3 & 2 \end{bmatrix} = [0a+0a+b0+b0c+0c+d0+d]\begin{bmatrix} 0 & a + 0 & a + b & 0 + b \\ 0 & c + 0 & c + d & 0 + d \end{bmatrix}

Comparing the corresponding value of equal matrix,
a = 3, b = 1
c = 1, d = 2

∴ T.M. = [abcd]\begin{bmatrix} a & b \ c & d \end{bmatrix} = [3112]\begin{bmatrix} 3 & 1 \ 1 & 2 \end{bmatrix}

A square ABCD with vertices A(2, 0), B(5, 1), C(4, 4) and D(1, 3) is mapped onto a parallelogram A'B'C'D' by a 2 × 2 matrix so that the vertices of the parallelogram are A'(2, 2), B'(7, 3), C'(12, -4) and D'(7, -5). Find the 2 × 2 transformation matrix.

Let the 2 × 2 transformation matrix (T.M.) be [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}
Object = [25410143]\begin{bmatrix} 2 & 5 & 4 & 1 \\ 0 & 1 & 4 & 3 \end{bmatrix}
Image = [2712723−4−5]\begin{bmatrix} 2 & 7 & 12 & 7 \\ 2 & 3 & -4 & -5 \end{bmatrix}

Now, we know, Image = T.M. × Object or, [2712723−4−5]\begin{bmatrix} 2 & 7 & 12 & 7 \\ 2 & 3 & -4 & -5 \end{bmatrix} = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} [2712723−4−5]\begin{bmatrix} 2 & 7 & 12 & 7 \\ 2 & 3 & -4 & -5 \end{bmatrix}
or, [2712723−4−5]\begin{bmatrix} 2 & 7 & 12 & 7 \\ 2 & 3 & -4 & -5 \end{bmatrix} = [2a+05a+b4a+4ba+3b2c+05c+d4c+4dc+3d]\begin{bmatrix} 2a + 0 & 5a + b & 4a + 4b & a + 3b \\ 2c + 0 & 5c + d & 4c + 4d & c + 3d \end{bmatrix}

Comparing the corresponding value of equal matrix,

2 = 2a
∴ a = 1
2c = 2
∴ c = 1
5a + b = 7
∴ b = 2
5c + d = 3
∴ d = -2

∴ T.M. = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} = [121−2]\begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix}

Find a 2 × 2 matrix which transforms a △PQR with vertices P(4, 3), Q(6, 4) and R(8, 1) into P'(-3, -4), Q'(-4, -6) and R'(-1, -8).

Let the 2 × 2 transformation matrix (T.M.) be [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}
Object = [468341]\begin{bmatrix} 4 & 6 & 8 \\ 3 & 4 & 1 \end{bmatrix}
Image = [−3−4−1−4−6−8]\begin{bmatrix} -3 & -4 & -1 \\ -4 & -6 & -8 \end{bmatrix}

Now, we know, Image = T.M. × Object or, [−3−4−1−4−6−8]\begin{bmatrix} -3 & -4 & -1 \\ -4 & -6 & -8 \end{bmatrix} = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} [468341]\begin{bmatrix} 4 & 6 & 8 \\ 3 & 4 & 1 \end{bmatrix}
or, [−3−4−1−4−6−8]\begin{bmatrix} -3 & -4 & -1 \\ -4 & -6 & -8 \end{bmatrix} = [4a+3b6a+4b8a+b4c+3d6c+4d8c+d]\begin{bmatrix} 4a + 3b & 6a + 4b & 8a + b \\ 4c + 3d & 6c + 4d & 8c + d \end{bmatrix}

Comparing the corresponding value of equal matrix,
4a + 3b = -3 --- (i)
8a + b = -1 --- (ii)
4c + 3d = -4 --- (iii)
8c + d = -8 --- (iv)

