GSEB Class 12 Maths Solutions Chapter 3 શ્રેણિક Exercise 3.1

GSEB Solutions Class 12 Maths Chapter 3 શ્રેણિક Ex 3.1

 

Question 1. શ્રેણિક A = \(\begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}\), તો નીચેની બાબતો શોધો:
(i) શ્રેણિકની કક્ષા,
(ii) ઘટકોની સંખ્યા,
(iii) ઘટકો \(a_{13}\), \(a_{21}\), \(a_{33}\), \(a_{24}\), \(a_{23}\) શોધો.

Answer:The given matrix is: \[A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}\] To understand the elements and order, we can compare it with a general matrix:

\(a_{11}\)	\(a_{12}\)	\(a_{13}\)	\(a_{14}\)
\(a_{21}\)	\(a_{22}\)	\(a_{23}\)	\(a_{24}\)
\(a_{31}\)	\(a_{32}\)	\(a_{33}\)	\(a_{34}\)

(i) અહીં શ્રેણિક A માં 3 હાર તથા 4 સ્તંભ છે. Therefore, the order of matrix A is \(3 \times 4\).
(ii) શ્રેણિક A માં કુલ ઘટકો \(3 \times 4 = 12\) છે.
(iii) The specific elements are: \(a_{13} = 19\), \(a_{21} = 35\), \(a_{33} = -5\), \(a_{24} = 12\), and \(a_{23} = \frac{5}{2}\).
In simple words: This question asks for the size of a given matrix (rows by columns), the total number of items inside it, and the values of specific items at certain positions. We found it has 3 rows and 4 columns, making 12 items in total, and then listed the values for the requested positions.

🎯 Exam Tip: Understanding matrix order (rows x columns) and element indexing (\(a_{ij}\) for row i, column j) is fundamental for matrix operations and frequently tested.

 

Question 2. જો કોઈ શ્રેણિકને 24 ઘટકો હોય, તો તેની શક્ય કક્ષાઓ કઈ હોય ? જો તેને 13 ઘટકો હોય, તો તેની શક્ય કક્ષાઓ શું થશે ?

Answer:A matrix with 24 elements can have different possible orders. These orders are found by listing all pairs of factors of 24. The possible orders are: \(1 \times 24\), \(24 \times 1\), \(2 \times 12\), \(12 \times 2\), \(3 \times 8\), \(8 \times 3\), \(4 \times 6\), \(6 \times 4\). If a matrix has 13 elements, since 13 is a prime number, it can only be arranged in two ways: \(13 \times 1\) and \(1 \times 13\).
In simple words: If a matrix has 24 items, we can arrange them in different row and column combinations like 1 row and 24 columns, or 2 rows and 12 columns, and so on. If it has 13 items, which is a prime number, there are only two ways to arrange them: 1 row and 13 columns, or 13 rows and 1 column.

🎯 Exam Tip: Remember that the number of elements in a matrix is the product of its number of rows and columns. Prime numbers will have only two possible matrix orders.

 

Question 3. જો કોઈ શ્રેણિકને 18 ઘટકો હોય, તો તેની શક્ય કક્ષાઓ કઈ હોય ? જો તેને 5 ઘટકો હોય, તો તેની શક્ય કક્ષાઓ શું થાય ?

Answer:A matrix with 18 elements can have the following possible orders, which are all pairs of factors of 18: The possible orders are: \(1 \times 18\), \(18 \times 1\), \(2 \times 9\), \(9 \times 2\), \(3 \times 6\), \(6 \times 3\). If a matrix has 5 elements, since 5 is a prime number, it can only have two possible orders: \(5 \times 1\) and \(1 \times 5\).
In simple words: For a matrix with 18 items, you can arrange them in different row and column pairs that multiply to 18. For example, 1 row and 18 columns, or 2 rows and 9 columns. If a matrix has 5 items, because 5 is a prime number, it can only have 5 rows and 1 column, or 1 row and 5 columns.

