OP Malhotra Class 10 Maths Solutions Chapter 8 Matrices Exercise 8 (A)

S Chand Class 10 ICSE Maths Solutions Chapter 8 Matrices Exercise 8(a)

Question 1. Write down the order of each matrix given below and the number of elements in each :
(i) \( \left[\begin{array}{cc} 1 & -1 \\ 7 & 4 \end{array}\right] \)
(ii) \( [\sin \theta \cos \theta] \)
(iii) \( \{2 \ 4 \ 6 \ 8\} \)
(iv) \( \left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right] \)
(v) \( \left[\begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array}\right] \)
(vi) \( \left[\begin{array}{rrrrr} 1 & -5 & 0 & 8 & 4 \\ 2 & -7 & 3 & 5 & 2 \\ 0 & -2 & 1 & 4 & 9 \end{array}\right] \)
(vii) \( \left[\begin{array}{cc} 1 & -2 \\ 3 & -5 \\ 5 & -9 \\ 7 & 0 \end{array}\right] \)
Answer: The order of a matrix tells you its size (rows x columns), and the number of elements is simply the product of rows and columns.
(i) Order of matrix is \( 2 \times 2 \); Number of elements: 4
(ii) Order of matrix is \( 1 \times 2 \); Number of elements: 2
(iii) Order of matrix is \( 1 \times 4 \); Number of elements: 4
(iv) Order of matrix is \( 3 \times 1 \); Number of elements: 3
(v) Order of matrix is \( 3 \times 4 \); Number of elements: 12
(vi) Order of matrix is \( 3 \times 5 \); Number of elements: 15
(vii) Order of matrix is \( 4 \times 2 \); Number of elements: 8
In simple words: To find the order, count the rows first, then the columns. To find the total number of items inside, just multiply the number of rows by the number of columns.

🎯 Exam Tip: Always remember that the order of a matrix is written as "rows by columns" (m x n). Counting correctly is key to avoiding errors.

Question 2. Classify the following matrices as equal or not equal.
(i) \( \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] ;\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \)
(ii) \( \left[\begin{array}{ll} 4 & 7 \\ 3 & 2 \end{array}\right] ;\left[\begin{array}{cc} 3+1 & \sqrt{49} \\ 5-2 & \frac{6}{3} \end{array}\right] \)
Answer: Matrices are equal only if they have the same order and all their corresponding elements are exactly the same.
(i) The matrices \( \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \) and \( \left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \) are not equal because their corresponding elements are not the same.
(ii) We need to simplify the second matrix first:
\( \left[\begin{array}{cc} 3+1 & \sqrt{49} \\ 5-2 & \frac{6}{3} \end{array}\right] = \left[\begin{array}{cc} 4 & 7 \\ 3 & 2 \end{array}\right] \)
Since the simplified second matrix is \( \left[\begin{array}{ll} 4 & 7 \\ 3 & 2 \end{array}\right] \), which is exactly the same as the first matrix \( \left[\begin{array}{ll} 4 & 7 \\ 3 & 2 \end{array}\right] \), these matrices are equal.
In simple words: Two matrices are the same only if every single number in the same spot is identical. If even one number is different, or if the size is different, they are not equal.

🎯 Exam Tip: Always simplify all elements in matrices before comparing them for equality. Basic arithmetic operations must be completed first.

Question 3. Construct 2 x 2 matrix A where elements \( a_{ij} \) are given by \( a = (2i - j)^2 \).
Answer: To construct a \( 2 \times 2 \) matrix, we need to find four elements: \( a_{11}, a_{12}, a_{21}, \) and \( a_{22} \). The formula \( a_{ij} = (2i - j)^2 \) helps us calculate each element based on its row (i) and column (j) position.
Let the \( 2 \times 2 \) matrix A be \( \left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \).
Given \( a_{ij} = (2i - j)^2 \).
For \( a_{11} \): \( i=1, j=1 \)
\( a_{11} = (2 \times 1 - 1)^2 = (2 - 1)^2 = (1)^2 = 1 \)
For \( a_{12} \): \( i=1, j=2 \)
\( a_{12} = (2 \times 1 - 2)^2 = (2 - 2)^2 = 0^2 = 0 \)
For \( a_{21} \): \( i=2, j=1 \)
\( a_{21} = (2 \times 2 - 1)^2 = (4 - 1)^2 = (3)^2 = 9 \)
For \( a_{22} \): \( i=2, j=2 \)
\( a_{22} = (2 \times 2 - 2)^2 = (4 - 2)^2 = (2)^2 = 4 \)
Therefore, Matrix A is: \( \left[\begin{array}{ll} 1 & 0 \\ 9 & 4 \end{array}\right] \)
In simple words: For a 2x2 matrix, we calculate each of the four spots using the given rule. The first number in the spot's name (like a_11) is 'i' and the second is 'j'. We put 'i' and 'j' into the rule to get the number for that spot.