Multiplying (i) by 2 and subtracting it by (ii),
8a+6b=−68a+b=−15b=−5\begin{array}{cc} 8a + 6b = -6 \\ 8a + b = -1 \\ \hline 5b = -5 \end{array}
So, b = -1
Then, a = 0 (from 'i')

Again, multiplying (iii) by 2 and subtracting it by (iv),
8c+6d=−88c+d=−85d=0\begin{array}{cc} 8c + 6d = -8 \ 8c + d = -8 \ \hline 5d = 0 \end{array}
So, d = 0
Then, c = -1 (from 'iv')

∴ T.M. = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} = [0−1−10]\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}

The parallelogram QRST has the vertices W(-1, 1), R(-2, -1), S(2, -1) and T(3, 1). Transform the given parallelogram under the matrix [4320]\begin{bmatrix} 4 & 3 \\ 2 & 0 \end{bmatrix} and find the coordinates of vertices of its image.

Let the image of parallelogram QRST be [acegbdfh]\begin{bmatrix} a & c & e & g \\ b & d & f & h \end{bmatrix}
Object = [−1−2231−1−11]\begin{bmatrix} -1 & -2 & 2 & 3 \\ 1 & -1 & -1 & 1 \end{bmatrix}
Transformation matrix (T.M.) = [4320]\begin{bmatrix} 4 & 3 \\ 2 & 0 \end{bmatrix}

We know, Image = T.M. × Object or, [acegbdfh]\begin{bmatrix} a & c & e & g \\ b & d & f & h \end{bmatrix} = [4320]\begin{bmatrix} 4 & 3 \\ 2 & 0 \end{bmatrix} [−1−2231−1−11]\begin{bmatrix} -1 & -2 & 2 & 3 \\ 1 & -1 & -1 & 1 \end{bmatrix} or, [acegbdfh]\begin{bmatrix} a & c & e & g \\ b & d & f & h \end{bmatrix} = [−4+3−8−38−312+3−2+0−4−04−06+0]\begin{bmatrix} -4 + 3 & -8 - 3 & 8 - 3 & 12 + 3 \\ -2 + 0 & -4 - 0 & 4 - 0 & 6 + 0 \end{bmatrix} ∴ [acegbdfh]\begin{bmatrix} a & c & e & g \\ b & d & f & h \end{bmatrix} = [−1−11515−2−446]\begin{bmatrix} -1 & -11 & 5 & 15 \\ -2 & -4 & 4 & 6 \end{bmatrix}

Hence, the coordinates of vertices of the image are Q'(-1, 2), R'(-11, -4), S'(5, 4), and T'(15, 6)

The square WXYZ has the vertices W(0, 3), X(1, 1), Y(3, 2) and Z(2, 4). Transform the given square WXYZ under the matrix [0−1−10]\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} and find the coordinates if the vertices of its image.

Let the image of parallelogram WXYZ be [acegbdfh]\begin{bmatrix} a & c & e & g \\ b & d & f & h \end{bmatrix}
Object = [01323124]\begin{bmatrix} 0 & 1 & 3 & 2 \\ 3 & 1 & 2 & 4 \end{bmatrix}
Transformation matrix (T.M.) = [0−1−10]\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}

We know, Image = T.M. × Object or, [acegbdfh]\begin{bmatrix} a & c & e & g \\ b & d & f & h \end{bmatrix} = [0−1−10]\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} [01323124]\begin{bmatrix} 0 & 1 & 3 & 2 \\ 3 & 1 & 2 & 4 \end{bmatrix} ∴ [acegbdfh]\begin{bmatrix} a & c & e & g \\ b & d & f & h \end{bmatrix} = [−3−1−2−40−1−3−2]\begin{bmatrix} -3 & -1 & -2 & -4 \\ 0 & -1 & -3 & -2 \end{bmatrix}

Hence, the coordinates of vertices of the image are W'(-3, 0), X'(-1, -1), Y'(-2, -3), and Z'(-4, -2)

Which 2 × 2 matrix maps a unit square to parallelogram [003−15−22−1]\begin{bmatrix} 0 & 0 & 3 & -1 \\ 5 & -2 & 2 & -1 \end{bmatrix} ?

