GSEB Class 12 Maths Solutions Chapter 3 Matrices Exercise 3.1

Question 1. In the matrix \( A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix} \), write :
(i) The order of the matrix.
(ii) The number of elements.
(iii) Write the elements \( a_{13}, a_{21}, a_{33}, a_{24}, a_{23} \).
Answer:
(i) The order of the matrix is \( 3 \times 4 \). This means it has 3 rows and 4 columns.
(ii) The number of elements is \( 3 \times 4 = 12 \). You get this by multiplying the number of rows by the number of columns.
(iii) The elements are as follows:
\( a_{13} = 19 \)
\( a_{21} = 35 \)
\( a_{33} = -5 \)
\( a_{24} = 12 \)
\( a_{23} = \frac{5}{2} \)
In simple words: First, find out how many rows and columns the matrix has. Then, multiply those numbers to get the total elements. Finally, find the specific numbers at the listed positions.

Exam Tip: Remember that \( a_{ij} \) means the element in the \( i \)-th row and \( j \)-th column. Always count rows first, then columns when finding the order of a matrix.

Question 2. If a matrix has 24 elements, what are possible orders it can have? What, if it has 13 elements?
Answer:
If a matrix has 24 elements, the possible orders (combinations of rows and columns) are found by listing all pairs of factors of 24:
(i) For 24 elements:
\( 1 \times 24 \)
\( 24 \times 1 \)
\( 2 \times 12 \)
\( 12 \times 2 \)
\( 3 \times 8 \)
\( 8 \times 3 \)
\( 4 \times 6 \)
\( 6 \times 4 \)
Thus, there are 8 different matrices possible with 24 elements.
(ii) For 13 elements:
Since 13 is a prime number, its only factors are 1 and 13.
\( 1 \times 13 \)
\( 13 \times 1 \)
Therefore, there are only 2 possible matrices with 13 elements.
In simple words: To find the possible sizes of a matrix, list all the ways you can multiply two whole numbers to get the total number of elements. Each pair of numbers is a possible size.

Exam Tip: When finding matrix orders, remember to consider both \( m \times n \) and \( n \times m \) unless \( m = n \). Prime numbers will always have only two possible orders.

Question 3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Answer:
(i) If a matrix has 18 elements, the possible orders (combinations of rows and columns) are found by listing all pairs of factors of 18:
\( 1 \times 18 \)
\( 18 \times 1 \)
\( 2 \times 9 \)
\( 9 \times 2 \)
\( 3 \times 6 \)
\( 6 \times 3 \)
So, there are 6 distinct matrices possible with 18 elements.
(ii) If a matrix has 5 elements, the possible orders are determined by the factors of 5. Since 5 is a prime number, its only factors are 1 and 5.
\( 1 \times 5 \)
\( 5 \times 1 \)
Hence, there are two matrices that can have 5 elements.
In simple words: For a given number of elements, list all pairs of numbers that multiply to that total. Each pair gives a possible size for the matrix. For prime numbers, there are only two possibilities.

Exam Tip: Always list all factor pairs, including the number times one, and vice-versa. This ensures you cover all possible matrix dimensions.