🎯 Exam Tip: This question tests the understanding of matrix order and how it relates to the total number of elements. Always list all factor pairs (m x n) to find all possible orders.

 

Question 4. જો કોઈ \(2 \times 2\) શ્રેણિક A = \([a_{ij}]\) ના સભ્યો,
(i) \(a_{ij} = \frac{(i+j)^2}{2}\), (ii) \(a_{ij} = \frac{i}{j}\), (iii) \(a_{ij} = \frac{(i+2j)^2}{2}\) થી મળે, તો શ્રેણિક A ની રચના કરો.

Answer:A \(2 \times 2\) matrix A has elements \(a_{ij}\) where i (row) and j (column) can be 1 or 2. The general form of a \(2 \times 2\) matrix is: \[A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\] (i) For \(a_{ij} = \frac{(i+j)^2}{2}\): To find \(a_{11}\), we set \(i=1\) and \(j=1\):
\(a_{11} = \frac{(1+1)^2}{2} = \frac{2^2}{2} = \frac{4}{2} = 2\) To find \(a_{12}\), we set \(i=1\) and \(j=2\):
\(a_{12} = \frac{(1+2)^2}{2} = \frac{3^2}{2} = \frac{9}{2}\) To find \(a_{21}\), we set \(i=2\) and \(j=1\):
\(a_{21} = \frac{(2+1)^2}{2} = \frac{3^2}{2} = \frac{9}{2}\) To find \(a_{22}\), we set \(i=2\) and \(j=2\):
\(a_{22} = \frac{(2+2)^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8\) So, the matrix A is: \[A = \begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix}\] (ii) For \(a_{ij} = \frac{i}{j}\): To find \(a_{11}\), we set \(i=1\) and \(j=1\):
\(a_{11} = \frac{1}{1} = 1\) To find \(a_{12}\), we set \(i=1\) and \(j=2\):
\(a_{12} = \frac{1}{2}\) To find \(a_{21}\), we set \(i=2\) and \(j=1\):
\(a_{21} = \frac{2}{1} = 2\) To find \(a_{22}\), we set \(i=2\) and \(j=2\):
\(a_{22} = \frac{2}{2} = 1\) So, the matrix A is: \[A = \begin{bmatrix} 1 & \frac{1}{2} \\ 2 & 1 \end{bmatrix}\] (iii) For \(a_{ij} = \frac{(i+2j)^2}{2}\): To find \(a_{11}\), we set \(i=1\) and \(j=1\):
\(a_{11} = \frac{(1+2 \times 1)^2}{2} = \frac{(1+2)^2}{2} = \frac{3^2}{2} = \frac{9}{2}\) To find \(a_{12}\), we set \(i=1\) and \(j=2\):
\(a_{12} = \frac{(1+2 \times 2)^2}{2} = \frac{(1+4)^2}{2} = \frac{5^2}{2} = \frac{25}{2}\) To find \(a_{21}\), we set \(i=2\) and \(j=1\):
\(a_{21} = \frac{(2+2 \times 1)^2}{2} = \frac{(2+2)^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8\) To find \(a_{22}\), we set \(i=2\) and \(j=2\):
\(a_{22} = \frac{(2+2 \times 2)^2}{2} = \frac{(2+4)^2}{2} = \frac{6^2}{2} = \frac{36}{2} = 18\) So, the matrix A is: \[A = \begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix}\]
In simple words: This question asks us to build a \(2 \times 2\) matrix, which means it has 2 rows and 2 columns. We are given different rules for how to calculate each element in the matrix based on its row number 'i' and column number 'j'. We apply these rules for each position (\(a_{11}\), \(a_{12}\), \(a_{21}\), \(a_{22}\)) to find its value and then create the matrix.

🎯 Exam Tip: Pay close attention to the formula given for \(a_{ij}\). Substitute 'i' for the row number and 'j' for the column number carefully. Double-check calculations for each element to avoid errors.

