Exercise 3.1

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Question 1

In the matrix

$$ A= \begin{bmatrix} 2 & 5 & 19 & -7\\ 35 & -2 & \dfrac{5}{2} & 12\\ \sqrt3 & 1 & -5 & 17 \end{bmatrix}, $$

write:

1. The order of the matrix.
2. The number of elements.
3. The elements \(a_{13},\ a_{21},\ a_{33},\ a_{24},\ a_{23}\).

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Solution

(i) Order of the matrix

The given matrix has

$$ 3 $$

rows and

$$ 4 $$

columns.

Therefore, the order of the matrix is

$$ \boxed{3\times4.} $$

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(ii) Number of elements

The number of elements in a matrix is given by

$$ \text{Rows}\times\text{Columns}. $$

Hence,

$$ 3\times4=12. $$

Therefore, the matrix contains

$$ \boxed{12\text{ elements}.} $$

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(iii) Write the following elements

The element

$$ a_{13} $$

is the element in the first row and third column.

$$ a_{13}=19. $$

The element

$$ a_{21} $$

is the element in the second row and first column.

$$ a_{21}=35. $$

The element

$$ a_{33} $$

is the element in the third row and third column.

$$ a_{33}=-5. $$

The element

$$ a_{24} $$

is the element in the second row and fourth column.

$$ a_{24}=12. $$

The element

$$ a_{23} $$

is the element in the second row and third column.

$$ a_{23}=\frac52. $$

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Answer:

$$ \boxed{\text{Order}=3\times4} $$ $$ \boxed{\text{Number of elements}=12} $$ $$ \boxed{a_{13}=19,\quad a_{21}=35,\quad a_{33}=-5,\quad a_{24}=12,\quad a_{23}=\frac52} $$

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Question 2

If a matrix has

$$ 24 $$

elements, what are the possible orders it can have? What if it has

$$ 13 $$

elements?

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Solution

The total number of elements in a matrix is equal to

$$ \text{Number of rows}\times\text{Number of columns}. $$

Therefore, we find all possible factor pairs of the given number.

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(i) Matrix having 24 elements

The factor pairs of

$$ 24 $$

are

$$ 1\times24,\; 2\times12,\; 3\times8,\; 4\times6. $$

Since the order can also be obtained by interchanging rows and columns, the possible orders are

$$ 1\times24,\; 24\times1, $$ $$ 2\times12,\; 12\times2, $$ $$ 3\times8,\; 8\times3, $$ $$ 4\times6,\; 6\times4. $$

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(ii) Matrix having 13 elements

Since

$$ 13 $$

is a prime number, its only factor pairs are

$$ 1\times13 \quad\text{and}\quad 13\times1. $$

Hence, the possible orders are

$$ 1\times13 \quad\text{and}\quad 13\times1. $$

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Answer:

For 24 elements:

$$ \boxed{1\times24,\;24\times1,\;2\times12,\;12\times2,\;3\times8,\;8\times3,\;4\times6,\;6\times4} $$

For 13 elements:

$$ \boxed{1\times13,\;13\times1} $$

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Question 3

If a matrix has

$$ 18 $$

elements, what are the possible orders it can have? What if it has

$$ 5 $$

elements?

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Solution

The number of elements in a matrix is equal to

$$ \text{Number of rows}\times\text{Number of columns}. $$

Hence, we find all possible factor pairs of the given numbers.

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(i) Matrix having 18 elements

The factor pairs of

$$ 18 $$

are

$$ 1\times18,\; 2\times9,\; 3\times6. $$

Interchanging rows and columns also gives valid matrix orders.

Therefore, the possible orders are

$$ 1\times18,\; 18\times1, $$ $$ 2\times9,\; 9\times2, $$ $$ 3\times6,\; 6\times3. $$

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(ii) Matrix having 5 elements

Since

$$ 5 $$

is a prime number, its only factor pairs are

$$ 1\times5 \quad\text{and}\quad 5\times1. $$

Hence, the possible orders are

$$ 1\times5 \quad\text{and}\quad 5\times1. $$

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Answer:

For 18 elements:

$$ \boxed{1\times18,\;18\times1,\;2\times9,\;9\times2,\;3\times6,\;6\times3} $$

For 5 elements:

$$ \boxed{1\times5,\;5\times1} $$

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Question 4

Construct a

$$ 2\times2 $$

matrix

$$ A=[a_{ij}], $$

whose elements are given by:

1. $$ a_{ij}=\frac{(i+j)^2}{2} $$
2. $$ a_{ij}=\frac{i}{j} $$
3. $$ a_{ij}=\frac{(i+2j)^2}{2} $$

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(i)

Since the matrix is of order

$$ 2\times2, $$

we take

$$ i,j=1,2. $$

Now,

$$ a_{11}=\frac{(1+1)^2}{2}=\frac42=2, $$ $$ a_{12}=\frac{(1+2)^2}{2}=\frac92, $$ $$ a_{21}=\frac{(2+1)^2}{2}=\frac92, $$ $$ a_{22}=\frac{(2+2)^2}{2}=\frac{16}{2}=8. $$