Question 4. Construct \( 2 \times 2 \) matrix \( A = [a_{ij}] \) whose elements are given by
(i) \( a_{ij} = \frac{(i+j)^{2}}{2} \)
(ii) \( a_{ij} = \frac { i }{j} \)
(iii) \( a_{ij} = \frac{(i+2j)^{2}}{2} \)
Answer:
(i) Given \( a_{ij} = \frac{(i+j)^{2}}{2} \)
For a \( 2 \times 2 \) matrix, we need to calculate \( a_{11}, a_{12}, a_{21}, a_{22} \).
Put \( i = 1, j = 1 \): \( a_{11} = \frac{(1+1)^{2}}{2} = \frac{2^{2}}{2} = \frac{4}{2} = 2 \)
Put \( i = 1, j = 2 \): \( a_{12} = \frac{(1+2)^{2}}{2} = \frac{3^{2}}{2} = \frac{9}{2} \)
Put \( i = 2, j = 1 \): \( a_{21} = \frac{(2+1)^{2}}{2} = \frac{3^{2}}{2} = \frac{9}{2} \)
Put \( i = 2, j = 2 \): \( a_{22} = \frac{(2+2)^{2}}{2} = \frac{4^{2}}{2} = \frac{16}{2} = 8 \)
So, the matrix \( A \) is:
\[ A = \begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix} \]
(ii) Given \( a_{ij} = \frac { i }{j} \)
Put \( i = 1, j = 1 \): \( a_{11} = \frac{1}{1} = 1 \)
Put \( i = 1, j = 2 \): \( a_{12} = \frac{1}{2} \)
Put \( i = 2, j = 1 \): \( a_{21} = \frac{2}{1} = 2 \)
Put \( i = 2, j = 2 \): \( a_{22} = \frac{2}{2} = 1 \)
So, the matrix \( A \) is:
\[ A = \begin{bmatrix} 1 & \frac{1}{2} \\ 2 & 1 \end{bmatrix} \]
(iii) Given \( a_{ij} = \frac{(i+2j)^{2}}{2} \)
Put \( i = 1, j = 1 \): \( a_{11} = \frac{(1+2(1))^{2}}{2} = \frac{(1+2)^{2}}{2} = \frac{3^{2}}{2} = \frac{9}{2} \)
Put \( i = 1, j = 2 \): \( a_{12} = \frac{(1+2(2))^{2}}{2} = \frac{(1+4)^{2}}{2} = \frac{5^{2}}{2} = \frac{25}{2} \)
Put \( i = 2, j = 1 \): \( a_{21} = \frac{(2+2(1))^{2}}{2} = \frac{(2+2)^{2}}{2} = \frac{4^{2}}{2} = \frac{16}{2} = 8 \)
Put \( i = 2, j = 2 \): \( a_{22} = \frac{(2+2(2))^{2}}{2} = \frac{(2+4)^{2}}{2} = \frac{6^{2}}{2} = \frac{36}{2} = 18 \)
So, the matrix \( A \) is:
\[ A = \begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix} \]
In simple words: To build a matrix from a rule, use the row number (i) and column number (j) in the given formula to calculate each element. Then, arrange these calculated numbers into the correct row and column positions in your matrix.

Exam Tip: Double-check your calculations for each element. A single error can lead to an incorrect matrix. Pay close attention to the order of operations, especially with squares and fractions.

