# Exercise 3.1 Solutions

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Exercise 3.1 Solutions
1. In the matrix {A = \left[\begin{array}{cccc} 2 & 5 & 19 & -7 \\[5pt] 35 & -2 & \dfrac{5}{2} & 12 \\[10pt] \sqrt{3} & 1 & -5 & 17 \end{array} \right]}, write:
(i) The order of the matrix
(ii) The number of elements,
(iii) Write the elements a_{13}, a_{21}, a_{33}, a_{24}, a_{23}.
(i)
The given matrix A has 3 rows and 4 columns.
So, the order of the matrix A is 3 × 4
(ii)
The number of elements in the given matrix A are 3 × 4 = 12
(iii)
The following are the elements at the given positions in the matrix A
Element
Value
a_{13}
19
a_{21}
35
a_{33}
17
a_{24}
12
a_{23}
\dfrac{5}{2}

2. If a matrix has 24 elements, what are the possible orders it can have? What if it has 13 elements.
The order of the matrix containing 24 elements would be m × n with all possible values of m and n as follows.
It can be noted that the order of the matrix can be obtained by finding all those two factors of 24 whose product would yield 24 as follows:
24
=
1 × 24
=
2 × 12
=
3 × 8
=
4 × 6
=
6 × 4
=
8 × 3
=
12 × 2
=
24 × 1
So, the number of possible orders for a matrix that has 24 elements is 8.
Now, if we consider a matrix that has 13 elements, the number of possile orders would be as follows:
13
=
1 × 13
=
13 × 1
So, the number of possible orders for a matrix that has 13 elements is 2.
Note: Any matrix whose number of elements is a prime number would have only 2 possible orders. It can also be noted that these orders would be a row matrix and a column matrix.

3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
The order of the matrix containing 18 elements would be m × n with all possible values of m and n as follows.
It can be noted that the order of the matrix can be obtained by finding all those two factors of 18 whose product would yield 18 as follows:
18
=
1 × 18
=
2 × 9
=
3 × 6
=
6 × 3
=
9 × 2
=
18 × 1
So, the number of possible orders for a matrix that has 18 elements is 6.
Now, if we consider a matrix that has 5 elements, the number of possile orders would be as follows:
5
=
1 × 5
=
5 × 1
So, the number of possible orders for a matrix that has 5 elements is 2.
Note: Any matrix whose number of elements is a prime number would have only 2 possible orders. It can also be noted that these orders would be a row matrix and a column matrix.

4. Construct a 2 × 2 matrix, {\text{A} = [a_{ij}]}, whose elements are given by:
(i) {a_{ij} = \dfrac{(i + j)^2}{2}} (ii) {a_{ij} = \dfrac{i}{j}} (iii) {a_{ij} = \dfrac{(i + 2j)^2}{2}}
Any 2 x 2 matrix {\text{A} = [a_{ij}]} can be expressed as
{\text{A} = \left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]}
(i) To construct A when {a_{ij} = \dfrac{(i + j)^2}{2}}
i
j
a_{ij}
{\dfrac{(i + j)^2}{2}}
1
1
a_{11}
{\dfrac{(1 + 1)^2}{2} = \dfrac{2^2}{2} = 2}
1
2
a_{12}
{\dfrac{(1 + 2)^2}{2} = \dfrac{3^2}{2} = \dfrac{9}{2}}
2
1
a_{21}
{\dfrac{(2 + 1)^2}{2} = \dfrac{3^2}{2} = \dfrac{9}{2}}
2
2
a_{22}
{\dfrac{(2 + 2)^2}{2} = \dfrac{4^2}{2} = 8}
So, the constructed matrix will be
{\text{A} = \left[ \begin{array}{cc} a_{11} & a_{12} \\[5pt] a_{21} & a_{22} \end{array} \right] =} {\left[ \begin{array}{cc} 2 & \dfrac{9}{2} \\[10pt] \dfrac{9}{2} & 8 \end{array} \right]}
(ii) To construct A when {a_{ij} = \dfrac{i}{j}}
i
j
a_{ij}
{\dfrac{i}{j}}
1
1
a_{11}
{\dfrac{1}{1} = 1}
1
2
a_{12}
{\dfrac{1}{2} = \dfrac{1}{2}}
2
1
a_{21}
{\dfrac{2}{1} = 2}
2
2
a_{22}
{\dfrac{2}{2} = 1}
So, the constructed matrix will be
{\text{A} = \left[ \begin{array}{cc} a_{11} & a_{12} \\[5pt] a_{21} & a_{22} \end{array} \right] =} {\left[ \begin{array}{cc} 1 & \dfrac{1}{2} \\[10pt] 2 & 1 \end{array} \right]}
(iii) To construct A when {a_{ij} = \dfrac{(i + 2j)^2}{2}}
i
j
a_{ij}
{\dfrac{(i + 2j)^2}{2}}
1
1
a_{11}
{\dfrac{(1 + (2 × 1))^2}{2} = \dfrac{3^2}{2} = \dfrac{9}{2}}
1
2
a_{12}
{\dfrac{(1 + (2 × 2))^2}{2} = \dfrac{5^2}{2} = \dfrac{25}{2}}
2
1
a_{21}
{\dfrac{(2 + (2 × 1))^2}{2} = \dfrac{16^2}{2} = 8}
2
2
a_{22}
{\dfrac{(2 + (2 × 2))^2}{2} = \dfrac{6^2}{2} = \dfrac{36}{2} = 18}
So, the constructed matrix will be
{\text{A} = \left[ \begin{array}{cc} a_{11} & a_{12} \\[5pt] a_{21} & a_{22} \end{array} \right] =} {\left[ \begin{array}{cc} \dfrac{9}{2} & \dfrac{25}{2} \\[10pt] 8 & 18 \end{array} \right]}