Here,
Let the 2 × 2 transformation matrix (T.M.) be [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}
Object = The unit square, [01100011]\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}
Image = [003−15−22−1]\begin{bmatrix} 0 & 0 & 3 & -1 \\ 5 & -2 & 2 & -1 \end{bmatrix}

Now, we know, Image = T.M. × Object or, [003−15−22−1]\begin{bmatrix} 0 & 0 & 3 & -1 \\ 5 & -2 & 2 & -1 \end{bmatrix} = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} [01100011]\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}
or, [003−15−22−1]\begin{bmatrix} 0 & 0 & 3 & -1 \\ 5 & -2 & 2 & -1 \end{bmatrix} = [0a+0a+b0+b0c+0c+d0+d]\begin{bmatrix} 0 & a + 0 & a + b & 0 + b \\ 0 & c + 0 & c + d & 0 + d \end{bmatrix}

Comparing the corresponding value of equal matrix,
a = 0, b = -1
c = -2, d = -1

∴ T.M. = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} = [0−1−2−1]\begin{bmatrix} 0 & -1 \\ -2 & -1 \end{bmatrix}

Find the 2 × 2 transformation matrix which transforms a square ABCD with vertices A(2, 3), B(4, 3), C(4, 5) and D(2, 5) into a square A'B'C'D' with vertices A'(3, 2), B'(3, 4), C'(5, 4) and D'(5, 2).

Let the 2 × 2 transformation matrix (T.M.) be [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}
Object = [24423355]\begin{bmatrix} 2 & 4 & 4 & 2 \\ 3 & 3 & 5 & 5 \end{bmatrix}
Image = [33552442]\begin{bmatrix} 3 & 3 & 5 & 5 \\ 2 & 4 & 4 & 2 \end{bmatrix}

Now, we know, Image = T.M. × Object or, [33552442]\begin{bmatrix} 3 & 3 & 5 & 5 \\ 2 & 4 & 4 & 2 \end{bmatrix} = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} [24423355]\begin{bmatrix} 2 & 4 & 4 & 2 \\ 3 & 3 & 5 & 5 \end{bmatrix}
or, [33552442]\begin{bmatrix} 3 & 3 & 5 & 5 \\ 2 & 4 & 4 & 2 \end{bmatrix} = [2a+3b4a+3b4a+5b2a+5b2c+3d4c+4d4c+5d2c+5d]\begin{bmatrix} 2a + 3b & 4a + 3b & 4a + 5b & 2a + 5b \\ 2c + 3d & 4c + 4d & 4c + 5d & 2c + 5d \end{bmatrix}

Comparing the corresponding value of equal matrix,
4a + 3b = 3 --- (i)
4a + 5b = 5 --- (ii)
4c + 4d = 4 --- (iii)
4c + 5d = 4 --- (iv)

Subtracting (i) by (ii),
4a+3b=34a+5b=5−2b=−2\begin{array}{cc} 4a + 3b = 3 \\ 4a + 5b = 5 \\ \hline -2b = -2 \end{array}
So, b = 1
Then, a = 0 (from 'i')

Again, subtracting (iii) by (iv),
4c+4d=44c+5d=4d=0\begin{array}{cc} 4c + 4d = 4 \\ 4c + 5d = 4 \\ \hline d = 0 \end{array}
So, d = 0
Then, c = 1 (from 'iv')

∴ T.M. = [abcd]\begin{bmatrix} a & b \ c & d \end{bmatrix} = [0110]\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}

△ABC having the vertices A(3, 6), B(5, -3) and C(-4, 2) is transformed by a 2 × 2 matrix so that the coordinates of the vertices of its images are A'(-3, -6), B'(-5, 3) and C'(4, -2). Find the 2 × 2 matrix.