Question 5. Construct a \( 3 \times 4 \) matrix whose elements are given by:
(i) \( a_{ij} = \frac { 1 }{ 2 }|-3i+j| \)
(ii) \( a_{ij} = 2i - j \)
Answer:
(i) Given \( a_{ij} = \frac { 1 }{ 2 }|-3i+j| \)
For a \( 3 \times 4 \) matrix, we need 3 rows and 4 columns.
Row 1 (i = 1):
\( a_{11} = \frac{1}{2}|-3(1)+1| = \frac{1}{2}|-2| = \frac{1}{2}(2) = 1 \)
\( a_{12} = \frac{1}{2}|-3(1)+2| = \frac{1}{2}|-1| = \frac{1}{2}(1) = \frac{1}{2} \)
\( a_{13} = \frac{1}{2}|-3(1)+3| = \frac{1}{2}|0| = 0 \)
\( a_{14} = \frac{1}{2}|-3(1)+4| = \frac{1}{2}|1| = \frac{1}{2} \)
Row 2 (i = 2):
\( a_{21} = \frac{1}{2}|-3(2)+1| = \frac{1}{2}|-5| = \frac{1}{2}(5) = \frac{5}{2} \)
\( a_{22} = \frac{1}{2}|-3(2)+2| = \frac{1}{2}|-4| = \frac{1}{2}(4) = 2 \)
\( a_{23} = \frac{1}{2}|-3(2)+3| = \frac{1}{2}|-3| = \frac{1}{2}(3) = \frac{3}{2} \)
\( a_{24} = \frac{1}{2}|-3(2)+4| = \frac{1}{2}|-2| = \frac{1}{2}(2) = 1 \)
Row 3 (i = 3):
\( a_{31} = \frac{1}{2}|-3(3)+1| = \frac{1}{2}|-8| = \frac{1}{2}(8) = 4 \)
\( a_{32} = \frac{1}{2}|-3(3)+2| = \frac{1}{2}|-7| = \frac{1}{2}(7) = \frac{7}{2} \)
\( a_{33} = \frac{1}{2}|-3(3)+3| = \frac{1}{2}|-6| = \frac{1}{2}(6) = 3 \)
\( a_{34} = \frac{1}{2}|-3(3)+4| = \frac{1}{2}|-5| = \frac{1}{2}(5) = \frac{5}{2} \)
The resulting matrix \( A \) is:
\[ A = \begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix} \]
(ii) Given \( a_{ij} = 2i - j \)
Row 1 (i = 1):
\( a_{11} = 2(1) - 1 = 1 \)
\( a_{12} = 2(1) - 2 = 0 \)
\( a_{13} = 2(1) - 3 = -1 \)
\( a_{14} = 2(1) - 4 = -2 \)
Row 2 (i = 2):
\( a_{21} = 2(2) - 1 = 3 \)
\( a_{22} = 2(2) - 2 = 2 \)
\( a_{23} = 2(2) - 3 = 1 \)
\( a_{24} = 2(2) - 4 = 0 \)
Row 3 (i = 3):
\( a_{31} = 2(3) - 1 = 5 \)
\( a_{32} = 2(3) - 2 = 4 \)
\( a_{33} = 2(3) - 3 = 3 \)
\( a_{34} = 2(3) - 4 = 2 \)
The resulting matrix \( A \) is:
\[ A = \begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix} \]
In simple words: For each position in the 3x4 matrix, use its row (i) and column (j) numbers in the given formula to calculate the value for that spot. Be careful with absolute values and basic arithmetic.

Exam Tip: When constructing larger matrices, it's helpful to organize your calculations by row or by column to avoid errors. Always double-check the formula for each element, especially involving absolute values.

Question 6. Find the values of x, y and z from the following equations:
(i) \( \begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & 3 \\ 1 & 5 \end{bmatrix} \)
(ii) \( \begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 5 \\ 2 & 8 \end{bmatrix} \)
(iii) \( \begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix} \)
Answer:
(i) To find the values, we compare the matching elements of the two matrices:
\( 4 = y \)
\( 3 = 3 \)
\( x = 1 \)
\( 5 = 5 \)
Therefore, we get \( x = 1, y = 4 \), and \( z \) is not present in this equation.
(ii) Comparing the corresponding elements of the matrices, we get a system of equations:
1. \( x+y = 6 \)
2. \( 5+z = 5 \)
3. \( xy = 8 \)
From equation 1, we can express \( y = 6-x \).
Substitute this into equation 3:
\( x(6-x) = 8 \)
\( 6x - x^2 = 8 \)
\( x^2 - 6x + 8 = 0 \)
Factor this quadratic equation:
\( (x-4)(x-2) = 0 \)
This gives us two possible values for \( x \): \( x = 4 \) or \( x = 2 \).
If \( x = 4 \), then \( y = 6 - 4 = 2 \).
If \( x = 2 \), then \( y = 6 - 2 = 4 \).
From equation 2:
\( 5+z = 5 \)
\( z = 5 - 5 \)
\( z = 0 \)
So, there are two sets of solutions:
Set 1: \( x = 4, y = 2, z = 0 \)
Set 2: \( x = 2, y = 4, z = 0 \)
(iii) Equating the corresponding elements of the matrices gives us:
1. \( x+y+z = 9 \)
2. \( x+z = 5 \)
3. \( y+z = 7 \)
We can solve this system of linear equations:
Subtract equation 2 from equation 1:
\( (x+y+z) - (x+z) = 9 - 5 \)
\( y = 4 \)
Now, substitute \( y=4 \) into equation 3:
\( 4+z = 7 \)
\( z = 7 - 4 \)
\( z = 3 \)
Finally, substitute \( z=3 \) into equation 2:
\( x+3 = 5 \)
\( x = 5 - 3 \)
\( x = 2 \)
Thus, the values are \( x = 2, y = 4, z = 3 \).
In simple words: When two matrices are equal, their matching elements must also be equal. Write down equations for each pair of matching elements. Then, solve these equations to find the values of x, y, and z.