5. Construct a 3 × 4 matrix, whose elements are given by:
(i) {a_{ij} = \dfrac{1}{2}|-3i + j|} (ii) {a_{ij} = 2i - j}
Any 3 x 4 matrix {\text{A} = [a_{ij}]} can be expressed as
{\text{A} = \left[ \begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14} \\[5pt] a_{21} & a_{22} & a_{23} & a_{24} \\[5pt] a_{31} & a_{32} & a_{33} & a_{34} \end{array} \right]}
(i) To construct a matrix when {a_{ij} = \dfrac{1}{2}|-3i + j|}
i
j
a_{ij}
{\dfrac{1}{2}\Big|-3i + j\Big|}
1
1
a_{11}
{\dfrac{1}{2}\Big|(-3 × 1) + 1\Big| = \dfrac{1}{2}\Big|-3 + 1\Big| = \dfrac{1}{2} × 2 = 1}
1
2
a_{12}
{\dfrac{1}{2}\Big|(-3 × 1) + 2\Big| = \dfrac{1}{2}\Big|-3 + 2\Big| = \dfrac{1}{2} × 1 = \dfrac{1}{2}}
1
3
a_{13}
{\dfrac{1}{2}\Big|(-3 × 1) + 3\Big| = \dfrac{1}{2}\Big|-3 + 3\Big| = \dfrac{1}{2} × 0 = 0}
1
4
a_{14}
{\dfrac{1}{2}\Big|(-3 × 1) + 4\Big| = \dfrac{1}{2}\Big|-3 + 4\Big| = \dfrac{1}{2} × 1 = \dfrac{1}{2}}
2
1
a_{21}
{\dfrac{1}{2}\Big|(-3 × 2) + 1\Big| = \dfrac{1}{2}\Big|-6 + 1\Big| = \dfrac{1}{2} × 5 = \dfrac{5}{2}}
2
2
a_{22}
{\dfrac{1}{2}\Big|(-3 × 2) + 2\Big| = \dfrac{1}{2}\Big|-6 + 2\Big| = \dfrac{1}{2} × 4 = 2}
2
3
a_{23}
{\dfrac{1}{2}\Big|(-3 × 2) + 3\Big| = \dfrac{1}{2}\Big|-6 + 3\Big| = \dfrac{1}{2} × 3 = \dfrac{3}{2}}
2
4
a_{24}
{\dfrac{1}{2}\Big|(-3 × 2) + 4\Big| = \dfrac{1}{2}\Big|-6 + 4\Big| = \dfrac{1}{2} × 2 = 1}
3
1
a_{31}
{\dfrac{1}{2}\Big|(-3 × 3) + 1\Big| = \dfrac{1}{2}\Big|-9 + 1\Big| = \dfrac{1}{2} × 8 = 4}
3
2
a_{32}
{\dfrac{1}{2}\Big|(-3 × 3) + 2\Big| = \dfrac{1}{2}\Big|-9 + 2\Big| = \dfrac{1}{2} × 7 = \dfrac{7}{2}}
3
3
a_{33}
{\dfrac{1}{2}\Big|(-3 × 3) + 3\Big| = \dfrac{1}{2}\Big|-9 + 3\Big| = \dfrac{1}{2} × 6 = 3}
3
4
a_{34}
{\dfrac{1}{2}\Big|(-3 × 3) + 4\Big| = \dfrac{1}{2}\Big|-9 + 4\Big| = \dfrac{1}{2} × 5 = \dfrac{5}{2}}
So, the constructed matrix will be
{\text{A} = \left[ \begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14} \\[5pt] a_{21} & a_{22} & a_{23} & a_{24} \\[5pt] a_{31} & a_{32} & a_{33} & a_{34} \end{array} \right] =} {\left[ \begin{array}{cccc} 1 & \dfrac{1}{2} & 0 & \dfrac{1}{2} \\[10pt] \dfrac{5}{2} & 2 & \dfrac{3}{2} & 1 \\[10pt] 4 & \dfrac{7}{2} & 3 & \dfrac{5}{2} \end{array} \right]}
(ii) To construct a matrix when {a_{ij} = 2i - j}
i
j
a_{ij}
{2i - j}
1
1
a_{11}
(2 × 1) – 1 = 2 – 1 = 1
1
2
a_{12}
(2 × 1) – 2 = 2 – 2 = 0
1
3
a_{13}
(2 × 1) – 3 = 2 – 3 = -1
1
4
a_{14}
(2 × 1) – 4 = 2 – 4 = -2
2
1
a_{21}
(2 × 2) – 1 = 4 – 1 = 3
2
2
a_{22}
(2 × 2) – 2 = 4 – 2 = 2
2
3
a_{23}
(2 × 2) – 3 = 4 – 3 = 1
2
4
a_{24}
(2 × 2) – 4 = 4 – 4 = 0
3
1
a_{31}
(2 × 3) – 1 = 6 – 1 = 5
3
2
a_{32}
(2 × 3) – 2 = 6 – 2 = 4
3
3
a_{33}
(2 × 3) – 3 = 6 – 3 = 3
3
4
a_{34}
(2 × 3) – 4 = 6 – 4 = 2
So, the constructed matrix will be
{\text{A} = \left[ \begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14} \\[5pt] a_{21} & a_{22} & a_{23} & a_{24} \\[5pt] a_{31} & a_{32} & a_{33} & a_{34} \end{array} \right] =} {\left[ \begin{array}{cccc} 1 & 0 & -1 & -2 \\[5pt] 3 & 2 & 1 & 0 \\[5pt] 5 & 4 & 3 & 2 \end{array} \right]}