Hence,

$$ \boxed{ A= \begin{bmatrix} 2 & \dfrac92\\[2mm] \dfrac92 & 8 \end{bmatrix} } $$

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(ii)

Using

$$ a_{ij}=\frac{i}{j}, $$

we obtain

$$ a_{11}=1, $$ $$ a_{12}=\frac12, $$ $$ a_{21}=2, $$ $$ a_{22}=1. $$

Therefore,

$$ \boxed{ A= \begin{bmatrix} 1 & \dfrac12\\[2mm] 2 & 1 \end{bmatrix} } $$

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(iii)

Using

$$ a_{ij}=\frac{(i+2j)^2}{2}, $$

we get

$$ a_{11}=\frac{(1+2)^2}{2}=\frac92, $$ $$ a_{12}=\frac{(1+4)^2}{2}=\frac{25}{2}, $$ $$ a_{21}=\frac{(2+2)^2}{2}=8, $$ $$ a_{22}=\frac{(2+4)^2}{2}=18. $$

Hence,

$$ \boxed{ A= \begin{bmatrix} \dfrac92 & \dfrac{25}{2}\\[2mm] 8 & 18 \end{bmatrix} } $$

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Question 5

Construct a

$$ 3\times4 $$

matrix whose elements are given by:

1. $$ a_{ij}=\frac12|-3i+j| $$
2. $$ a_{ij}=2i-j $$

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(i)

Since the matrix is of order

$$ 3\times4, $$

we take

$$ i=1,2,3 \quad\text{and}\quad j=1,2,3,4. $$

Now calculate each element.

$$ \begin{aligned} a_{11}&=\frac12|-3(1)+1| =\frac12|-2| =1,\\[2mm] a_{12}&=\frac12|-3(1)+2| =\frac12|-1| =\frac12,\\[2mm] a_{13}&=\frac12|-3(1)+3| =0,\\[2mm] a_{14}&=\frac12|-3(1)+4| =\frac12. \end{aligned} $$ $$ \begin{aligned} a_{21}&=\frac12|-3(2)+1| =\frac12|-5| =\frac52,\\[2mm] a_{22}&=\frac12|-3(2)+2| =\frac12|-4| =2,\\[2mm] a_{23}&=\frac12|-3(2)+3| =\frac12|-3| =\frac32,\\[2mm] a_{24}&=\frac12|-3(2)+4| =\frac12|-2| =1. \end{aligned} $$ $$ \begin{aligned} a_{31}&=\frac12|-3(3)+1| =\frac12|-8| =4,\\[2mm] a_{32}&=\frac12|-3(3)+2| =\frac12|-7| =\frac72,\\[2mm] a_{33}&=\frac12|-3(3)+3| =\frac12|-6| =3,\\[2mm] a_{34}&=\frac12|-3(3)+4| =\frac12|-5| =\frac52. \end{aligned} $$

Hence, the required matrix is

$$ \boxed{ A= \begin{bmatrix} 1 & \dfrac12 & 0 & \dfrac12\\[2mm] \dfrac52 & 2 & \dfrac32 & 1\\[2mm] 4 & \dfrac72 & 3 & \dfrac52 \end{bmatrix} } $$

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(ii)

Using

$$ a_{ij}=2i-j, $$

we calculate each element.

$$ \begin{aligned} a_{11}&=2(1)-1=1,\\ a_{12}&=2(1)-2=0,\\ a_{13}&=2(1)-3=-1,\\ a_{14}&=2(1)-4=-2. \end{aligned} $$ $$ \begin{aligned} a_{21}&=2(2)-1=3,\\ a_{22}&=2(2)-2=2,\\ a_{23}&=2(2)-3=1,\\ a_{24}&=2(2)-4=0. \end{aligned} $$ $$ \begin{aligned} a_{31}&=2(3)-1=5,\\ a_{32}&=2(3)-2=4,\\ a_{33}&=2(3)-3=3,\\ a_{34}&=2(3)-4=2. \end{aligned} $$

Therefore, the required matrix is

$$ \boxed{ A= \begin{bmatrix} 1 & 0 & -1 & -2\\ 3 & 2 & 1 & 0\\ 5 & 4 & 3 & 2 \end{bmatrix} } $$

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Question 6 (i)

Find the values of

$$ x,\;y,\;\text{and}\;z $$

from the equation

$$ \begin{bmatrix} 4 & 3\\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z\\ 1 & 5 \end{bmatrix}. $$

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Solution

Two matrices are equal if and only if their corresponding elements are equal.

Comparing the corresponding elements, we get

$$ y=4, $$ $$ z=3, $$

and

$$ x=1. $$

The remaining element

$$ 5=5 $$

is already equal.