 

Question 5. જો 3 \(\times\) 4 શ્રેણિકના સભ્યો (i) \(a_{ij} = \frac{1}{2}|-3i + j|\), (ii) \(a_{ij} = 2i - j\) દ્વારા મળે, તો તે શ્રેણિકની રચના કરો.

Answer:A \(3 \times 4\) matrix has 3 rows and 4 columns. Its general form is: \[A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}\] (i) For \(a_{ij} = \frac{1}{2}|-3i + j|\): For Row 1 (i=1):
\(a_{11} = \frac{1}{2}|-3(1) + 1| = \frac{1}{2}|-3 + 1| = \frac{1}{2}|-2| = \frac{1}{2}(2) = 1\)
\(a_{12} = \frac{1}{2}|-3(1) + 2| = \frac{1}{2}|-3 + 2| = \frac{1}{2}|-1| = \frac{1}{2}(1) = \frac{1}{2}\)
\(a_{13} = \frac{1}{2}|-3(1) + 3| = \frac{1}{2}|-3 + 3| = \frac{1}{2}|0| = 0\)
\(a_{14} = \frac{1}{2}|-3(1) + 4| = \frac{1}{2}|-3 + 4| = \frac{1}{2}|1| = \frac{1}{2}(1) = \frac{1}{2}\) For Row 2 (i=2):
\(a_{21} = \frac{1}{2}|-3(2) + 1| = \frac{1}{2}|-6 + 1| = \frac{1}{2}|-5| = \frac{1}{2}(5) = \frac{5}{2}\)
\(a_{22} = \frac{1}{2}|-3(2) + 2| = \frac{1}{2}|-6 + 2| = \frac{1}{2}|-4| = \frac{1}{2}(4) = 2\)
\(a_{23} = \frac{1}{2}|-3(2) + 3| = \frac{1}{2}|-6 + 3| = \frac{1}{2}|-3| = \frac{1}{2}(3) = \frac{3}{2}\)
\(a_{24} = \frac{1}{2}|-3(2) + 4| = \frac{1}{2}|-6 + 4| = \frac{1}{2}|-2| = \frac{1}{2}(2) = 1\) For Row 3 (i=3):
\(a_{31} = \frac{1}{2}|-3(3) + 1| = \frac{1}{2}|-9 + 1| = \frac{1}{2}|-8| = \frac{1}{2}(8) = 4\)
\(a_{32} = \frac{1}{2}|-3(3) + 2| = \frac{1}{2}|-9 + 2| = \frac{1}{2}|-7| = \frac{1}{2}(7) = \frac{7}{2}\)
\(a_{33} = \frac{1}{2}|-3(3) + 3| = \frac{1}{2}|-9 + 3| = \frac{1}{2}|-6| = \frac{1}{2}(6) = 3\)
\(a_{34} = \frac{1}{2}|-3(3) + 4| = \frac{1}{2}|-9 + 4| = \frac{1}{2}|-5| = \frac{1}{2}(5) = \frac{5}{2}\) So, the 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) For \(a_{ij} = 2i - j\): For Row 1 (i=1):
\(a_{11} = 2(1) - 1 = 2 - 1 = 1\)
\(a_{12} = 2(1) - 2 = 2 - 2 = 0\)
\(a_{13} = 2(1) - 3 = 2 - 3 = -1\)
\(a_{14} = 2(1) - 4 = 2 - 4 = -2\) For Row 2 (i=2):
\(a_{21} = 2(2) - 1 = 4 - 1 = 3\)
\(a_{22} = 2(2) - 2 = 4 - 2 = 2\)
\(a_{23} = 2(2) - 3 = 4 - 3 = 1\)
\(a_{24} = 2(2) - 4 = 4 - 4 = 0\) For Row 3 (i=3):
\(a_{31} = 2(3) - 1 = 6 - 1 = 5\)
\(a_{32} = 2(3) - 2 = 6 - 2 = 4\)
\(a_{33} = 2(3) - 3 = 6 - 3 = 3\)
\(a_{34} = 2(3) - 4 = 6 - 4 = 2\) So, the matrix A is: \[A = \begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}\]
In simple words: Here, we need to create a matrix with 3 rows and 4 columns. We are given two different rules to find the value of each item in the matrix. For each rule, we put the row number 'i' and column number 'j' into the formula to find the value for every spot in the matrix, then we write down the complete matrix.