Let the 2 × 2 transformation matrix (T.M.) be [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}
Object = [35−46−32]\begin{bmatrix} 3 & 5 & -4 \\ 6 & -3 & 2 \end{bmatrix}
Image = [−3−54−63−2]\begin{bmatrix} -3 & -5 & 4 \\ -6 & 3 & -2 \end{bmatrix}

Now, we know, Image = T.M. × Object or, [−3−54−63−2]\begin{bmatrix} -3 & -5 & 4 \\ -6 & 3 & -2 \end{bmatrix} = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} [35−46−32]\begin{bmatrix} 3 & 5 & -4 \\ 6 & -3 & 2 \end{bmatrix}
or, [−3−54−63−2]\begin{bmatrix} -3 & -5 & 4 \\ -6 & 3 & -2 \end{bmatrix} = [3a+6b5a−3b−4a+2b3c+6d5c−3d−4c+2d]\begin{bmatrix} 3a + 6b & 5a - 3b & -4a + 2b \\ 3c + 6d & 5c - 3d & -4c + 2d \end{bmatrix}

Comparing the corresponding value of equal matrix,
3a + 6b = -3 --- (i)
5a - 3b = -5 --- (ii)
3c + 6d = -6 --- (iii)
5c - 3d = 3 --- (iv)

Multiplying (ii) by 2 and adding it with (i),
10a−6b=−103a+6b=−313a=−13\begin{array}{cc} 10a - 6b = -10 \ 3a + 6b = -3 \ \hline 13a = -13 \end{array}
So, a = -1
Then, b = 0 (from 'i')

Again, multiplying (iv) by 2 abd adding with (iii),
10c−6d=63c+6d=−613c=0\begin{array}{cc} 10c - 6d = 6 \\ 3c + 6d = -6 \\ \hline 13c = 0 \end{array}
So, c = 0
Then, d = -1 (from 'iv')

∴ T.M. = [abcd]\begin{bmatrix} a & b \ c & d \end{bmatrix} = [−100−1]\begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}

A square PQRS with vertices P(-2, 0), Q(-5, 1), R(-4, 4) and S(-1, 3) is mapped onto a parallelogram P'Q'R'S' by a 2 × 2 matrix so that the vertices of the parallelogram are P'(-2, -2), Q'(-3, -7), R'(4, -12) and S'(5, -7). Find the 2 × 2 transformation matrix.

Let the 2 × 2 transformation matrix (T.M.) be [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}
Object = [−2−5−4−10143]\begin{bmatrix} -2 & -5 & -4 & -1 \\ 0 & 1 & 4 & 3 \end{bmatrix}
Image = [−2−345−2−7−12−7]\begin{bmatrix} -2 & -3 & 4 & 5 \\ -2 & -7 & -12 & -7 \end{bmatrix}

Now, we know, Image = T.M. × Object or, [−2−345−2−7−12−7]\begin{bmatrix} -2 & -3 & 4 & 5 \\ -2 & -7 & -12 & -7 \end{bmatrix} = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} [−2−5−4−10143]\begin{bmatrix} -2 & -5 & -4 & -1 \\ 0 & 1 & 4 & 3 \end{bmatrix} or, [−2−345−2−7−12−7]\begin{bmatrix} -2 & -3 & 4 & 5 \\ -2 & -7 & -12 & -7 \end{bmatrix} = [−2a−5a+b−4a+4b−a+3b−2c−5c+d−4c+4d−c+3d]\begin{bmatrix} -2a & -5a + b & -4a + 4b & -a + 3b \\ -2c & -5c + d & -4c + 4d & -c + 3d \end{bmatrix}

Comparing the corresponding value of equal matrix,

-2 = -2a
∴ a = 1
-2c = -2
∴ c = 1
-5a + b = -3
∴ b = 2
-5c + d = 7
∴ d = -2

∴ T.M. = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} = [121−2]\begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix}

A square ABCD whose vertices are A(0, 3), B(1, 1), C(3, 2) and D(2, 4) is mapped to the parallelogram A'B'C'D' by a 2 × 2 matrix so that the vertices of the parallelogram are A'(6, -6), B'(3, -1), C(7, -1) and D'(10, -6). Find 2 × 2 matrix.