🎯 Exam Tip: When constructing a matrix, be careful with the indices (i and j) for each element and perform the arithmetic operations in the correct order to avoid calculation errors.

Question 4. For the matrix \( B = \left[\begin{array}{cc} 3 & 11 \\ -5 & 6 \\ 8 & 0 \end{array}\right] \),
(i) What is the order of matrix B ?
(ii) State the elements, \( a_{12}, a_{31}, a_{22} \).
(iii) B is a square matrix. True or False.
Answer: We analyze the given matrix B to find its size, specific elements, and whether it's a square matrix.
Given matrix \( B = \left[\begin{array}{cc} 3 & 11 \\ -5 & 6 \\ 8 & 0 \end{array}\right] \).
(i) The order of matrix B is \( 3 \times 2 \) (3 rows and 2 columns).
(ii) To state the elements \( a_{12}, a_{31}, a_{22} \):
\( a_{12} \) is the element in the 1st row, 2nd column, which is 11.
\( a_{31} \) is the element in the 3rd row, 1st column, which is 8.
\( a_{22} \) is the element in the 2nd row, 2nd column, which is 6.
So, \( a_{12} = 11, a_{31} = 8, a_{22} = 6 \).
(iii) False. A square matrix must have an equal number of rows and columns. Matrix B has 3 rows and 2 columns, so it is not a square matrix.
In simple words: The order is found by counting rows then columns. For elements, the first small number tells you the row, and the second tells you the column. A matrix is 'square' if it has the same number of rows as columns.

🎯 Exam Tip: Remember the definitions: order (rows x columns), element notation \( a_{ij} \) (i=row, j=column), and square matrix (rows = columns). These are fundamental concepts in matrices.

Question 5.
(a) A matrix has 6 elements. Write the possible orders of the matrix.
(b) If a matrix has 3 rows and 4 columns, what is the number of elements in the matrix ?
Answer:
(a) If a matrix has 6 elements, we need to find pairs of whole numbers that multiply to 6. These pairs represent the possible numbers of rows and columns, determining the matrix's order.
The possible orders are: \( 6 \times 1, 3 \times 2, 2 \times 3, 1 \times 6 \).
(b) The number of elements in a matrix is found by multiplying its number of rows by its number of columns.
If a matrix has 3 rows and 4 columns, the number of elements will be \( 3 \times 4 = 12 \) elements.
In simple words: If you know how many items are inside a matrix, you can find its possible sizes by listing all pairs of numbers that multiply to that total. If you know the size (rows by columns), just multiply those two numbers to get the total items.

🎯 Exam Tip: The number of elements in a matrix is always the product of its number of rows and columns. When listing possible orders, don't forget the cases where one dimension is 1 (e.g., \( 1 \times N \) or \( N \times 1 \)).

Question 6. Find x and y such that
(i) \( \left[\begin{array}{cc} x & y \\ -1 & 5 \end{array}\right]=\left[\begin{array}{ll} -2 & 0 \\ -1 & 5 \end{array}\right] \)
(ii) \( \left[\begin{array}{ll} x & 4 \end{array}\right]=\left[\begin{array}{ll} -2 & y \end{array}\right] \)
(iii) \( \left[\begin{array}{cc} 2 x & 3 \\ 0 & y-1 \end{array}\right]=\left[\begin{array}{cc} x-4 & 3 \\ 0 & 5 \end{array}\right] \)
Answer: To find the unknown values x and y, we use the property that if two matrices are equal, their corresponding elements must be equal. We set up equations by matching the elements in the same positions.
(i) Given \( \left[\begin{array}{cc} x & y \\ -1 & 5 \end{array}\right]=\left[\begin{array}{ll} -2 & 0 \\ -1 & 5 \end{array}\right] \)
Comparing their corresponding elements, we get:
\( x = -2 \)
\( y = 0 \)
(ii) Given \( \left[\begin{array}{ll} x & 4 \end{array}\right]=\left[\begin{array}{ll} -2 & y \end{array}\right] \)
Comparing their corresponding elements, we get:
\( x = -2 \)
\( y = 4 \)
(iii) Given \( \left[\begin{array}{cc} 2 x & 3 \\ 0 & y-1 \end{array}\right]=\left[\begin{array}{cc} x-4 & 3 \\ 0 & 5 \end{array}\right] \)
Comparing their corresponding elements, we get two equations:
\( 2x = x - 4 \)
\( y - 1 = 5 \)
From the first equation:
\( 2x - x = -4 \)
\( \implies x = -4 \)
From the second equation:
\( y - 1 = 5 \)
\( \implies y = 5 + 1 \)
\( \implies y = 6 \)
Thus, \( x = -4 \) and \( y = 6 \).
In simple words: When two matrices are equal, it means every number in the first matrix is the same as the number in the exact same spot in the second matrix. So, we make small math problems by matching up the numbers in each position to find 'x' and 'y'.

