{"id":69013,"date":"2023-09-11T12:37:00","date_gmt":"2023-09-11T07:07:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=69013"},"modified":"2023-11-08T10:28:45","modified_gmt":"2023-11-08T04:58:45","slug":"rd-sharma-solutions-class-12-maths-chapter-6-exercise-6-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-6-exercise-6-1\/","title":{"rendered":"RD Sharma Class 12 Solutions Chapter 6 Determinants Exercise 6.1 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-119879\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-6-Exercise-6.1.png\" alt=\"RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-6-Exercise-6.1.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-6-Exercise-6.1-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1:\u00a0<\/strong>The key topics covered in exercise 6.1 are the determinant of a square matrix of order 1, 2, 3, and 4, singular matrix, minors, and cofactors of specified determinants. Students can use this exercise as a model of reference to strengthen their conceptual knowledge and comprehend the various methods utilized to solve problems. <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-6\/\"><strong>RD Sharma Solutions Class 12 Maths Chapter 6<\/strong><\/a> Exercise 6.1 is available in PDF format. The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-12-solutions\/\"><strong>RD Sharma Class 12 Solutions<\/strong><\/a> were designed by Kopykitab&#8217;s Math experts in order to boost students&#8217; confidence, which is important in board exams.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d0113abd5bc\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-6-exercise-6-1\/#download-rd-sharma-solutions-class-12-maths-chapter-6-exercise-61-free-pdf\" title=\"Download RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1 Free PDF\">Download RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1 Free PDF<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-6-exercise-6-1\/#access-answers-to-maths-rd-sharma-solutions-for-class-12-chapter-6-%e2%80%93-determinants-exercise-61-important-questions-with-solution\" title=\"Access answers to Maths RD Sharma Solutions For Class 12 Chapter 6 \u2013 Determinants Exercise 6.1 Important Questions With Solution\">Access answers to Maths RD Sharma Solutions For Class 12 Chapter 6 \u2013 Determinants Exercise 6.1 Important Questions With Solution<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-6-exercise-6-1\/#faqs-on-rd-sharma-solutions-class-12-maths-chapter-6-exercise-61\" title=\"FAQs on RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1\">FAQs on RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-6-exercise-6-1\/#what-are-the-advantages-of-using-rd-sharma-solutions-class-12-maths-chapter-6-exercise-61\" title=\"What are the advantages of using RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1?\">What are the advantages of using RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-6-exercise-6-1\/#where-can-i-get-rd-sharma-solutions-class-12-maths-chapter-6-exercise-61-free-pdf\" title=\"Where can I get RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1 Free PDF?