Let the 2 × 2 transformation matrix (T.M.) be [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}
Object = [01323124]\begin{bmatrix} 0 & 1 & 3 & 2 \\ 3 & 1 & 2 & 4 \end{bmatrix}
Image = [63710−6−1−1−6]\begin{bmatrix} 6 & 3 & 7 & 10 \\ -6 & -1 & -1 & -6 \end{bmatrix}

Now, we know, Image = T.M. × Object or, [63710−6−1−1−6]\begin{bmatrix} 6 & 3 & 7 & 10 \\ -6 & -1 & -1 & -6 \end{bmatrix} = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} [01323124]\begin{bmatrix} 0 & 1 & 3 & 2 \\ 3 & 1 & 2 & 4 \end{bmatrix}
or, [63710−6−1−1−6]\begin{bmatrix} 6 & 3 & 7 & 10 \\ -6 & -1 & -1 & -6 \end{bmatrix} = [3ba+b3a+2b2a+4b3dc+d3c+2d2c+4d]\begin{bmatrix} 3b & a + b & 3a + 2b & 2a + 4b \\ 3d & c + d & 3c + 2d & 2c + 4d \end{bmatrix}

Comparing the corresponding value of equal matrix,

6 = 3b
∴ b = 2
-6 = 3d
∴ d = -2
a + b = 3
∴ a = 1
c + d = -1
∴ c = 1

∴ T.M. = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} = [121−2]\begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix}

A unit square MNOP having vertices M(0, 0), N(1, 0), O(1, 1) and P(0, 1) are transformed under transformation matrix which represents y = x, find the coordinates of the images of quadrilateral M'N'O'P' so formed.

Given vertices of unit square MNOP are M(0, 0), N(1, 0), O(1, 1) and P(0, 1)
Let, Object = [xy]\begin{bmatrix} x \\ y \end{bmatrix} and Transformation matrix (T.M.) = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}
Image on reflection on y = -x is [−y−x]\begin{bmatrix} -y \\ -x \end{bmatrix}

Now, we know, Image = T.M. × Object --- (I) or, [−y−x]\begin{bmatrix} -y \\ -x \end{bmatrix} = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix} × [xy]\begin{bmatrix} x \\ y \end{bmatrix}
or, [−y−x]\begin{bmatrix} -y \\ -x \end{bmatrix} = [ax+bycx+dy]\begin{bmatrix} ax + by \\ cx + dy \end{bmatrix}

Above condition will be true if,
a = 0, b = -1
c = -1, d = 0

So, Transformation matrix = [0−1−10]\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}

Now, using (I) at given points,
M' = [0−1−10]\begin{bmatrix} 0 & -1 \ -1 & 0 \end{bmatrix} [00]\begin{bmatrix} 0 \ 0 \end{bmatrix} = [00]\begin{bmatrix} 0 \ 0 \end{bmatrix}
N' = [0−1−10]\begin{bmatrix} 0 & -1 \ -1 & 0 \end{bmatrix} [10]\begin{bmatrix} 1 \ 0 \end{bmatrix} = [0−1]\begin{bmatrix} 0 \ -1 \end{bmatrix}
O' = [0−1−10]\begin{bmatrix} 0 & -1 \ -1 & 0 \end{bmatrix} [11]\begin{bmatrix} 1 \ 1 \end{bmatrix} = [−1−1]\begin{bmatrix} -1 \ -1 \end{bmatrix}
P' = [0−1−10]\begin{bmatrix} 0 & -1 \ -1 & 0 \end{bmatrix} [01]\begin{bmatrix} 0 \ 1 \end{bmatrix} = [−10]\begin{bmatrix} -1 \ 0 \end{bmatrix}

Hence, the coordinates of the images of quadrilateral M'N'O'P' are M'(0, 0), N'(0, -1), O'(-1, -1), P'(-1, 0).

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