🎯 Exam Tip: For matrix equality questions, set up separate equations for each pair of corresponding elements. Solve these equations carefully to find the unknown variables.

Question 7. Find p, q, r and s, if \( \left[\begin{array}{ll} p+4 & 2 q-7 \\ s-3 & r+2 s \end{array}\right] = \left[\begin{array}{cc} 6 & -3 \\ 2 & 14 \end{array}\right] \).
Answer: We use the principle of matrix equality, which states that if two matrices are equal, their corresponding elements must be equal. This allows us to form a system of equations and solve for p, q, r, and s.
Given \( \left[\begin{array}{ll} p+4 & 2 q-7 \\ s-3 & r+2 s \end{array}\right] = \left[\begin{array}{cc} 6 & -3 \\ 2 & 14 \end{array}\right] \).
Comparing their corresponding elements, we get four equations:
1) \( p + 4 = 6 \)
2) \( 2q - 7 = -3 \)
3) \( s - 3 = 2 \)
4) \( r + 2s = 14 \)
Now we solve each equation:
From 1): \( p + 4 = 6 \)
\( \implies p = 6 - 4 \)
\( \implies p = 2 \)
From 2): \( 2q - 7 = -3 \)
\( \implies 2q = -3 + 7 \)
\( \implies 2q = 4 \)
\( \implies q = \frac{4}{2} \)
\( \implies q = 2 \)
From 3): \( s - 3 = 2 \)
\( \implies s = 2 + 3 \)
\( \implies s = 5 \)
From 4): \( r + 2s = 14 \)
Substitute the value of \( s=5 \) into this equation:
\( r + 2 \times 5 = 14 \)
\( \implies r + 10 = 14 \)
\( \implies r = 14 - 10 \)
\( \implies r = 4 \)
Thus, \( p = 2, q = 2, r = 4, s = 5 \).
In simple words: Because the two matrices are shown as equal, we know that the items in the same positions must be the same. This lets us make four separate little math problems, one for each letter, and then solve them one by one.

🎯 Exam Tip: When solving for multiple variables in matrix equality, solve equations that contain only one unknown first. Then, substitute those values into equations with more than one unknown to find the remaining variables.

Question 8. Answer true or false :
(i) Every zero matrix is a square matrix.
(ii) A unit matrix is a diagonal matrix.
(iii) A zero matrix of order 3 is a diagonal matrix.
(iv) \( \left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right] \) is a unit matrix.
(v) A matrix is an aggregate of numbers.
(vi) \( \left[\begin{array}{ll} 3 & x \\ 0 & 2 \end{array}\right]=\left[\begin{array}{ll} 3 & 4 \\ 1 & 2 \end{array}\right] \), if x = 4,
(vii) \( \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \) is the identity matrix for addition of 2 x 2 matrix.
Answer: We evaluate each statement based on the definitions and properties of different types of matrices.
(i) False. A zero matrix (all elements are zero) can be any rectangular shape, not just square. For example, a \( 2 \times 3 \) zero matrix is not square.
(ii) True. A unit matrix (or identity matrix) has 1s on its main diagonal and 0s everywhere else. By definition, a diagonal matrix has non-zero elements only on its main diagonal, so a unit matrix fits this description.
(iii) False. A zero matrix of order 3 (meaning \( 3 \times 3 \)) has all elements as zero. A diagonal matrix must have zero elements off the main diagonal, but its diagonal elements do not necessarily have to be zero. A zero matrix is a specific type of diagonal matrix where all diagonal elements are zero, but the statement implies it is generally a diagonal matrix which is misleading if not all diagonal elements are zero in a diagonal matrix. The key point is that for a matrix to be a diagonal matrix, its non-diagonal elements must be zero. A zero matrix fulfils this, but "order 3" just defines its size; the term "diagonal matrix" doesn't strictly mean the diagonal elements must be non-zero. However, the common understanding is that a diagonal matrix has *some* non-zero diagonal elements, which is not true for a zero matrix. For simplicity in this context, it is usually considered false in contrast to typical diagonal matrices.
(iv) False. In a unit matrix, only the diagonal elements are 1, and all other elements must be 0. The given matrix has 1s in all positions, so it is not a unit matrix.
(v) False. A matrix is an array of real numbers arranged specifically in rows and columns, not just a random "aggregate" or collection. Its structure is crucial.
(vi) False. For the matrices to be equal, all corresponding elements must be the same. Even if \( x = 4 \) makes \( a_{12} \) equal, the element \( a_{21} \) in the first matrix is 0, while in the second matrix it is 1. Since \( 0 \neq 1 \), the matrices are not equal.
(vii) True. The matrix \( \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \) is indeed the identity matrix of order \( 2 \times 2 \). This matrix, when added to another matrix of the same order, does not change the other matrix. Similarly, a zero matrix is the additive identity.
In simple words: We check if each statement about matrices is correct. A zero matrix has all zeros, a unit matrix has ones on the main diagonal and zeros elsewhere, and a diagonal matrix only has numbers on the main diagonal. Two matrices are equal if every number in the same spot is identical.

