NCERT Solutions for Class 12 Maths<\/a> Chapter 4 Determinants Ex 4.6 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-4-ex-4-6\/<\/p>\nNCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.6<\/h2>\n <\/p>\n
Ex 4.6 Class 12 NCERT Solutions Question 1.<\/strong> \nx + 2y = 2 \n2x + 3y = 3 \nSolution: \nx + 2y = 2, \n2x + 3y = 3 \n\u21d2 \\(\\begin{bmatrix} 1 & 2 \\\\ 2 & 3 \\end{bmatrix}\\left[ \\begin{matrix} x \\\\ y \\end{matrix} \\right] =\\left[ \\begin{matrix} 2 \\\\ 3 \\end{matrix} \\right] \\) \n\u21d2 AX = B \nNow |A| = \\(\\begin{vmatrix} 1 & 2 \\\\ 2 & 3 \\end{vmatrix}\\) \n= 3 – 4 \n= – 1 \u2260 0. \nHence, equations are consistent.<\/p>\nExercise 4.6 Class 12 NCERT Solutions Question 2.<\/strong> \n2x – y = 5 \nx + y = 4 \nSolution: \n2x – y = 5, \nx + y = 4 \n\u21d2 \\(\\begin{bmatrix} 2 & -1 \\\\ 1 & 1 \\end{bmatrix}\\left[ \\begin{matrix} x \\\\ y \\end{matrix} \\right] =\\left[ \\begin{matrix} 5 \\\\ 4 \\end{matrix} \\right] \\) \n\u21d2 AX = B \nNow |A| = \\(\\begin{vmatrix} 2 & -1 \\\\ 1 & 1 \\end{vmatrix}\\) \n= 2 + 1 \n= 3 \u2260 0. \nHence, equations are consistent.<\/p>\nExercise 4.6 Maths Class 12 Question 3.<\/strong> \nx + 3y = 5, \n2x + 6y = 8 \nSolution: \nx + 3y = 5, \n2x + 6y = 8 \n \nHence solution does not exists and the system is inconsistent.<\/p>\n <\/p>\n
Exercise 4.6 Class 12 Maths Question 4.<\/strong> \nx + y + z = 1 \n2x + 3y + 2z = 2 \nax + ay + 2az = 4 \nSolution: \nThe solution can be written as AX = B where A = \\(\\left[\\begin{array}{ccc} \n1 & 1 & 1 \\\\ \n2 & 3 & 2 \\\\ \na & a & 2 a \n\\end{array}\\right]\\); X = \\(\\left[\\begin{array}{l} \nx \\\\ \ny \\\\ \nz \n\\end{array}\\right]\\); B = \\(\\left[\\begin{array}{l} \n1 \\\\ \n2 \\\\ \n4 \n\\end{array}\\right]\\) \n|A| = \\(\\left|\\begin{array}{ccc} \n1 & 1 & 1 \\\\ \n2 & 3 & 2 \\\\ \na & a & 2 a \n\\end{array}\\right|\\) \n= 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a) = 4a – 2a – a = a \u2260 0 \n\u2234 A is non singular and has a unique solution. \nHence the system is consistent (if a \u2260 0)<\/p>\n4.6 Class 12 NCERT Solutions Question 5.<\/strong> \n3x – y – 2z = 2 \n2y – z = – 1 \n3x – 5y = 3 \nSolution: \nThe system can be written as AX = B where A = \\(\\left[\\begin{array}{ccc} \n3 & -1 & -2 \\\\ \n0 & 2 & -1 \\\\ \n3 & -5 & 0 \n\\end{array}\\right]\\); X = \\(\\left[\\begin{array}{l} \nx \\\\ \ny \\\\ \nz \n\\end{array}\\right]\\); B = \\(\\left[\\begin{array}{l} \n2 \\\\ \n-1 \\\\ \n3 \n\\end{array}\\right]\\) \n \n\u2234 The solution does not exists and the system is inconsistent.<\/p>\n <\/p>\n
Class 12 Maths Ex 4.6 Solutions Question 6.<\/strong> \n5x – y + 4z = 5 \n2x + 3y + 5z = 2 \n5x – 2y + 6z = – 1 \nSolution: \nGiven \n5x – y + 4z = 5 \n2x + 3y + 5z = 2 \n5x – 2y + 6z = -1 \n\\(\\left[ \\begin{matrix} 5 & -1 & 4 \\\\ 2 & 3 & 5 \\\\ 5 & -2 & 6 \\end{matrix} \\right] \\left[ \\begin{matrix} x \\\\ y \\\\ z \\end{matrix} \\right] =\\left[ \\begin{matrix} 5 \\\\ 2 \\\\ -1 \\end{matrix} \\right] \\) \n\\(AX=B|A|=\\left[ \\begin{matrix} 5 & -1 & 4 \\\\ 2 & 3 & 5 \\\\ 5 & -2 & 6 \\end{matrix} \\right] \\) \n= 5(18 + 10)+1(12 – 25)+4(-4-15) \n= 140-13-76 \n= 51 \u2260 0 \n\u2234 Hence equations are consistent with a unique solution.<\/p>\nEx 4.6 Class 12 Maths NCERT Solutions Question 7.<\/strong> \n5x + 2y = 4 \n7x + 3y = 5 \nSolution: \nThe given system of equations can be written as \n <\/p>\nClass 12 Maths Chapter 4 Exercise 4.6 Solutions Question 8.<\/strong> \n2x – y = – 2 \n3x + 3y = 3 \nSolution: \nThe given system of equations can be written \n <\/p>\n <\/p>\n