\">Where can I get RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1 Free PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-6-exercise-6-1\/#how-many-questions-are-there-in-rd-sharma-solutions-class-12-maths-chapter-6-exercise-61\" title=\"How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1?\">How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-12-maths-chapter-6-exercise-61-free-pdf\"><\/span>Download RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Maths-Chapter-6-Ex-6.1-1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Maths-Chapter-6-Ex-6.1-1.pdf\">RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-maths-rd-sharma-solutions-for-class-12-chapter-6-%e2%80%93-determinants-exercise-61-important-questions-with-solution\"><\/span>Access answers to Maths RD Sharma Solutions For Class 12 Chapter 6 \u2013 Determinants Exercise 6.1 Important Questions With Solution<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:<\/strong><\/p>\n<p><img class=\"alignnone  wp-image-119853\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/1.png\" alt=\"1\" width=\"299\" height=\"505\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Let M<sub>ij<\/sub>\u00a0and C<sub>ij<\/sub> represent the minor and cofactor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then find the absolute value of the matrix newly formed.<\/p>\n<p>Also, C<sub>ij<\/sub>\u00a0= (\u20131)<sup>i+j<\/sup>\u00a0\u00d7 M<sub>ij<\/sub><\/p>\n<p>Given,<\/p>\n<p><img class=\"alignnone size-full wp-image-119854\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/1a.png\" alt=\"1a\" width=\"103\" height=\"40\" \/><\/p>\n<p>From the given matrix we have,<\/p>\n<p>M<sub>11<\/sub>\u00a0= \u20131<\/p>\n<p>M<sub>21<\/sub>\u00a0= 20<\/p>\n<p>C<sub>11<\/sub>\u00a0= (\u20131)<sup>1+1<\/sup>\u00a0\u00d7 M<sub>11<\/sub><\/p>\n<p>= 1 \u00d7 \u20131<\/p>\n<p>= \u20131<\/p>\n<p>C<sub>21<\/sub>\u00a0= (\u20131)<sup>2+1<\/sup>\u00a0\u00d7 M<sub>21<\/sub><\/p>\n<p>= 20 \u00d7 \u20131<\/p>\n<p>= \u201320<\/p>\n<p>Now expanding along the first column we get<\/p>\n<p>|A| = a<sub>11<\/sub>\u00a0\u00d7 C<sub>11<\/sub>\u00a0+ a<sub>21<\/sub>\u00d7 C<sub>21<\/sub><\/p>\n<p>= 5\u00d7 (\u20131) + 0 \u00d7 (\u201320)<\/p>\n<p>= \u20135<\/p>\n<p>(ii) Let M<sub>ij<\/sub>\u00a0and C<sub>ij<\/sub> represent the minor and co\u2013factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n<p>Also, C<sub>ij<\/sub>\u00a0= (\u20131)<sup>i+j<\/sup>\u00a0\u00d7 M<sub>ij<\/sub><\/p>\n<p>Given<\/p>\n<p><img class=\"alignnone size-full wp-image-119855\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/1b.png\" alt=\"1b\" width=\"103\" height=\"39\" \/><\/p>\n<p>From the above matrix we have<\/p>\n<p>M<sub>11<\/sub>\u00a0= 3<\/p>\n<p>M<sub>21<\/sub>\u00a0= 4<\/p>\n<p>C<sub>11<\/sub>\u00a0= (\u20131)<sup>1+1<\/sup>\u00a0\u00d7 M<sub>11<\/sub><\/p>\n<p>= 1 \u00d7 3<\/p>\n<p>= 3<\/p>\n<p>C<sub>21<\/sub>\u00a0= (\u20131)<sup>2+1<\/sup>\u00a0\u00d7 4<\/p>\n<p>= \u20131 \u00d7 4<\/p>\n<p>= \u20134<\/p>\n<p>Now expanding along the first column we get<\/p>\n<p>|A| = a<sub>11<\/sub>\u00a0\u00d7 C<sub>11<\/sub>\u00a0+ a<sub>21<\/sub>\u00d7 C<sub>21<\/sub><\/p>\n<p>= \u20131\u00d7 3 + 2 \u00d7 (\u20134)<\/p>\n<p>= \u201311<\/p>\n<p>(iii) Let M<sub>ij<\/sub>\u00a0and C<sub>ij<\/sub>\u00a0represents the minor and co\u2013factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n<p>Also, C<sub>ij<\/sub>\u00a0= (\u20131)<sup>i+j<\/sup>\u00a0\u00d7 M<sub>ij<\/sub><\/p>\n<p>Given,<\/p>\n<p><img class=\"alignnone size-full wp-image-119856\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/1c.png\" alt=\"1c\" width=\"222\" height=\"399\" \/><\/p>\n<p>M<sub>31<\/sub>\u00a0= \u20133 \u00d7 2 \u2013 (\u20131) \u00d7 2<\/p>\n<p>M<sub>31<\/sub>\u00a0= \u20134<\/p>\n<p>C<sub>11<\/sub>\u00a0= (\u20131)<sup>1+1<\/sup>\u00a0\u00d7 M<sub>11<\/sub><\/p>\n<p>= 1 \u00d7 \u201312<\/p>\n<p>= \u201312<\/p>\n<p>C<sub>21<\/sub>\u00a0= (\u20131)<sup>2+1<\/sup>\u00a0\u00d7 M<sub>21<\/sub><\/p>\n<p>= \u20131 \u00d7 \u201316<\/p>\n<p>= 16<\/p>\n<p>C<sub>31<\/sub>\u00a0= (\u20131)<sup>3+1<\/sup>\u00a0\u00d7 M<sub>31<\/sub><\/p>\n<p>= 1 \u00d7 \u20134<\/p>\n<p>= \u20134<\/p>\n<p>Now expanding along the first column we get<\/p>\n<p>|A| = a<sub>11<\/sub>\u00a0\u00d7 C<sub>11<\/sub>\u00a0+ a<sub>21<\/sub>\u00d7 C<sub>21<\/sub>+ a<sub>31<\/sub>\u00d7 C<sub>31<\/sub><\/p>\n<p>= 1\u00d7 (\u201312) + 4 \u00d7 16 + 3\u00d7 (\u20134)<\/p>\n<p>= \u201312 + 64 \u201312<\/p>\n<p>= 40<\/p>\n<p>(iv) Let M<sub>ij<\/sub>\u00a0and C<sub>ij<\/sub>\u00a0represents the minor and co\u2013factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n<p>Also, C<sub>ij<\/sub>\u00a0= (\u20131)<sup>i+j<\/sup>\u00a0\u00d7 M<sub>ij<\/sub><\/p>\n<p>Given,<\/p>\n<p><img class=\"alignnone size-full wp-image-119857\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/1d.png\" alt=\"1d\" width=\"168\" height=\"362\" \/><\/p>\n<p>M<sub>31<\/sub>\u00a0= a \u00d7 c a \u2013 b \u00d7 bc<\/p>\n<p>M<sub>31<\/sub>\u00a0= a<sup>2<\/sup>c \u2013 b<sup>2<\/sup>c<\/p>\n<p>C<sub>11<\/sub>\u00a0= (\u20131)<sup>1+1<\/sup>\u00a0\u00d7 M<sub>11<\/sub><\/p>\n<p>= 1 \u00d7 (ab<sup>2<\/sup>\u00a0\u2013 ac<sup>2<\/sup>)<\/p>\n<p>= ab<sup>2<\/sup>\u00a0\u2013 ac<sup>2<\/sup><\/p>\n<p>C<sub>21<\/sub>\u00a0= (\u20131)<sup>2+1<\/sup>\u00a0\u00d7 M<sub>21<\/sub><\/p>\n<p>= \u20131 \u00d7 (a<sup>2<\/sup>b \u2013 c<sup>2<\/sup>b)<\/p>\n<p>= c<sup>2<\/sup>b \u2013 a<sup>2<\/sup>b<\/p>\n<p>C<sub>31<\/sub>\u00a0= (\u20131)<sup>3+1<\/sup>\u00a0\u00d7 M<sub>31<\/sub><\/p>\n<p>= 1 \u00d7 (a<sup>2<\/sup>c \u2013 b<sup>2<\/sup>c)<\/p>\n<p>= a<sup>2<\/sup>c \u2013 b<sup>2<\/sup>c<\/p>\n<p>Now expanding along the first column we get<\/p>\n<p>|A| = a<sub>11<\/sub>\u00a0\u00d7 C<sub>11<\/sub>\u00a0+ a<sub>21<\/sub>\u00d7 C<sub>21<\/sub>+ a<sub>31<\/sub>\u00d7 C<sub>31<\/sub><\/p>\n<p>= 1\u00d7 (ab<sup>2<\/sup>\u00a0\u2013 ac<sup>2<\/sup>) + 1 \u00d7 (c<sup>2<\/sup>b \u2013 a<sup>2<\/sup>b) + 1\u00d7 (a<sup>2<\/sup>c \u2013 b<sup>2<\/sup>c)<\/p>\n<p>= ab<sup>2<\/sup>\u00a0\u2013 ac<sup>2<\/sup>\u00a0+ c<sup>2<\/sup>b \u2013 a<sup>2<\/sup>b + a<sup>2<\/sup>c \u2013 b<sup>2<\/sup>c<\/p>\n<p>(v) Let M<sub>ij<\/sub>\u00a0and C<sub>ij<\/sub>\u00a0represents the minor and