Exercise 3.1 Practice
Matrices: Order, Construction, and Equality
Q1
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In the matrix \( A = \begin{bmatrix} 3 & -1 & 5 \\ 2 & 4 & 0 \\ \sqrt{2} & 7 & -3 \\ 1 & 9 & 6 \end{bmatrix} \),
write:
(i) The order of the matrix,
(ii) The number of elements,
(iii) Write the elements \( a_{13}, a_{21}, a_{32}, a_{43}, a_{22} \).
(i) The order of the matrix,
(ii) The number of elements,
(iii) Write the elements \( a_{13}, a_{21}, a_{32}, a_{43}, a_{22} \).
Part (i): Order
The matrix has 4 rows and 3 columns.
Order = 4 × 3
Order = 4 × 3
Part (ii): Number of Elements
Total elements = Rows × Columns = 4 × 3 = 12.
Part (iii): Specific Elements
\( a_{13} \) (Row 1, Col 3) = 5
\( a_{21} \) (Row 2, Col 1) = 2
\( a_{32} \) (Row 3, Col 2) = 7
\( a_{43} \) (Row 4, Col 3) = 6
\( a_{22} \) (Row 2, Col 2) = 4
\( a_{21} \) (Row 2, Col 1) = 2
\( a_{32} \) (Row 3, Col 2) = 7
\( a_{43} \) (Row 4, Col 3) = 6
\( a_{22} \) (Row 2, Col 2) = 4
Q2
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If a matrix has 12 elements, what are the possible orders it can have? What, if it has
7 elements?
Case 1: 12 Elements
We need pairs of natural numbers (m, n) such that \( m \times n = 12 \).
Pairs are: (1, 12), (12, 1), (2, 6), (6, 2), (3, 4), (4, 3).
Possible Orders: 1×12, 12×1, 2×6, 6×2, 3×4, 4×3.
Pairs are: (1, 12), (12, 1), (2, 6), (6, 2), (3, 4), (4, 3).
Possible Orders: 1×12, 12×1, 2×6, 6×2, 3×4, 4×3.
Case 2: 7 Elements
7 is a prime number. Factors are only 1 and 7.
Possible Orders: 1×7, 7×1.
Possible Orders: 1×7, 7×1.
Q3
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If a matrix has 20 elements, what are the possible orders it can have? What, if it has
11 elements?
Case 1: 20 Elements
Factors of 20: 1, 2, 4, 5, 10, 20.
Pairs: (1, 20), (20, 1), (2, 10), (10, 2), (4, 5), (5, 4).
Pairs: (1, 20), (20, 1), (2, 10), (10, 2), (4, 5), (5, 4).
Case 2: 11 Elements
11 is prime.
Pairs: (1, 11), (11, 1).
Pairs: (1, 11), (11, 1).
Q4
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Construct a \( 2 \times 2 \) matrix, \( A = [a_{ij}] \), whose elements are given by:
(i) \( a_{ij} = \frac{(i - j)^2}{2} \)
(ii) \( a_{ij} = 3i - j \)
(i) \( a_{ij} = \frac{(i - j)^2}{2} \)
(ii) \( a_{ij} = 3i - j \)
Part (i): \( a_{ij} = \frac{(i - j)^2}{2} \)
\( a_{11} = (1-1)^2/2 = 0 \)
\( a_{12} = (1-2)^2/2 = 1/2 \)
\( a_{21} = (2-1)^2/2 = 1/2 \)
\( a_{22} = (2-2)^2/2 = 0 \)
Matrix: \( \begin{bmatrix} 0 & 1/2 \\ 1/2 & 0 \end{bmatrix} \)
\( a_{12} = (1-2)^2/2 = 1/2 \)
\( a_{21} = (2-1)^2/2 = 1/2 \)
\( a_{22} = (2-2)^2/2 = 0 \)
Matrix: \( \begin{bmatrix} 0 & 1/2 \\ 1/2 & 0 \end{bmatrix} \)
Part (ii): \( a_{ij} = 3i - j \)
\( a_{11} = 3(1)-1 = 2 \)
\( a_{12} = 3(1)-2 = 1 \)
\( a_{21} = 3(2)-1 = 5 \)
\( a_{22} = 3(2)-2 = 4 \)
Matrix: \( \begin{bmatrix} 2 & 1 \\ 5 & 4 \end{bmatrix} \)
\( a_{12} = 3(1)-2 = 1 \)
\( a_{21} = 3(2)-1 = 5 \)
\( a_{22} = 3(2)-2 = 4 \)
Matrix: \( \begin{bmatrix} 2 & 1 \\ 5 & 4 \end{bmatrix} \)
Q5
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Construct a \( 3 \times 2 \) matrix whose elements are given by \( a_{ij} = \frac{1}{2}|2i - 3j| \).
