Matrices Solved ex - 3.1 by PCM Encyclopedia.

1. Chapter - 3 MATRICES (Ex – 3.1)
PCM ENCYCLOPEDIA (www.pcmpedia.blogspot.in) 1
1. In the Matrix A=[ ⁄
√
], Write:
(i) The order of the matrix,
(ii) The number of elements,
(iii) Write the elements of a13, a21, a33, a24, a23.
Solution 1:
(i) In the given matrix, the number of rows is 3 and the number of
columns is 4. Therefore, the order of the matrix is 3 × 4.
(ii) Since the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements
in it.
(iii) a13 = 19, a21 = 35, a33 = −5, a24 = 12, a23 = 5/2
2. If a matrix has 24 elements, what are the possible orders it can have?
What, if it has 13 elements?
Solution 2:
We know that if a matrix is of the order m × n, it has mn elements.So,
here we have to find out the possible orders of a matrix having 24
elements.
Hence, the possible orders of a matrix having 24 elements are:
1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4 and the
possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.
3. If a matrix has 18 elements, what are the possible orders it can have?
What, if it has 5 elements?
Solution 3:
We know that if a matrix is of the order m × n, it has mn elements.So,
here we have to find out the possible orders of a matrix having 18

2. Chapter - 3 MATRICES (Ex – 3.1)
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elements.
Hence, the possible orders of a matrix having 18 elements are:
1 × 18, 2 × 9, 3 × 6, 6 × 3, 9 × 2, and 18 × 1 and the possible orders of
a matrix having 5 elements are 1 × 5 and 5 × 1.
4. Construct a 2X2 matrix, A=[aij], whose elements are given by:
(i) aij =
( )
(ii) aij = (iii) aij =
( )
Solution 4:
(i) Given aij =
( )
We Know, a 2 x 2 matrix is given by A = [ ]
So,
a11 =
( )
= = 2 a12 =
( )
=
a21 =
( )
= a22 =
( )
= = 8
Therefore, the required Matrix is
A = [ ]
(ii) Given aij =
We Know, a 2 x 2 matrix is given by A = [ ]
So,
a11 = = 1 a12 =
a21 = = 2 a22 = = 1

3. Chapter - 3 MATRICES (Ex – 3.1)
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Therefore, the required Matrix is
A = [ ]
(iii) Given aij =
( )
We Know, a 2 x 2 matrix is given by A = [ ]
So,
a11 =
( )
= a12 =
( )
=
a21 =
( )
= = 8 a22 =
( )
= = 18
Therefore, the required Matrix is
A = [ ]
5. Construct a 3X4 Matrix, whose elements are given by:
(i) aij = | | (ii) aij = 2i – j
Solution 5:
(i) aij = | |
We Know, a 3 x 4 matrix is given by A = [ ]
a11 = | | = 1 a12 = | | =
a13 = | | = 0 a14 = | | =

4. Chapter - 3 MATRICES (Ex – 3.1)
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a21 = | | = a22 = | | = 2
a23 = | | = a24 = | | = 1
a31 = | | = 4 a32 = | | =
a33 = | | = 3 a34 = | | =
Hence, the required matrix is:
A =
[ ]
(ii) aij = 2i – j
We Know, a 3 x 4 matrix is given by A = [ ]
a11 = 2 × 1 − 1 = 1 a12 = 2 × 1 − 2 = 0
a13 = 2 × 1 − 3 = −1 a14 = 2 × 1 − 4 = −2
a21 = 2 × 2 − 1 = 3 a22 = 2 × 2 − 2 = 2
a23 = 2 × 2 − 3 = 1 a24 = 2 × 2 − 4 = 0
a31 = 2 × 3 − 1 = 5 a32 = 2 × 3 − 2 = 4
a33 = 2 × 3 − 3 = 3 a34 = 2 × 3 − 4 = 2
Hence, the required matrix is:
A = [ ]

5. Chapter - 3 MATRICES (Ex – 3.1)
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6. Find the values of x, y and z from the following equations:
(i) [ ]=[ ] (ii) [ ]=[ ]
(ii) [ ]=[ ]
Solution 6:
(i) [ ] = [ ]
Since both are matrices are equal, their corresponding elements are also
equal.
Comparing corresponding elements, we get
y = 4, z = 3 and x = 1
(ii) [ ]=[ ]
Since both are matrices are equal, their corresponding elements are also
equal.
Comparing corresponding elements, we get
x + y = 6, 5 + z = 5 and xy = 8
Now, 5 + z = 5 z = 0
Here, x + y = 6 ----------------------- eq. 1
xy = 8 -------------------------- eq. 2
From eq. 1
y = 6 – x -------------------------------- eq .3
Put this value of y in eq. 2, we get
x (6 – x) = 8
6x – x2 = 8 or x2 6x + 8 = 0
By Factorization,
x2 2x – 4x + 8 = 0
x(x 2) 4(x 2) = 0
(x 4)(x – 2) = 0

6. Chapter - 3 MATRICES (Ex – 3.1)
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x = 4 or x = 2
Put these values in eq. 3
From x = 4, we get From x = 2, we get
y = 6 4 = 2 y = 6 – 2 = 4
x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0
(iii) [ ]=[ ]
Since both are matrices are equal, their corresponding elements are also
equal.
Comparing corresponding elements, we get
x y z = ……………….. ( )
x z = ……………………….( )
y z = ……………………….( )
From (1) and (2), we have
y + 5 = 9
⇒ y = 4
Put this value of y in (3)
4 + z = 7
⇒ z = 3
Also,
x + z = 5
x + 3 = 5
⇒ x = 2
∴ x = 2, y = 4 and z = 3
7. Find the Value of a, b, c and d from the equation:
[ ] = [ ]

7. Chapter - 3 MATRICES (Ex – 3.1)
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Solution 7:
Since both are matrices are equal, their corresponding elements are also
equal.
Comparing corresponding elements, we get:
= … … … … … ( )
= … … … … … ( )
= … … … … … ( )
= … … … … … ( )
From (2) we have:
= … … … … … . ( )
From (1) & (5),
=
=
= and =
Now from (3), we have:
=
⇒ =
Now from (4), we get:
=
=
Hence, = = = =
8. = is a square matrix, if
(A) m<n (B) m>n (C) m=n (D) None of these
Solution 8:
Correct Option is (C).
To be a Square Matrix m should be equal to n. i.e No. of rows should be
equal to No. of Columns.
9. Which of the given values of x and y make the following pair of matrices
equal
[ ] = [ ]

8. Chapter - 3 MATRICES (Ex – 3.1)
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(A) = = (B) Not Possible to find
(C) = = (D) = =
Solution 9:
Correct Option is (B).
Given that [ ] = [ ]
Equating the Corresponding element of the given Matrix.
3 = = and 5 = = 7
= = 7 and = =
On Comparing the Corresponding elements of the two matrices, we get
two different values of , which is not possible.
Hence, it is not possible to find the values of and y for which the given
matrices are equal.
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
Solution 10:
Correct Option is (D).
The given matrix of the order 3 × 3 has 9 elements and each of these
elements can be either 0 or 1.
Now, each of the 9 elements can be filled in two possible ways.
Therefore, by the multiplication principle, the required number of
possible matrices is 29 = 512.