Exam Tip: For problems involving equal matrices, always set up equations for each corresponding element. When solving systems of equations, substitution or elimination methods can be used. Be careful when a quadratic equation arises, as it may lead to multiple possible solutions.

Question 7. Find the values of a, b, c and d from the equation
\( \begin{bmatrix} a-b & 3a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} \)
Answer:
By comparing the corresponding elements of the two matrices, we get the following system of equations:
1. \( a - b = -1 \)
2. \( 2a - b = 0 \)
3. \( 3a + c = 5 \)
4. \( 3c + d = 13 \)
From equation 2, we can write \( b = 2a \).
Substitute \( b = 2a \) into equation 1:
\( a - (2a) = -1 \)
\( -a = -1 \)
\( a = 1 \)
Now, find \( b \) using \( b = 2a \):
\( b = 2(1) \)
\( b = 2 \)
Substitute \( a = 1 \) into equation 3:
\( 3(1) + c = 5 \)
\( 3 + c = 5 \)
\( c = 5 - 3 \)
\( c = 2 \)
Substitute \( c = 2 \) into equation 4:
\( 3(2) + d = 13 \)
\( 6 + d = 13 \)
\( d = 13 - 6 \)
\( d = 7 \)
Thus, the values are \( a = 1, b = 2, c = 2, d = 7 \).
In simple words: Since the matrices are equal, set up equations by matching the elements in the same positions. Then, solve these equations step-by-step to find the value of each variable.

Exam Tip: Be careful with signs when substituting values or rearranging equations. Solve for variables in a logical order (e.g., solve for 'a' and 'b' first, then 'c', then 'd').

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 the options
Answer: (d) None of the options
In simple words: A square matrix is one where the number of rows (m) is exactly the same as the number of columns (n). So, m must equal n for a matrix to be square.

Exam Tip: Know the fundamental definitions in matrix algebra. A square matrix is defined by having an equal number of rows and columns, i.e., \( m = n \).

Question 9. Which of the given values of x and y make the following pair of matrices equal.
\( \begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix} = \begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix} \)
(a) \( x = - \frac { 1 }{ 3 }, y = 7 \)
(b) Not Possible to find
(c) \( y = 7, x = - \frac { 2 }{ 3 } \)
(d) \( x = - \frac { 1 }{ 3 }, y = - \frac { 2 }{ 3 } \)
Answer: (b) Not Possible to find
In simple words: When you set the matching parts of the matrices equal, you get different values for x from different equations. This means no single value of x can make both matrices equal, so it's not possible to find a solution.

Exam Tip: Always solve for all variables based on all corresponding elements. If you get contradictory values for a single variable from different equations, it means no solution exists for making the matrices equal.

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
Answer: (d) 512
In simple words: A 3x3 matrix has 9 spots for numbers. If each spot can be either a 0 or a 1 (2 options), then you multiply 2 by itself 9 times to get all the possible different matrices.

Exam Tip: For matrices where each entry has 'k' possible choices, and the matrix is of order \( m \times n \), the total number of such matrices is \( k^{mn} \).