co\u2013factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n<p>Also, C<sub>ij<\/sub>\u00a0= (\u20131)<sup>i+j<\/sup>\u00a0\u00d7 M<sub>ij<\/sub><\/p>\n<p>Given,<\/p>\n<p><img class=\"alignnone size-full wp-image-119858\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/1e.png\" alt=\"1e\" width=\"254\" height=\"397\" \/><\/p>\n<p>M<sub>31<\/sub>\u00a0= 2\u00d70 \u2013 5\u00d76<\/p>\n<p>M<sub>31<\/sub>\u00a0= \u201330<\/p>\n<p>C<sub>11<\/sub>\u00a0= (\u20131)<sup>1+1<\/sup>\u00a0\u00d7 M<sub>11<\/sub><\/p>\n<p>= 1 \u00d7 5<\/p>\n<p>= 5<\/p>\n<p>C<sub>21<\/sub>\u00a0= (\u20131)<sup>2+1<\/sup>\u00a0\u00d7 M<sub>21<\/sub><\/p>\n<p>= \u20131 \u00d7 \u201340<\/p>\n<p>= 40<\/p>\n<p>C<sub>31<\/sub>\u00a0= (\u20131)<sup>3+1<\/sup>\u00a0\u00d7 M<sub>31<\/sub><\/p>\n<p>= 1 \u00d7 \u201330<\/p>\n<p>= \u201330<\/p>\n<p>Now expanding along the first column we get<\/p>\n<p>|A| = a<sub>11<\/sub>\u00a0\u00d7 C<sub>11<\/sub>\u00a0+ a<sub>21<\/sub>\u00d7 C<sub>21<\/sub>+ a<sub>31<\/sub>\u00d7 C<sub>31<\/sub><\/p>\n<p>= 0\u00d7 5 + 1 \u00d7 40 + 3\u00d7 (\u201330)<\/p>\n<p>= 0 + 40 \u2013 90<\/p>\n<p>= 50<\/p>\n<p>(vi) Let M<sub>ij<\/sub>\u00a0and C<sub>ij<\/sub>\u00a0represents the minor and co\u2013factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n<p>Also, C<sub>ij<\/sub>\u00a0= (\u20131)<sup>i+j<\/sup>\u00a0\u00d7 M<sub>ij<\/sub><\/p>\n<p>Given,<\/p>\n<p><img class=\"alignnone size-full wp-image-119859\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/1f.png\" alt=\"1f\" width=\"272\" height=\"409\" \/><\/p>\n<p>M<sub>31<\/sub>\u00a0= h \u00d7 f \u2013 b \u00d7 g<\/p>\n<p>M<sub>31<\/sub>\u00a0= hf \u2013 bg<\/p>\n<p>C<sub>11<\/sub>\u00a0= (\u20131)<sup>1+1<\/sup>\u00a0\u00d7 M<sub>11<\/sub><\/p>\n<p>= 1 \u00d7 (bc\u2013 f<sup>2<\/sup>)<\/p>\n<p>= bc\u2013 f<sup>2<\/sup><\/p>\n<p>C<sub>21<\/sub>\u00a0= (\u20131)<sup>2+1<\/sup>\u00a0\u00d7 M<sub>21<\/sub><\/p>\n<p>= \u20131 \u00d7 (hc \u2013 fg)<\/p>\n<p>= fg \u2013 hc<\/p>\n<p>C<sub>31<\/sub>\u00a0= (\u20131)<sup>3+1<\/sup>\u00a0\u00d7 M<sub>31<\/sub><\/p>\n<p>= 1 \u00d7 (hf \u2013 bg)<\/p>\n<p>= hf \u2013 bg<\/p>\n<p>Now expanding along the first column we get<\/p>\n<p>|A| = a<sub>11<\/sub>\u00a0\u00d7 C<sub>11<\/sub>\u00a0+ a<sub>21<\/sub>\u00d7 C<sub>21<\/sub>+ a<sub>31<\/sub>\u00d7 C<sub>31<\/sub><\/p>\n<p>= a\u00d7 (bc\u2013 f<sup>2<\/sup>) + h\u00d7 (fg \u2013 hc) + g\u00d7 (hf \u2013 bg)<\/p>\n<p>= abc\u2013 af<sup>2<\/sup>\u00a0+ hgf \u2013 h<sup>2<\/sup>c +ghf \u2013 bg<sup>2<\/sup><\/p>\n<p>(vii) Let M<sub>ij<\/sub>\u00a0and C<sub>ij<\/sub>\u00a0represents the minor and co\u2013factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n<p>Also, C<sub>ij<\/sub>\u00a0= (\u20131)<sup>i+j<\/sup>\u00a0\u00d7 M<sub>ij<\/sub><\/p>\n<p>Given,<\/p>\n<p><img class=\"alignnone size-full wp-image-119860\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/1g.png\" alt=\"1g\" width=\"533\" height=\"486\" \/><\/p>\n<p>M<sub>31<\/sub>\u00a0= \u20131(1 \u00d7 0 \u2013 5 \u00d7 (\u20132)) \u2013 0(0 \u00d7 0 \u2013 (\u20131) \u00d7 (\u20132)) + 1(0 \u00d7 5 \u2013 (\u20131) \u00d7 1)<\/p>\n<p>M<sub>31<\/sub>\u00a0= \u20139<\/p>\n<p><img class=\"alignnone size-full