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If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
24 = 1 X 24
= 2 X 12
= 3 X 8
= 4 X 6
= 6 X 4
= 8 X 3
= 12 X 2
= 24 X 1
So, there are 8 orders possible.
Similarly 13 = 1 X 13 and 13 X1, So, only 2 orders are possible.
In the matrix
, write:
(i) The order of the matrix (ii) The number of elements,
(iii) Write the elements
i) Order is 3×4 because there are 3 rows and 4 columns in A.
ii) Number of elements is 12. as there are 3 × 4 = 12 elements in it.
(iii) a13 = 19, a21 = 35, a33 = −5, a24 = 12, a23 = 
If a matrix has 18 elements, what are the possible orders it can have? What, if it
has 5 elements?
18= 1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3
Therefore, there are 6 posiible orders.
And for 5 it is 1 X 5 and 5 X 1, So, it has 2 possible orders
Construct a 2 × 2 matrix,
, whose elements are given by:
(i) 
(ii) 
(iii) 
For 2 X 2 matrix the elements are a11, a12, a21 and
(i)
Therefore, the required matrix is
(ii)
Therefore, the required matrix is
(iii)
Therefore, the required matrix is
Construct a 3 × 4 matrix, whose elements are given by
(i) 
(ii) 
In general, a 3 × 4 matrix is given by
(i) 
Therefore, the required matrix is
(ii) 
Therefore, the required matrix is
Find the value of a, b, c, and d from the equation:
As the two matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:
a − b = −1 … (1)
2a − b = 0 … (2)
2a + c = 5 … (3)
3c + d = 13 … (4)
a - b = -1 and 2a - b = 0; solving a =1 and b = 2
2a + c = 5, substituting a = 1, c =
3c + d = 13, substituting c = ; d =
is a square matrix, if
(A) m < n
(B) m > n
(C) m = n
(D) None of these
(C)
If m and n are the rows and columns, then in a square matrix m=n.
Find the value of x, y, and z from the following equation:
(i) 
(ii) 
(iii) 
(i)
As the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:
x = 1, y = 4, and z = 3
(ii)
As the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:
x + y = 6, xy = 8, 5 + z = 5
Now, 5 + z = 5 ⇒ z = 0
We know that:
(x − y)2 = (x + y)2 − 4xy
⇒ (x − y)2 = 36 − 32 = 4
⇒ x − y = ±2
Now, when x − y = 2 and x + y = 6, we get x = 4 and y = 2
When x − y = − 2 and x + y = 6, we get x = 2 and y = 4
∴x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0
(iii)
As the two matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:
x + y + z = 9 … (1)
x + z = 5 … (2)
y + z = 7 … (3)
From (1) and (2), we have:
y + 5 = 9
⇒ y = 4
Then, from (3), we have:
4 + z = 7
⇒ z = 3
∴ x + z = 5
⇒ x = 2
∴ x = 2, y = 4, and z = 3
Which of the given values of x and y make the following pair of matrices equal
(A) 
(B) Not possible to find
(C) 
(D) 
The correct answer is B.
It is given that
Equating the corresponding elements, we get:
We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible.
Hence, it is not possible to find the values of x and y for which the given matrices are equal.
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 correct answer 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
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