Row 1 (i=1)
\( a_{11} = \frac{1}{2}|2(1)-3(1)| = \frac{1}{2}|-1| = 1/2 \)
\( a_{12} = \frac{1}{2}|2(1)-3(2)| = \frac{1}{2}|-4| = 2 \)
\( a_{12} = \frac{1}{2}|2(1)-3(2)| = \frac{1}{2}|-4| = 2 \)
Row 2 (i=2)
\( a_{21} = \frac{1}{2}|2(2)-3(1)| = \frac{1}{2}|1| = 1/2 \)
\( a_{22} = \frac{1}{2}|2(2)-3(2)| = \frac{1}{2}|-2| = 1 \)
\( a_{22} = \frac{1}{2}|2(2)-3(2)| = \frac{1}{2}|-2| = 1 \)
Row 3 (i=3)
\( a_{31} = \frac{1}{2}|2(3)-3(1)| = \frac{1}{2}|3| = 3/2 \)
\( a_{32} = \frac{1}{2}|2(3)-3(2)| = \frac{1}{2}|0| = 0 \)
\( a_{32} = \frac{1}{2}|2(3)-3(2)| = \frac{1}{2}|0| = 0 \)
Matrix: \( \begin{bmatrix} 1/2 & 2 \\ 1/2 & 1 \\ 3/2 & 0 \end{bmatrix} \)
Q6
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Find the values of x, y, and z from the following equation:
\( \begin{bmatrix} x+y & 3 \\ 5 & x-y \end{bmatrix} = \begin{bmatrix} 8 & 3 \\ 5 & 2 \end{bmatrix} \)
\( \begin{bmatrix} x+y & 3 \\ 5 & x-y \end{bmatrix} = \begin{bmatrix} 8 & 3 \\ 5 & 2 \end{bmatrix} \)
Step 1: Compare Elements
\( x + y = 8 \) ... (i)
\( x - y = 2 \) ... (ii)
(3=3 and 5=5 are consistent).
\( x - y = 2 \) ... (ii)
(3=3 and 5=5 are consistent).
Step 2: Solve System
Add (i) and (ii): \( 2x = 10 \Rightarrow x = 5 \).
Subtract (ii) from (i): \( 2y = 6 \Rightarrow y = 3 \).
Subtract (ii) from (i): \( 2y = 6 \Rightarrow y = 3 \).
\( x = 5, y = 3 \) (No z variable in this version).
Q7
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Find the values of a, b, c, and d from the equation:
\( \begin{bmatrix} 2a+b & a-2b \\ 5c-d & 4c+3d \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 11 & 24 \end{bmatrix} \)
\( \begin{bmatrix} 2a+b & a-2b \\ 5c-d & 4c+3d \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 11 & 24 \end{bmatrix} \)
Step 1: Set up Equations
1) \( 2a + b = 4 \)
2) \( a - 2b = -3 \)
3) \( 5c - d = 11 \)
4) \( 4c + 3d = 24 \)
2) \( a - 2b = -3 \)
3) \( 5c - d = 11 \)
4) \( 4c + 3d = 24 \)
Step 2: Solve for a, b
From (2), \( a = 2b - 3 \). Substitute into (1):
\( 2(2b-3) + b = 4 \Rightarrow 4b - 6 + b = 4 \Rightarrow 5b = 10 \Rightarrow b = 2 \).
\( a = 2(2) - 3 = 1 \).
\( 2(2b-3) + b = 4 \Rightarrow 4b - 6 + b = 4 \Rightarrow 5b = 10 \Rightarrow b = 2 \).
\( a = 2(2) - 3 = 1 \).
Step 3: Solve for c, d
From (3), \( d = 5c - 11 \). Substitute into (4):
\( 4c + 3(5c - 11) = 24 \Rightarrow 4c + 15c - 33 = 24 \Rightarrow 19c = 57 \Rightarrow c = 3 \).
\( d = 5(3) - 11 = 4 \).
\( 4c + 3(5c - 11) = 24 \Rightarrow 4c + 15c - 33 = 24 \Rightarrow 19c = 57 \Rightarrow c = 3 \).
\( d = 5(3) - 11 = 4 \).
\( a=1, b=2, c=3, d=4 \)
Q8
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\( A = [a_{ij}]_{m \times n} \) is a square matrix if:
(A) \( m < n \)
(B) \( m > n \)
(C) \( m = n \)
(D) None of these
(A) \( m < n \)
(B) \( m > n \)
(C) \( m = n \)
(D) None of these
Definition
A matrix is a square matrix if the number of rows equals the number of columns.
Correct Option: (C) \( m = n \)
Q9
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Which values of x and y make the following matrices equal?
\( \begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix} = \begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix} \)
\( \begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix} = \begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix} \)
Step 1: Check x
From \( a_{11} \): \( 3x + 7 = 0 \Rightarrow x = -7/3 \).
From \( a_{22} \): \( 2 - 3x = 4 \Rightarrow -3x = 2 \Rightarrow x = -2/3 \).
Since \( x \) has two different values, equality is impossible.
From \( a_{22} \): \( 2 - 3x = 4 \Rightarrow -3x = 2 \Rightarrow x = -2/3 \).
Since \( x \) has two different values, equality is impossible.
Answer: Not possible to find.
Q10
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The number of all possible matrices of order \( 2 \times 2 \) with each entry 1 or 2 is:
(A) 27 (B) 18 (C) 16 (D) 81
(A) 27 (B) 18 (C) 16 (D) 81
Calculation
Total elements in \( 2 \times 2 \) matrix = 4.
Each element has 2 choices (1 or 2).
Total matrices = \( 2^4 = 16 \).
Each element has 2 choices (1 or 2).
Total matrices = \( 2^4 = 16 \).
Correct Option: (C) 16