Question 1. In the matrix
A = \begin{bmatrix} 2 & 5 & -19 & 7 \\\\ 35 & -2 & \dfrac{5}{2} & 12 \\\\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}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}
Solution:
(i) The order of the matrix is 3 \times 4
(ii) The number of elements is 3 \times 4 = 12
(iii) Elements:
a_{13} = -19,\quad a_{21} = 35,\quad a_{33} = -5,\quad a_{24} = 12,\quad a_{23} = \dfrac{5}{2}
Question 2. If a matrix has 24 elements, what are the possible orders it can have?
What, if it has 13 elements?
Solution:
If a matrix has 24 elements, we list all possible ordered pairs (m, n) such that m \times n = 24:
(1,24),\ (2,12),\ (3,8),\ (4,6),\ (6,4),\ (8,3),\ (12,2),\ (24,1)
If the matrix has 13 elements:
Since 13 is a prime number, only possible orders are 1 \times 13\ \text{and}\ 13 \times 1
Question 3. If a matrix has 18 elements, what are the possible orders it can have?
What, if it has 5 elements?
Solution:
If the matrix has 18 elements, we find all pairs (m, n) such that
m \times n = 18
Possible orders:
(1,18),\ (2,9),\ (3,6),\ (6,3),\ (9,2),\ (18,1)
If the matrix has 5 elements (and 5 is a prime number), possible orders are:
(1,5)\ \text{and}\ (5,1)
Question 4. Construct a 2 \times 2 matrix 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}
Solution
(i)
A = \begin{bmatrix}
\dfrac{(1+1)^2}{2} & \dfrac{(1+2)^2}{2} \\\\
\dfrac{(2+1)^2}{2} & \dfrac{(2+2)^2}{2}
\end{bmatrix}
=
\begin{bmatrix}
2 & \dfrac{9}{2} \\\\
\dfrac{9}{2} & 8
\end{bmatrix}
(ii)
A = \begin{bmatrix}
\dfrac{1}{1} & \dfrac{1}{2} \\\\
\dfrac{2}{1} & \dfrac{2}{2}
\end{bmatrix}
=
\begin{bmatrix}
1 & \dfrac{1}{2} \\\\
2 & 1
\end{bmatrix}
(iii) A = \begin{bmatrix} \dfrac{(1 + 2)^2}{2} & \dfrac{(1 + 4)^2}{2} \\\\ \dfrac{(2 + 2)^2}{2} & \dfrac{(2 + 4)^2}{2} \end{bmatrix} = \begin{bmatrix} \dfrac{9}{2} & \dfrac{25}{2} \\\\ 8 & 18 \end{bmatrix}
Question 5. Construct a 3 \times 4 matrix, whose elements are given by:
(i) a_{ij} = \dfrac{1}{2} \left| -3i + j \right|
(ii) a_{ij} = 2i - j
Solution:
(i)
A = \begin{bmatrix}
\dfrac{1}{2}|-3(1)+1| & \dfrac{1}{2}|-3(1)+2| & \dfrac{1}{2}|-3(1)+3| & \dfrac{1}{2}|-3(1)+4| \\\\
\dfrac{1}{2}|-3(2)+1| & \dfrac{1}{2}|-3(2)+2| & \dfrac{1}{2}|-3(2)+3| & \dfrac{1}{2}|-3(2)+4| \\\\
\dfrac{1}{2}|-3(3)+1| & \dfrac{1}{2}|-3(3)+2| & \dfrac{1}{2}|-3(3)+3| & \dfrac{1}{2}|-3(3)+4|
\end{bmatrix}
= \\\\
\begin{bmatrix}
1 & \dfrac{1}{2} & 0 & \dfrac{1}{2} \\\\
\dfrac{5}{2} & 2 & \dfrac{3}{2} & 1 \\\\
4 & \dfrac{7}{2} & 3 & \dfrac{5}{2}
\end{bmatrix}
(ii) A = \begin{bmatrix} 2(1) - 1 & 2(1) - 2 & 2(1) - 3 & 2(1) - 4 \\\\ 2(2) - 1 & 2(2) - 2 & 2(2) - 3 & 2(2) - 4 \\\\ 2(3) - 1 & 2(3) - 2 & 2(3) - 3 & 2(3) - 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}
Question 6. Find the values of x, y, z from the following equations:
(i) \begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}
(ii) \begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}
(iii) \begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}
Solution: (i) Comparing corresponding elements of the matrices:
4 = y \Rightarrow y = 4
3 = z \Rightarrow z = 3
x = 1 (from the bottom-left elements)
(ii) Comparing matrices:
x + y = 6 \quad \text{(1)}
z + 5 = 5 \Rightarrow z = 0 \quad \text{(2)}
xy = 8 \quad \text{(3)}
From (1):
y = 6 - x
Now, substitute into (3):
x(6 - x) = 8 \\\\
x^2 - 6x + 8 = 0
Solving the quadratic:
x = 2 \quad and \quad x = 4
y = 6 - 2 = 4 \quad and \quad y = 6 - 4 =2
Therefore, possible solutions:
(x, y, z) = (2, 4, 0) \text{ or } (4, 2, 0)
(iii) Given:
x + y + z = 9 \quad \text{(i)}
x + z = 5 \quad \text{(ii)}
y + z = 7 \quad \text{(iii)}
From (i) and (ii):
(x + y + z) - (x + z) = 9 - 5 \\\\ \Rightarrow y = 4Now using y in (iii): 4 + z = 7 \Rightarrow z = 3
Then from (ii): x = 5 - z = 2
Final solution: x = 2, y = 4, z = 3
Question 7. Find the value of a, b, c, 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}
Solution: Compare elements:
a - b = -1 \quad (1)
2a + c = 5 \quad (2)
2a - b = 0 \quad (3)
3c + d = 13 \quad (4)
From (1) and (3):
Subtract (1) from (3):
(2a - b) - (a - b) = 0 - (-1) \Rightarrow a = 1
Then
From (1): 1 - b = -1 \Rightarrow b = 2
From (2): 2(1) + c = 5 \Rightarrow c = 3
From (4): 3(3) + d = 13 \Rightarrow d = 4
Final values: a = 1, b = 2, c = 3, d = 4
Question 8. A = [a_{ij}]_{m \times n} is a square matrix, if:
(A) m < n
(B) m > n
(C) m = n
(D) None of these
Solution: A square matrix is defined as a matrix with equal number of rows and columns.
So, the correct condition is: m = n
✅ Correct Option: (C)
Question 9. Which of the given values of x and 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}Solution: Compare elements:
3x + 7 = 0 \quad (1)
\Rightarrow x = -\dfrac{7}{3}
y + 1 = 8 \quad (2)
\Rightarrow y = 7
y - 2 = 5 \quad (3)
\Rightarrow y = 7 \quad (OK)
2 - 3x = 4 \quad (4)
\Rightarrow x = -\dfrac{2}{3} \quad (Contradiction)
Equations (1) and (4) are inconsistent. Hence,
❌ No consistent values of x and y satisfy all conditions.
✅ Correct Option: (B) Not possible to find
Question 10. The number of all possible matrices of order 3 \times 3 with each entry 0 or 1 is:
(A) 27 (B) 18 (C) 81 (D) 512
Solution: Each element has 2 possibilities: 0 or 1
A 3 \times 3 matrix has 9 elements.
So total number of such matrices = 2^9 = 512
✅ Correct Option: (D) 512