wp-image-119861\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/2a.png\" alt=\"2a\" width=\"199\" height=\"60\" \/><\/p>\n<p>M<sub>41<\/sub>\u00a0= \u20131(1\u00d71 \u2013 (\u20131) \u00d7 (\u20132)) \u2013 0(0 \u00d7 1 \u2013 1 \u00d7 (\u20132)) + 1(0 \u00d7 (\u20131) \u2013 1 \u00d7 1)<\/p>\n<p>M<sub>41<\/sub>\u00a0= 0<\/p>\n<p>C<sub>11<\/sub>\u00a0= (\u20131)<sup>1+1<\/sup>\u00a0\u00d7 M<sub>11<\/sub><\/p>\n<p>= 1 \u00d7 (\u20139)<\/p>\n<p>= \u20139<\/p>\n<p>C<sub>21<\/sub>\u00a0= (\u20131)<sup>2+1<\/sup>\u00a0\u00d7 M<sub>21<\/sub><\/p>\n<p>= \u20131 \u00d7 9<\/p>\n<p>= \u20139<\/p>\n<p>C<sub>31<\/sub>\u00a0= (\u20131)<sup>3+1<\/sup>\u00a0\u00d7 M<sub>31<\/sub><\/p>\n<p>= 1 \u00d7 \u20139<\/p>\n<p>= \u20139<\/p>\n<p>C<sub>41<\/sub>\u00a0= (\u20131)<sup>4+1<\/sup>\u00a0\u00d7 M<sub>41<\/sub><\/p>\n<p>= \u20131 \u00d7 0<\/p>\n<p>= 0<\/p>\n<p>Now expanding along the first column we get<\/p>\n<p>|A| = a<sub>11<\/sub>\u00a0\u00d7 C<sub>11<\/sub>\u00a0+ a<sub>21<\/sub>\u00d7 C<sub>21<\/sub>+ a<sub>31<\/sub>\u00d7 C<sub>31<\/sub>\u00a0+ a<sub>41<\/sub>\u00d7 C<sub>41<\/sub><\/p>\n<p>= 2 \u00d7 (\u20139) + (\u20133) \u00d7 \u20139 + 1 \u00d7 (\u20139) + 2 \u00d7 0<\/p>\n<p>= \u2013 18 + 27 \u20139<\/p>\n<p>= 0<\/p>\n<p><strong>2. Evaluate the following determinants:<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-119862\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/22a.png\" alt=\"22a\" width=\"1499\" height=\"413\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/22a.png 1499w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/22a-768x212.png 768w\" sizes=\"(max-width: 1499px) 100vw, 1499px\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>(i) Given<\/p>\n<p><img class=\"alignnone size-full wp-image-119863\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/2b.png\" alt=\"2b\" width=\"111\" height=\"52\" \/><\/p>\n<p>\u21d2\u00a0|A| = x (5x + 1) \u2013 (\u20137) x<\/p>\n<p>|A| = 5x<sup>2<\/sup>\u00a0+ 8x<\/p>\n<p>(ii) Given<\/p>\n<p><img class=\"alignnone size-full wp-image-119864\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/2c.png\" alt=\"2c\" width=\"140\" height=\"49\" \/><\/p>\n<p>\u21d2\u00a0|A| = cos \u03b8 \u00d7 cos \u03b8 \u2013 (\u2013sin \u03b8) x sin \u03b8<\/p>\n<p>|A| = cos<sup>2<\/sup>\u03b8 + sin<sup>2<\/sup>\u03b8<\/p>\n<p>We know that cos<sup>2<\/sup>\u03b8 + sin<sup>2<\/sup>\u03b8 = 1<\/p>\n<p>|A| = 1<\/p>\n<p>(iii) Given<\/p>\n<p><img class=\"alignnone size-full wp-image-119865\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/2d.gif\" alt=\"2d\" width=\"186\" height=\"46\" \/><\/p>\n<p>\u21d2\u00a0|A| = cos15\u00b0 \u00d7 cos75\u00b0 + sin15\u00b0 x sin75\u00b0<\/p>\n<p>We know that cos (A \u2013 B) = cos A cos B + Sin A sin B<\/p>\n<p>By substituting this we get, |A| = cos (75 \u2013 15)\u00b0<\/p>\n<p>|A| = cos60\u00b0<\/p>\n<p>|A| = 0.5<\/p>\n<p>(iv) Given<\/p>\n<p><img class=\"alignnone size-full wp-image-119866\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/2e.gif\" alt=\"2e\" width=\"186\" height=\"46\" \/><\/p>\n<p>\u21d2\u00a0|A| = (a + ib) (a \u2013 ib) \u2013 (c + id) (\u2013c + id)<\/p>\n<p>= (a + ib) (a \u2013 ib) + (c + id) (c \u2013 id)<\/p>\n<p>= a<sup>2<\/sup>\u00a0\u2013 i<sup>2<\/sup>\u00a0b<sup>2<\/sup>\u00a0+ c<sup>2<\/sup>\u00a0\u2013 i<sup>2<\/sup>\u00a0d<sup>2<\/sup><\/p>\n<p>We know that i<sup>2<\/sup>\u00a0= -1<\/p>\n<p>= a<sup>2<\/sup>\u00a0\u2013 (\u20131) b<sup>2<\/sup>\u00a0+ c<sup>2<\/sup>\u00a0\u2013 (\u20131) d<sup>2<\/sup><\/p>\n<p>= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0+ c<sup>2<\/sup>\u00a0+ d<sup>2<\/sup><\/p>\n<p><strong>3. Evaluate:<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-119867\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/3-2.gif\" alt=\"3\" width=\"109\" height=\"71\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Since |AB|= |A||B|<\/p>\n<p><img class=\"alignnone size-full wp-image-119868\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/3a.png\" alt=\"3a\" width=\"341\" height=\"102\" \/><\/p>\n<p>= 2(17 \u00d7 12 \u2013 5 \u00d7 20) \u2013 3(13 \u00d7 12 \u2013 5 \u00d7 15) + 7(13 \u00d7 20 \u2013 15 \u00d7 17)<\/p>\n<p>= 2 (204 \u2013 100) \u2013 3 (156 \u2013 75) + 7 (260 \u2013 255)<\/p>\n<p>= 2\u00d7104 \u2013 3\u00d781 + 7\u00d75<\/p>\n<p>= 208 \u2013 243 +35<\/p>\n<p>= 0<\/p>\n<p>Now |A|<sup>2<\/sup>\u00a0= |A|\u00d7|A|<\/p>\n<p>|A|<sup>2<\/sup>= 0<\/p>\n<p><strong>4. Show that<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-119869\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/4-2.gif\" alt=\"4\" width=\"150\" height=\"46\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given<\/p>\n<p><img class=\"alignnone size-full wp-image-119869\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/4-2.gif\" alt=\"4\" width=\"150\" height=\"46\" \/><\/p>\n<p>Let the given determinant as A<\/p>\n<p>Using sin (A+B) = sin A \u00d7 cos B + cos A \u00d7 sin B<\/p>\n<p>\u21d2\u00a0|A| = sin 10\u00b0 \u00d7 cos 80\u00b0 + cos 10\u00b0 x sin 80\u00b0<\/p>\n<p>|A| = sin (10 + 80)\u00b0<\/p>\n<p>|A| = sin90\u00b0<\/p>\n<p>|A| = 1<\/p>\n<p>Hence Proved<\/p>\n<p><img class=\"alignnone  wp-image-119870\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/5.png\" alt=\"5\" width=\"386\" height=\"288\" \/><\/p>\n<p>= 2(1 \u00d7 1 \u2013 4 \u00d7 (\u20132)) \u2013 3(7 \u00d7 1 \u2013 (\u20132) \u00d7 (\u20133)) \u2013 5(7 \u00d7 4 \u2013 1 \u00d7 (\u20133))<\/p>\n<p>= 2(1 + 8) \u2013 3(7 \u2013 6) \u2013 5(28 + 3)<\/p>\n<p>= 2 \u00d7 9 \u2013 3 \u00d7 1 \u2013 5 \u00d7 31<\/p>\n<p>= 18 \u2013 3 \u2013 155<\/p>\n<p>= \u2013140<\/p>\n<p>Now by expanding along the second column<\/p>\n<p><img class=\"alignnone size-full wp-image-119871\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/5a.png\" alt=\"5a\" width=\"317\" height=\"39\" \/><\/p>\n<p>= 2(1 \u00d7 1 \u2013 4 \u00d7 (\u20132)) \u2013 7(3 \u00d7 1 \u2013 4 \u00d7 (\u20135)) \u2013 3(3 \u00d7 (\u20132) \u2013 1 \u00d7 (\u20135))<\/p>\n<p>= 2 (1 + 8) \u2013 7 (3 + 20) \u2013 3 (\u20136 + 5)<\/p>\n<p>= 2 \u00d7 9 \u2013 7 \u00d7 23 \u2013 3 \u00d7 (\u20131)<\/p>\n<p>= 18 \u2013 161 +3<\/p>\n<p>= \u2013140<\/p>\n<p><img class=\"alignnone size-full wp-image-119872\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/6-1.gif\" alt=\"6\" width=\"397\" height=\"68\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given<\/p>\n<p><img