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Answer:

$$ \boxed{x=1,\qquad y=4,\qquad z=3} $$

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Question 6 (ii)

Find the values of

$$ x,\;y,\;\text{and}\;z $$

from the equation

$$ \begin{bmatrix} x+y & 2\\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2\\ 5 & 8 \end{bmatrix}. $$

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Solution

Since equal matrices have equal corresponding elements, we obtain

$$ x+y=6, $$ $$ 5+z=5, $$ $$ xy=8. $$

From

$$ 5+z=5, $$

we get

$$ z=0. $$

Now solve

$$ x+y=6 $$

and

$$ xy=8. $$

The numbers satisfying these equations are

$$ x=2,\qquad y=4, $$

or

$$ x=4,\qquad y=2. $$

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Answer:

$$ \boxed{z=0} $$

and

$$ \boxed{(x,y)=(2,4)\ \text{or}\ (4,2).} $$

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Question 6 (iii)

Find the values of

$$ x,\;y,\;\text{and}\;z $$

from the equation

$$ \begin{bmatrix} x+y+z\\ x+z\\ y+z \end{bmatrix} = \begin{bmatrix} 9\\ 5\\ 7 \end{bmatrix}. $$

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Solution

Comparing the corresponding elements, we obtain

$$ x+y+z=9, $$ $$ x+z=5, $$ $$ y+z=7. $$

Subtracting the second equation from the first,

$$ (x+y+z)-(x+z)=9-5, $$

we get

$$ y=4. $$

Substituting

$$ y=4 $$

into

$$ y+z=7, $$

we obtain

$$ 4+z=7, $$

which gives

$$ z=3. $$

Finally, substituting

$$ z=3 $$

into

$$ x+z=5, $$

we get

$$ x+3=5, $$

so

$$ x=2. $$

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Answer:

$$ \boxed{x=2,\qquad y=4,\qquad z=3} $$

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Question 7

Find the values of

$$ a,\;b,\;c\;\text{and}\;d $$

from the equation

$$ \begin{bmatrix} a-b & 2a+c\\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5\\ 0 & 13 \end{bmatrix}. $$

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Solution

Since two matrices are equal if and only if their corresponding elements are equal, we obtain the following equations:

$$ a-b=-1 $$ $$ 2a+c=5 $$ $$ 2a-b=0 $$ $$ 3c+d=13. $$

From

$$ 2a-b=0, $$

we get

$$ b=2a. $$

Substituting

$$ b=2a $$

into

$$ a-b=-1, $$

we obtain

$$ a-2a=-1. $$

Hence,

$$ -a=-1, $$

so

$$ a=1. $$

Therefore,

$$ b=2a=2. $$

Now substitute

$$ a=1 $$

into

$$ 2a+c=5. $$

We get

$$ 2+c=5, $$

which gives

$$ c=3. $$

Finally, substitute

$$ c=3 $$

into

$$ 3c+d=13. $$

Thus,

$$ 9+d=13, $$

which gives

$$ d=4. $$

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Answer:

$$ \boxed{a=1,\qquad b=2,\qquad c=3,\qquad d=4} $$

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Question 8

A matrix

$$ A=[a_{ij}]_{m\times n} $$

is a square matrix if

(A) \(m (B) \(m>n\)
(C) \(m=n\)
(D) None of these

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Solution

A matrix is called a square matrix when the number of rows is equal to the number of columns.

Hence, if the order of the matrix is

$$ m\times n, $$

then for a square matrix, we must have

$$ m=n. $$

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Answer:

$$ \boxed{\text{Option (C): }m=n} $$

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Question 9

Which of the given values of

$$ x\quad\text{and}\quad y $$

make the following pair of matrices equal?

$$ \begin{bmatrix} 3x+7 & 5\\ y+1 & 2-3x \end{bmatrix} = \begin{bmatrix} 0 & y-2\\ 8 & 4 \end{bmatrix} $$

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Solution

Two matrices are equal if their corresponding elements are equal.

Comparing the corresponding elements, we get

$$ 3x+7=0, $$ $$ 5=y-2, $$ $$ y+1=8, $$ $$ 2-3x=4. $$

From

$$ 5=y-2, $$

we obtain

$$ y=7. $$

Also,

$$ y+1=8 $$

gives

$$ y=7, $$

which is consistent.

Now, from

$$ 3x+7=0, $$

we get

$$ x=-\frac73. $$

But from

$$ 2-3x=4, $$

we obtain

$$ -3x=2, $$

or

$$ x=-\frac23. $$

Since

$$ -\frac73\ne-\frac23, $$

the equations for

$$ x $$

are inconsistent.

Therefore, no values of

$$ x $$

and

$$ y $$

satisfy all four equations simultaneously.

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Answer:

$$ \boxed{\text{Option (B): Not possible to find}} $$

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Question 10

The number of all possible matrices of order

$$ 3\times3 $$

with each entry either

$$ 0 $$

or

$$ 1 $$

is

(A) 27
(B) 18
(C) 81
(D) 512

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Solution

A matrix of order

$$ 3\times3 $$

contains

$$ 3\times3=9 $$

elements.

Each element can be chosen in

$$ 2 $$

ways, namely

$$ 0 \quad\text{or}\quad 1. $$

Hence, the total number of different matrices is

$$ 2^9=512. $$

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Answer:

$$ \boxed{\text{Option (D): }512} $$

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