{"id":18364,"date":"2020-01-07T11:08:37","date_gmt":"2020-01-07T05:38:37","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=18364"},"modified":"2021-07-02T15:30:44","modified_gmt":"2021-07-02T10:00:44","slug":"2nd-puc-maths-question-bank-chapter-3-ex-3-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-3-ex-3-3\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 3 Matrices Ex 3.3"},"content":{"rendered":"

Students can Download Maths Chapter 3 Matrices Ex 3.3 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Maths Question Bank Chapter 3 Matrices Ex 3.3<\/h2>\n

2nd PUC Maths Matrices NCERT Text Book Questions and Answers
\nEx 3.3<\/h3>\n

Question 1.
\n(i)
\n\\(\\left[\\begin{array}{c}{5} \\\\{\\frac{1}{2}} \\\\{-1}\\end{array}\\right]\\)
\n(ii)
\n\\(\\left[\\begin{array}{cc}{1} & {-1} \\\\{2} & {3}\\end{array}\\right]\\)
\n(iii)
\n\\(\\left[\\begin{array}{ccc}{-1} & {5} & {6} \\\\{\\sqrt{3}} & {5} & {6} \\\\{2} & {3} & {-1}\\end{array}\\right]\\)
\nAnswer:
\nQuestion 2.
\n\\(\\mathbf{A}=\\left[\\begin{array}{ccc}{-1} & {2} & {3} \\\\{5} & {7} & {9} \\\\{-2} & {1} & {1}\\end{array}\\right] \\text { and } \\mathbf{B}=\\left[\\begin{array}{ccc}{-4} & {1} & {-5} \\\\{1} & {2} & {0} \\\\{1} & {3} & {1}\\end{array}\\right]\\)
\nthen verify that
\n(i) (A + B)’ = A’ + B’,
\n(ii) (A – B)’ = A’ – B’
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\n\\(\\mathbf{A}^{\\prime}=\\left[\\begin{array}{rr}{3} & {4} \\\\{-1} & {2} \\\\{0} & {1}\\end{array}\\right] \\text { and } \\mathbf{B}=\\left[\\begin{array}{rrr}{-1} & {2} & {1} \\\\{1} & {2} & {3}\\end{array}\\right]\\)
\n(i) (A + B)’ = A’ + B’,
\n(ii) (A – B)’ = A’ – B’
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 4.
\nIf \\(A^{\\prime}=\\left[\\begin{array}{cc}{-2} & {3} \\\\{1} & {2}\\end{array}\\right] \\text { and } B=\\left[\\begin{array}{cc}{-1} & {0} \\\\{1} & {2}\\end{array}\\right]\\)
\nthen Find (A + 2B)’
\nAnswer:
\n\"2nd<\/p>\n