🎯 Exam Tip: When constructing a matrix, carefully follow the given formula for \(a_{ij}\). Be especially cautious with absolute values (as in part i) and negative signs, as small errors can lead to incorrect matrix elements.

 

Question 6. નીચેનાં સમીકરણોમાંથી x, y અને z નાં મૂલ્ય શોધો :
(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) \(\begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}\)

Answer:(i) Given: \(\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}\) Using the property of equality of matrices (corresponding elements are equal):
\(y = 4\)
\(z = 3\)
\(x = 1\) Thus, we find \(x=1\), \(y=4\), and \(z=3\). (ii) Given: \(\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}\) Using the property of equality of matrices:
\(x+y = 6\) ..........(1)
\(5+z = 5 \implies z = 0\)
\(xy = 8\) ..........(2) From (1), substitute \(y = 6-x\) into (2):
\(x(6-x) = 8\)
\(6x - x^2 = 8\)
\(x^2 - 6x + 8 = 0\) Factor the quadratic equation:
\((x-4)(x-2) = 0\) So, \(x=4\) or \(x=2\). If \(x=4\), then from \(y=6-x\), \(y = 6-4 = 2\). If \(x=2\), then from \(y=6-x\), \(y = 6-2 = 4\). Therefore, the solutions are:
1. \(x=4, y=2, z=0\)
2. \(x=2, y=4, z=0\) (iii) Given: \(\begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}\) Using the property of equality of matrices:
\(x+y+z = 9\) ..........(1)
\(x+z = 5\) ..........(2)
\(y+z = 7\) ..........(3) Subtract equation (2) from equation (1):
\((x+y+z) - (x+z) = 9 - 5\)
\(y = 4\) Subtract equation (3) from equation (1):
\((x+y+z) - (y+z) = 9 - 7\)
\(x = 2\) Substitute \(x=2\) and \(y=4\) into equation (1):
\(2+4+z = 9\)
\(6+z = 9\)
\(z = 3\) Therefore, \(x=2\), \(y=4\), and \(z=3\).
In simple words: This problem asks us to find the values of \(x\), \(y\), and \(z\) by using the rule that if two matrices are equal, then their matching elements must be equal. We set up equations from these matching elements and then solve them using simple algebra to find the unknown values.

🎯 Exam Tip: When matrices are equal, their corresponding elements must be equal. This allows you to form a system of linear equations. Solve these equations carefully, substituting values to find all variables.

 

Question 7. સમીકરણ \(\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}\) માંથી a, b, c અને d નાં મૂલ્ય શોધો.

Answer:Using the property of equality of matrices (corresponding elements are equal), we can set up a system of equations:
\(a-b = -1\) ..........(1)
\(2a-b = 0\) ..........(2)
\(2a+c = 5\) ..........(3)
\(3c+d = 13\) ..........(4) From equation (2):
\(2a = b\)
\(b = 2a\) Substitute \(b=2a\) into equation (1):
\(a - 2a = -1\)
\(-a = -1\)
\(a = 1\) Now that we have \(a=1\), we can find \(b\):
\(b = 2a = 2(1) = 2\) Substitute \(a=1\) into equation (3):
\(2(1) + c = 5\)
\(2 + c = 5\)
\(c = 5 - 2 = 3\) Substitute \(c=3\) into equation (4):
\(3(3) + d = 13\)
\(9 + d = 13\)
\(d = 13 - 9 = 4\) Thus, the values are \(a=1\), \(b=2\), \(c=3\), and \(d=4\).
In simple words: We have two matrices that are equal, so we make a set of four equations by comparing the items at the same spots in both matrices. We solve these equations one by one, starting with the simplest ones, to find the values for \(a\), \(b\), \(c\), and \(d\).

