{"id":29804,"date":"2022-03-29T11:00:20","date_gmt":"2022-03-29T05:30:20","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=29804"},"modified":"2022-03-29T11:46:24","modified_gmt":"2022-03-29T06:16:24","slug":"ncert-solutions-for-class-12-maths-chapter-4-ex-4-6","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-4-ex-4-6\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6"},"content":{"rendered":"

These 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>\n

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.6<\/h2>\n

\"NCERT<\/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>\n

Exercise 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>\n

Exercise 4.6 Maths Class 12 Question 3.<\/strong>
\nx + 3y = 5,
\n2x + 6y = 8
\nSolution:
\nx + 3y = 5,
\n2x + 6y = 8
\n\"Ex
\nHence solution does not exists and the system is inconsistent.<\/p>\n

\"NCERT<\/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>\n

4.6 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\"Exercise
\n\u2234 The solution does not exists and the system is inconsistent.<\/p>\n

\"NCERT<\/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>\n

Ex 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\"Exercise<\/p>\n

Class 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\"Exercise<\/p>\n

\"NCERT<\/p>\n

Ex4.6 Class 12 NCERT Solutions Question 9.<\/strong>
\n4x – 3y = 3
\n3x – 5y = 7
\nSolution:
\n\"4.6<\/p>\n

Ex 4.6 Class 12 Maths Ncert Solutions Question 10.<\/strong>
\n5x + 2y = 3
\n3x + 2y = 5
\nSolution:
\n\"NCERT<\/p>\n

4.6 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\"Class<\/p>\n

Exercise 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\"Ex<\/p>\n

\"NCERT<\/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\"Class<\/p>\n

Ncert 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\"Ex4.6<\/p>\n

\"NCERT<\/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\"Ex<\/p>\n

Solution 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\"Class
\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<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-4-ex-4-6\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 - 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