Question 5.
\nFor the matrices A and B, verify that (AB)’ = B’A’, where
\n(i)
\n\\(A=\\left[\\begin{array}{r}{1} \\\\{-4} \\\\{3}\\end{array}\\right], B=\\left[\\begin{array}{lll}{-1} & {2} & {1}\\end{array}\\right]\\)
\n(ii)
\n\\(\\mathbf{A}=\\left[\\begin{array}{l}{\\mathbf{0}} \\\\{\\mathbf{1}} {\\mathbf{2}}\\end{array}\\right], \\mathbf{B}=\\left[\\begin{array}{lll}{\\mathbf{1}} &{\\mathbf{5}} & {\\mathbf{7}}\\end{array}\\right]\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nIf (i)
\n\\(\\mathbf{A}=\\left[\\begin{array}{cc}{\\cos \\alpha} & {\\sin \\alpha} \\\\{-\\sin \\alpha} & {\\cos \\alpha}\\end{array}\\right]\\),then verify that A’ A = 1
\n(ii)
\n\\(\\mathbf{A}=\\left[\\begin{array}{cc}{\\sin \\alpha} & {\\cos \\alpha} \\\\{-\\cos \\alpha} & {\\sin \\alpha}\\end{array}\\right]\\),then verify that A’ A = 1
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 7.
\n(i) Show that the matrix \\(\\mathbf{A}=\\left[\\begin{array}{rrr}{\\mathbf{1}} & {\\mathbf{1}} & {\\mathbf{5}} \\\\{-\\mathbf{1}} & {\\mathbf{2}} & {\\mathbf{1}} \\\\{\\mathbf{5}} & {\\mathbf{1}} & {\\mathbf{3}}\\end{array}\\right]\\) is a symmentric matrix.
\n(ii) Show that the matrix
\n\\(\\mathbf{A}=\\left[\\begin{array}{rrr}{0} & {1} & {-1} \\\\{-1} & {0} & {1} \\\\{1} & {-1} & {0}\\end{array}\\right]\\)
\nis a skew symmentric matrix.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nFor the matrix \\(\\mathbf{A}=\\left[\\begin{array}{ll}{\\mathbf{1}} & {\\mathbf{5}} \\\\{\\mathbf{6}} & {7}\\end{array}\\right]\\) verify that.
\n(i) ( A + A’) is a symmetric matrix
\n(ii)(A – A’) is a skew symmetric matrix
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 9.
\nFind \\(\\frac{1}{2}\\left(A+A^{\\prime}\\right) \\text { and } \\frac{1}{2}(A-A)\\),when
\n\\(\\mathbf{A}=\\left[\\begin{array}{rrr}{0} & {\\mathbf{a}} & {\\mathbf{b}} \\\\{-\\mathbf{a}} & {\\mathbf{0}} & {\\mathbf{c}} \\\\{-\\mathbf{b}} & {-\\mathbf{c}} &{\\mathbf{0}}\\end{array}\\right]\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 10.
\nExpress the following matrices as the sum of a symmetric and a skew symmetric matrix:
\n(i)
\n\\(\\left[\\begin{array}{cc}{3} & {5} \\\\{1} & {-1}\\end{array}\\right]\\)
\n(ii)
\n\\(\\left[\\begin{array}{rrr}{6} & {-2} & {2} \\\\{-2} & {3} & {-1} \\\\{2} & {-1} & {3}\\end{array}\\right]\\)
\n(iii)
\n\\(\\left[\\begin{array}{rrr}{3} & {3} & {-1} \\\\{-2} & {-2} & {1} \\\\{-4} & {-5} & {2}\\end{array}\\right]\\)
\n(iv)
\n\\(\\left[\\begin{array}{rr}{1} & {5} \\\\{-1} & {2}\\end{array}\\right]\\)
\nAnswer:
\n\"2nd
\n\"2nd
\n\"2nd
\n\"2nd
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Choose the correct answer in the Exercise 11 and 12.<\/p>\n

Question 11.
\nIf A, B are symmetric matrices of same order, then AB – BA is a
\n(A) Skew symmetric matrix
\n(B) Symmetric matrix
\n(C) Zero matrix
\n(D) Identity matrix
\nAns:
\nA & B are symmetric matrices A’ = A B’ = B
\n\"2nd<\/p>\n

Question 12.
\n\\(\\mathbf{A}=\\left[\\begin{array}{cc}{\\cos \\alpha} & {-\\sin \\alpha} \\\\{\\sin \\alpha} & {\\cos \\alpha}\\end{array}\\right]\\),then A + A’ = I,, if the value of \u03b1 is
\n(A) \\(\\frac{\\pi}{6}\\)
\n(B) \\(\\frac{\\pi}{3}\\)
\n(C) \u03c0
\n(D) \\(\\frac{3 \\pi}{2}\\)
\nAnswer:
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 3 Matrices Ex 3.3 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Maths Question Bank Chapter 3 Matrices Ex 3.3 2nd PUC Maths Matrices NCERT Text Book …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/18364"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=18364"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/18364\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=18364"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=18364"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=18364"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}