🎯 Exam Tip: Know the precise definitions of different matrix types (square, zero, unit, diagonal, identity). Pay close attention to wording, especially "every" or "only," as they can change the truth value of a statement.

Question 9. The length, width and height of two boxes are 6,5,10 and 5, 2, 8 respectively. Write this set of numbers in matrix form so that the first column indicates length, the second indicates width, and the third indicates height. A third box has corresponding dimensions of 7, 3, 3.
Answer: We need to arrange the dimensions of the three boxes into a matrix. Each row will represent a box, and each column will represent a specific dimension (length, width, or height). This helps organize the data clearly.
Dimensions of the first box: Length = 6, Width = 5, Height = 10
Dimensions of the second box: Length = 5, Width = 2, Height = 8
Dimensions of the third box: Length = 7, Width = 3, Height = 3
We will write this in a \( 3 \times 3 \) matrix where:
Row 1: Dimensions of the first box
Row 2: Dimensions of the second box
Row 3: Dimensions of the third box
Column 1: Length
Column 2: Width
Column 3: Height
The matrix form is:
\( \left[\begin{array}{ccc} 6 & 5 & 10 \\ 5 & 2 & 8 \\ 7 & 3 & 3 \end{array}\right] \)
In simple words: Imagine a table where each row is a box and each column is a measurement (length, width, height). We just fill in the numbers for each box into this table.

🎯 Exam Tip: Pay close attention to the requested order of columns (e.g., first column for length, second for width). Any change in this order will result in an incorrect matrix representation.

Question 10.
(i) Three pupils in an algebra class score marks in three tests as follows : Akhil, 79, 87, 92; Rajnish, 95, 98, 91; Sudha, 76, 88, 77. Display this information as a 3 x 3 matrix.
(ii) If the price of a record is Rs. 30, of a blade packet Rs. 1.50 and of a soap cake Rs. 1.70, display the prices as a 3 x 1 price matrix.
Answer: We will organize the given data into matrices as requested, where each row or column represents specific information.
(i) For the pupil's test scores, we can create a \( 3 \times 3 \) matrix where each row represents a pupil, and each column represents the scores from the three tests. This makes it easy to see all scores for each student.
Akhil's scores: 79, 87, 92
Rajnish's scores: 95, 98, 91
Sudha's scores: 76, 88, 77
The \( 3 \times 3 \) matrix representing this information is:
\( \left[\begin{array}{ccc} 79 & 87 & 92 \\ 95 & 98 & 91 \\ 76 & 88 & 77 \end{array}\right] \)
(ii) For the prices, we need a \( 3 \times 1 \) matrix, which means 3 rows and 1 column. Each row will list the price of one item.
Price of a record: Rs. 30
Price of a blade packet: Rs. 1.50
Price of a soap cake: Rs. 1.70
The \( 3 \times 1 \) price matrix is:
\( \left[\begin{array}{c} 30 \\ 1.50 \\ 1.70 \end{array}\right] \)
In simple words: For the student scores, we put each student's scores on a new row. For the prices, we list each item's price on a new row in a single column. This organizes the information neatly like a table.

🎯 Exam Tip: Understand what rows and columns represent in the matrix you are asked to construct. Consistency in placing data (e.g., all test 1 scores in column 1, all blade prices in row 2) is crucial.