class=\"alignnone size-full wp-image-119873\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/6a.png\" alt=\"6a\" width=\"556\" height=\"163\" \/><\/p>\n<p>\u21d2\u00a0|A| = 0 (0 \u2013 sin\u03b2 (\u2013sin\u03b2)) \u2013sin\u03b1 (\u2013sin\u03b1 \u00d7 0 \u2013 sin\u03b2 cos\u03b1) \u2013 cos\u03b1 ((\u2013sin\u03b1) (\u2013sin\u03b2) \u2013 0 \u00d7 cos\u03b1)<\/p>\n<p>|A| = 0 + sin\u03b1 sin\u03b2 cos\u03b1 \u2013 cos\u03b1 sin\u03b1 sin\u03b2<\/p>\n<p>|A| = 0<\/p>\n<p>We have provided complete details of\u00a0RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 12 Exam, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-12-maths-chapter-6-exercise-61\"><\/span>FAQs on RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1629968026266\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-advantages-of-using-rd-sharma-solutions-class-12-maths-chapter-6-exercise-61\"><\/span>What are the advantages of using RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Referring to RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1 will provide students with a good understanding of the type of questions that might be asked in the Class 12 board exam from the chapter.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629968114804\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-solutions-class-12-maths-chapter-6-exercise-61-free-pdf\"><\/span>Where can I get RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629968182561\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-solutions-class-12-maths-chapter-6-exercise-61\"><\/span>How many questions are there in RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are a total of 12 questions in RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 12 Maths Chapter 6 Exercise 6.1:\u00a0The key topics covered in exercise 6.1 are the determinant of a square matrix of order 1, 2, 3, and 4, singular matrix, minors, and cofactors of specified determinants. Students can use this exercise as a model of reference to strengthen their conceptual knowledge and comprehend &#8230; <a title=\"RD Sharma Class 12 Solutions Chapter 6 Determinants Exercise 6.1 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-6-exercise-6-1\/\" aria-label=\"More on RD Sharma Class 12 Solutions Chapter 6 Determinants Exercise 6.1 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":119879,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3429,73223,61598,73664,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/69013"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=69013"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/69013\/revisions"}],"predecessor-version":[{"id":504093,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/69013\/revisions\/504093"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/119879"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=69013"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=69013"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=69013"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}