🎯 Exam Tip: This problem involves solving a system of linear equations derived from matrix equality. Organize your equations and solve them systematically, for instance, by substitution or elimination, to avoid errors and ensure all variables are correctly determined.

પ્રશ્ન 8, 9 તથા 10 માં વિધાન સાચું બને તે રીતે આપેલ વિક્લ્પોમાંથી યોગ્ય વિક્લ્પ પસંદ કરો :

 

Question 8. A = \([a_{ij}]_{m \times n}\) એ ચોરસ શ્રેણિક હોય તો,.....
(A) \(m < n\)
(B) \(m > n\)
(C) \(m = n\)
(D) આમાંથી એક પણ નહીં

Answer: (C) \(m = n\)If matrix A = \([a_{ij}]_{m \times n}\) is a square matrix, then the number of rows (m) must be equal to the number of columns (n).
In simple words: A square matrix is a special kind of matrix where the number of rows is exactly the same as the number of columns. So, if a matrix has 'm' rows and 'n' columns, for it to be square, 'm' must be equal to 'n'.

🎯 Exam Tip: The definition of a square matrix is crucial: number of rows equals number of columns. This is a fundamental concept in matrix algebra and is often tested in basic knowledge questions.

 

Question 9. x, y ની જે કિંમતો માટે શ્રેણિક જોડ \(\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix}\) અને \(\begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}\) સમાન થાય તેવી આપેલી x અને y ની કિંમત....
(A) \(x = -\frac{1}{3}\)
(B) શોધવું શક્ય નથી.

Answer: (B) શોધવું શક્ય નથી.For the two matrices to be equal, their corresponding elements must be equal:
\[\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix} = \begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}\] From the first row, first column:
\(3x+7 = 0\)
\(3x = -7\)
\(x = -\frac{7}{3}\) From the second row, second column:
\(2-3x = 4\)
\(-3x = 4-2\)
\(-3x = 2\)
\(x = -\frac{2}{3}\) Here, the value of x obtained from the first equation (\(x = -\frac{7}{3}\)) is different from the value of x obtained from the second equation (\(x = -\frac{2}{3}\)). Since 'x' cannot have two different values simultaneously, it is not possible for these two matrices to be equal. Therefore, values for x and y cannot be found for which these matrices are equal.
In simple words: We are given two matrices and asked to find values of \(x\) and \(y\) that make them equal. When we set up equations by matching items in the same spot, we get two different values for \(x\). Since \(x\) cannot be two different numbers at once, it means there are no such \(x\) and \(y\) values that can make these matrices equal.

🎯 Exam Tip: Always check for consistency in the values of variables derived from different element equalities. If you get contradictory values for a single variable, it means no solution exists for the matrices to be equal.

 

Question 10. પ્રત્યેક ઘટક 0 અથવા 1 હોય તેવા \(3 \times 3\) કક્ષાવાળા શ્રેણિકની સંખ્યા.....
(A) 27
(B) 18
(C) 81
(D) 512

Answer: (D) 512A \(3 \times 3\) matrix has:
Number of elements = \(3 \times 3 = 9\). Each element in the matrix can be either 0 or 1. This means there are 2 choices for each element. Since there are 9 elements, and each element has 2 independent choices, the total number of possible matrices is:
Total number of matrices = \(2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^9\)
\(2^9 = 512\)
In simple words: A matrix with 3 rows and 3 columns has 9 empty spots. For each spot, you can either put a '0' or a '1', which means there are 2 choices for each spot. Since there are 9 spots, and each choice is independent, we multiply 2 by itself 9 times (\(2^9\)) to find the total number of different matrices you can make, which is 512.

🎯 Exam Tip: For counting problems involving choices for each position (like elements in a matrix), use the multiplication principle. If there are 'N' elements and 'k' choices for each element, the total number of possibilities is \(k^N\).