6. Find the values of x, y and z from the following equations:
(i) {\left[ \begin{array}{cc} 4 & 3 \\[5pt] x & 5 \end{array} \right] = \left[ \begin{array}{cc} y & z \\[5pt] 1 & 5 \end{array} \right]}
(ii) {\left[ \begin{array}{cc} x + y & 2 \\[5pt] 5 + z & xy \end{array} \right] = \left[ \begin{array}{cc} 6 & 2 \\[5pt] 5 & 8 \end{array} \right]}
(iii) {\left[ \begin{array}{c} x + y + z \\[5pt] x + z \\[5pt] y + z \end{array} \right] = \left[ \begin{array}{cc} 9 \\[5pt] 5 \\[5pt] 7 \end{array} \right]}
(i) To find the values of x, y and z from the equation {\left[ \begin{array}{cc} 4 & 3 \\[5pt] x & 5 \end{array} \right] = \left[ \begin{array}{cc} y & z \\[5pt] 1 & 5 \end{array} \right]}
As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get
{x = 1,} {y = 4} and {z = 3}
(ii) To find the values of x, y and z from the equation {\left[ \begin{array}{cc} x + y & 2 \\[5pt] 5 + z & xy \end{array} \right] = \left[ \begin{array}{cc} 6 & 2 \\[5pt] 5 & 8 \end{array} \right]}
As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get
x + y
=
6 ———-❶
5 + z
=
5 ———-❷
xy
=
8 ———-❸
From equation ❸ we have {y = \dfrac{8}{x}}
Substituting this in equation ❶, we get
{x + \dfrac{8}{x} = 6}
{⇒ \dfrac{x^2 + 8}{x} = 6}
{⇒ x^2 + 8 = 6x}
{⇒ x^2 - 6x + 8 = 0}
{⇒ x^2 - 4x - 2x + 8 = 0}
{⇒ x(x - 4) - 2(x - 4) = 0}
{⇒ (x - 2)(x - 4) = 0}
{⇒ (x - 2) = 0} or {x - 4 = 0}
{⇒ x = 2} or {x = 4}
Now,
when {x = 2}
y
=
{\dfrac{8}{x}}
=
{\dfrac{8}{2}}
=
4
when {x = 4}
y
=
{\dfrac{8}{x}}
=
{\dfrac{8}{4}}
=
2
From equation ❷, we have
{z = 0}
So, the values of {x,} y and z are
{x = 2,} {y = 4} and {z = 0}
OR
{x = 4,} {y = 2} and {z = 0}
(iii) To find the values of x, y and z from the equation {\left[ \begin{array}{c} x + y + z \\[5pt] x + z \\[5pt] y + z \end{array} \right] = \left[ \begin{array}{cc} 9 \\[5pt] 5 \\[5pt] 7 \end{array} \right]}
As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get
x + y + z
=
9 ———-❶
x + z
=
5 ———-❷
y + z
=
7 ———-❸
Substituting ❸ into ❶, we get
{x + 7 = 9}
{⇒ x = 2}
Substituting the value of x into ❷, we get
{2 + z = 5}
{⇒ z = 3}
Substituting the value of z into ❸, we get
{y + 3 = 7}
{⇒ y = 4}
So, the values of {x,} y and z are
{x = 2,} {y = 4} and {z = 3}

