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Matrices solved ex 3.1

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Matrices Solved ex - 3.1 by PCM Encyclopedia.

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Matrices solved ex 3.1

  1. 1. Chapter - 3 MATRICES (Ex – 3.1) PCM ENCYCLOPEDIA (www.pcmpedia.blogspot.in) 1 1. In the Matrix A=[ ⁄ √ ], Write: (i) The order of the matrix, (ii) The number of elements, (iii) Write the elements of a13, a21, a33, a24, a23. Solution 1: (i) In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 × 4. (ii) Since the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements in it. (iii) a13 = 19, a21 = 35, a33 = −5, a24 = 12, a23 = 5/2 2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements? Solution 2: We know that if a matrix is of the order m × n, it has mn elements.So, here we have to find out the possible orders of a matrix having 24 elements. Hence, the possible orders of a matrix having 24 elements are: 1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4 and the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1. 3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements? Solution 3: We know that if a matrix is of the order m × n, it has mn elements.So, here we have to find out the possible orders of a matrix having 18
  2. 2. Chapter - 3 MATRICES (Ex – 3.1) PCM ENCYCLOPEDIA (www.pcmpedia.blogspot.in) 2 elements. Hence, the possible orders of a matrix having 18 elements are: 1 × 18, 2 × 9, 3 × 6, 6 × 3, 9 × 2, and 18 × 1 and the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1. 4. Construct a 2X2 matrix, A=[aij], whose elements are given by: (i) aij = ( ) (ii) aij = (iii) aij = ( ) Solution 4: (i) Given aij = ( ) We Know, a 2 x 2 matrix is given by A = [ ] So, a11 = ( ) = = 2 a12 = ( ) = a21 = ( ) = a22 = ( ) = = 8 Therefore, the required Matrix is A = [ ] (ii) Given aij = We Know, a 2 x 2 matrix is given by A = [ ] So, a11 = = 1 a12 = a21 = = 2 a22 = = 1
  3. 3. Chapter - 3 MATRICES (Ex – 3.1) PCM ENCYCLOPEDIA (www.pcmpedia.blogspot.in) 3 Therefore, the required Matrix is A = [ ] (iii) Given aij = ( ) We Know, a 2 x 2 matrix is given by A = [ ] So, a11 = ( ) = a12 = ( ) = a21 = ( ) = = 8 a22 = ( ) = = 18 Therefore, the required Matrix is A = [ ] 5. Construct a 3X4 Matrix, whose elements are given by: (i) aij = | | (ii) aij = 2i – j Solution 5: (i) aij = | | We Know, a 3 x 4 matrix is given by A = [ ] a11 = | | = 1 a12 = | | = a13 = | | = 0 a14 = | | =
  4. 4. Chapter - 3 MATRICES (Ex – 3.1) PCM ENCYCLOPEDIA (www.pcmpedia.blogspot.in) 4 a21 = | | = a22 = | | = 2 a23 = | | = a24 = | | = 1 a31 = | | = 4 a32 = | | = a33 = | | = 3 a34 = | | = Hence, the required matrix is: A = [ ] (ii) aij = 2i – j We Know, a 3 x 4 matrix is given by A = [ ] a11 = 2 × 1 − 1 = 1 a12 = 2 × 1 − 2 = 0 a13 = 2 × 1 − 3 = −1 a14 = 2 × 1 − 4 = −2 a21 = 2 × 2 − 1 = 3 a22 = 2 × 2 − 2 = 2 a23 = 2 × 2 − 3 = 1 a24 = 2 × 2 − 4 = 0 a31 = 2 × 3 − 1 = 5 a32 = 2 × 3 − 2 = 4 a33 = 2 × 3 − 3 = 3 a34 = 2 × 3 − 4 = 2 Hence, the required matrix is: A = [ ]
  5. 5. Chapter - 3 MATRICES (Ex – 3.1) PCM ENCYCLOPEDIA (www.pcmpedia.blogspot.in) 5 6. Find the values of x, y and z from the following equations: (i) [ ]=[ ] (ii) [ ]=[ ] (ii) [ ]=[ ] Solution 6: (i) [ ] = [ ] Since both are matrices are equal, their corresponding elements are also equal. Comparing corresponding elements, we get y = 4, z = 3 and x = 1 (ii) [ ]=[ ] Since both are matrices are equal, their corresponding elements are also equal. Comparing corresponding elements, we get x + y = 6, 5 + z = 5 and xy = 8 Now, 5 + z = 5 z = 0 Here, x + y = 6 ----------------------- eq. 1 xy = 8 -------------------------- eq. 2 From eq. 1 y = 6 – x -------------------------------- eq .3 Put this value of y in eq. 2, we get x (6 – x) = 8 6x – x2 = 8 or x2 6x + 8 = 0 By Factorization, x2 2x – 4x + 8 = 0 x(x 2) 4(x 2) = 0 (x 4)(x – 2) = 0
  6. 6. Chapter - 3 MATRICES (Ex – 3.1) PCM ENCYCLOPEDIA (www.pcmpedia.blogspot.in) 6 x = 4 or x = 2 Put these values in eq. 3 From x = 4, we get From x = 2, we get y = 6 4 = 2 y = 6 – 2 = 4 x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0 (iii) [ ]=[ ] Since both are matrices are equal, their corresponding elements are also equal. Comparing corresponding elements, we get x y z = ……………….. ( ) x z = ……………………….( ) y z = ……………………….( ) From (1) and (2), we have y + 5 = 9 ⇒ y = 4 Put this value of y in (3) 4 + z = 7 ⇒ z = 3 Also, x + z = 5 x + 3 = 5 ⇒ x = 2 ∴ x = 2, y = 4 and z = 3 7. Find the Value of a, b, c and d from the equation: [ ] = [ ]
  7. 7. Chapter - 3 MATRICES (Ex – 3.1) PCM ENCYCLOPEDIA (www.pcmpedia.blogspot.in) 7 Solution 7: Since both are matrices are equal, their corresponding elements are also equal. Comparing corresponding elements, we get: = … … … … … ( ) = … … … … … ( ) = … … … … … ( ) = … … … … … ( ) From (2) we have: = … … … … … . ( ) From (1) & (5), = = = and = Now from (3), we have: = ⇒ = Now from (4), we get: = = Hence, = = = = 8. = is a square matrix, if (A) m<n (B) m>n (C) m=n (D) None of these Solution 8: Correct Option is (C). To be a Square Matrix m should be equal to n. i.e No. of rows should be equal to No. of Columns. 9. Which of the given values of x and y make the following pair of matrices equal [ ] = [ ]
  8. 8. Chapter - 3 MATRICES (Ex – 3.1) PCM ENCYCLOPEDIA (www.pcmpedia.blogspot.in) 8 (A) = = (B) Not Possible to find (C) = = (D) = = Solution 9: Correct Option is (B). Given that [ ] = [ ] Equating the Corresponding element of the given Matrix. 3 = = and 5 = = 7 = = 7 and = = On Comparing the Corresponding elements of the two matrices, we get two different values of , which is not possible. Hence, it is not possible to find the values of and y for which the given matrices are equal. 10.The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is: (A)27 (B) 18 (C) 81 (D) 512 Solution 10: Correct Option is (D). The given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1. Now, each of the 9 elements can be filled in two possible ways. Therefore, by the multiplication principle, the required number of possible matrices is 29 = 512.

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