Ex4.6 Class 12 NCERT Solutions Question 9.<\/strong> \n4x – 3y = 3 \n3x – 5y = 7 \nSolution: \n <\/p>\nEx 4.6 Class 12 Maths Ncert Solutions Question 10.<\/strong> \n5x + 2y = 3 \n3x + 2y = 5 \nSolution: \n <\/p>\n4.6 Maths Class 12 NCERT Solutions Question 11.<\/strong> \n2x + y + z = 1, \nx – 2y – z = 3\/2 \n3y – 5z = 9 \nSolution: \nThe system can be written as AX = B \n <\/p>\nExercise 4.6 Class 12 Maths Ncert Solutions Question 12.<\/strong> \nx – y + z = 4 \n2x + y – 3z = 0 \nx + y + z = 2. \nSolution: \nThe system can be written as AX = B where A = \\(\\left[\\begin{array}{ccc} \n1 & -1 & 1 \\\\ \n2 & 1 & -3 \\\\ \n1 & 1 & 1 \n\\end{array}\\right]\\); X = \\(\\left[\\begin{array}{l} \nx \\\\ \ny \\\\ \nz \n\\end{array}\\right]\\); B = \\(\\left[\\begin{array}{l} \n4 \\\\ \n0 \\\\ \n2 \n\\end{array}\\right]\\) \n|A| = \\(\\left|\\begin{array}{ccc} \n1 & -1 & 1 \\\\ \n2 & 1 & -3 \\\\ \n1 & 1 & 1 \n\\end{array}\\right|\\) = 1(1 + 3) +1(2 + 3) + 1(2 – 1) = 10 \u2260 0 \n\u2234 A is non singular and has a unique solution. \n <\/p>\n <\/p>\n
Class 12 Ex 4.6 NCERT Solutions Question 13.<\/strong> \n2x + 3y + 3z = 5 \nx – 2y + z = – 4 \n3x – y – 2z = 3 \nSolution: \nThe system can be written as AX = B where A = \\(\\left[\\begin{array}{ccc} \n2 & 3 & 3 \\\\ \n1 & -2 & 1 \\\\ \n3 & -1 & -2 \n\\end{array}\\right]\\); X = \\(\\left[\\begin{array}{l} \nx \\\\ \ny \\\\ \nz \n\\end{array}\\right]\\); B = \\(\\left[\\begin{array}{l} \n5 \\\\ \n– 4 \\\\ \n3 \n\\end{array}\\right]\\) \n|A| = \\(\\left|\\begin{array}{ccc} \n2 & 3 & 3 \\\\ \n1 & -2 & 1 \\\\ \n3 & -1 & -2 \n\\end{array}\\right|\\) = 2(5) – 3(- 5) + 2(5) = 40 \u2260 0 \n\u2234 A is non singular and has a unique solution. \n <\/p>\nNcert Solutions For Class 12 Maths Chapter 4 Exercise 4.6 Question 14.<\/strong> \nx – y + 2z = 7 \n3x + 4y – 5z = – 5 \n2x – y + 3z = 12. \nSolution: \nThe system can be written as AX = B where A = \\(\\left[\\begin{array}{ccc} \n1 & -1 & 2 \\\\ \n3 & 4 & -5 \\\\ \n2 & -1 & 3 \n\\end{array}\\right]\\); X = \\(\\left[\\begin{array}{l} \nx \\\\ \ny \\\\ \nz \n\\end{array}\\right]\\); B = \\(\\left[\\begin{array}{l} \n7 \\\\ \n– 5 \\\\ \n12 \n\\end{array}\\right]\\) \n|A| = \\(\\left|\\begin{array}{ccc} \n1 & -1 & 2 \\\\ \n3 & 4 & -5 \\\\ \n2 & -1 & 3 \n\\end{array}\\right|\\) = 1(7) + 1(19) + 2(-11) = 4 \u2260 0 \n\u2234 A is non singular and has a unique solution. \n <\/p>\n <\/p>\n
Class 12 Maths Ex 4.6 NCERT Solutions Question 15.<\/strong> \nIf A = \\(\\left[ \\begin{matrix} 2 & -3 & 5 \\\\ 3 & 2 & -4 \\\\ 1 & 1 & -2 \\end{matrix} \\right] \\) Find A-1<\/sup>. Using A-1<\/sup>. Solve the following system of linear equations 2x – 3y + 5z = 11, 3x + 2y – 4z = – 5, x + y – 2z = – 3 \nSolution: \n <\/p>\nSolution Of Exercise 4.6 Class 12 Question 16.<\/strong> \nThe cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 69. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method \nSolution: \nLet cost of 1 kg onion = Rs x \nand cost of 1 kg wheat = Rs y \nand cost of 1 kg rice = Rs z \n4x + 3y + 2z = 60 \n2x + 4y + 6z = 90 \n6x + 2y + 3z = 70 \n \ni.e., x = 5; y = 8; z = 8 \ni.e., Price of onion = \u20b9 5\/kg \nPrice of wheat = \u20b9 8\/kg \nPrice of rice = \u20b9 8\/kg<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-4-ex-4-6\/ NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.6 Ex 4.6 Class 12 NCERT Solutions Question 1. x + 2y = 2 2x + 3y = 3 Solution: x …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n