7. Find the value of {a,} {b,} c and d from the eqution:
{\left[\begin{array}{cc} a - b & 2a + c \\[5pt] 2a - b & 3c + d \end{array}\right] = \left[ \begin{array}{cc} -1 & 5 \\[5pt] 0 & 13 \end{array}\right]}
As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get
a - b
=
-1 ———-❶
2a + c
=
5 ———-❷
2a - b
=
0 ———-❸
3c + d
=
13 ———-❹
Subtracting equation ❶ from ❸, we get
{(2a - b) - (a - b) = 0 -(-1)}
{⇒ 2a - b - a + b = 0 + 1}
{⇒ a = 1}
Substituting {a = 1} into the equation ❶, we have
{1 - b = -1}
{⇒ -b = -1 - 1}
{⇒ -b = -2}
{⇒ b = 2}
Similarly, substituting {a = 1} into the equation ❷, we have
{(2 × 1) + c = 5}
{⇒ c = 5 - 2}
{⇒ c = 3}
Now, substituting {c = 3} into the equation ❹, we have
{(3 × 3) + d = 13}
{⇒ d = 13 - 9}
{⇒ d = 4}
So, the values of {a,} {b,} c and d are
{a = 1,} {b = 2,} {c = 3} and {d = 4}

8. {\text{A} = {[a_{ij}]}_{m × n}} is a square matrix, if
(A) {m \lt n}
(B) {m \gt n}
(C) {m = n}
(D) None of these
The following is the definition of a square matrix.
Square Matrix: A matrix in which the number of rows are equal to the number of columns, is said to be square matrix. Thus an {m × n} matrix is said to be a square matrix if {m = n} and is known as a square matrix of the order {'n'}
This implies that option C is the correct answer.

9. Which of the given values of x and y make the following pair of matrices equal {\left[ \begin{array}{cc} 3x + 7 & 5 \\[5pt] y + 1 & 2 - 3x \end{array}\right],} {\left[ \begin{array}{cc} 0 & y - 2 \\[5pt] 8 & 4 \end{array}\right]}
(A) {x = \dfrac{-1}{3}, y = 7}
(B) Not possible to find ✔
(C) {y = 7, x = \dfrac{-2}{3}}
(D) {x = \dfrac{-1}{3}, y = \dfrac{-2}{3}}
For the given matrices to be equal, their corresponding elements must be equal. Comparing the corresponding elements, we get
3x + 7
=
0 ———-❶
y - 2
=
5 ———-❷
y + 1
=
8 ———-❸
2 - 3x
=
4 ———-❹
On solving all the equations,
On solving ❶, we get {x = -\dfrac{7}{3}}
On solving ❷, we get {y = 7}
On solving ❸, we get {y = 7}
On solving ❹, we get {x = -\dfrac{2}{3}}
The results obtained by equations ❷ and ❸ are consistent. However, the results obtained by solving both the equations ❶ and ❹ are not consistent.
So, it is not possible to find the values of both x and y which would make both the matrices equal.
Hence the option B which states that it is “Not possible to find” the values of x and y is the correct answer.

10. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
(A) 27
(B) 18
(C) 81
(D) 512 ✔
The number of elements in a 3 × 3 matrix are 3 × 3 = 9
So, there are 9 positions to be filled in with either 0 or 1.
So, the number of possibilities are:
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
=
{2^9}
=
512
So, option D is the correct answer.