{"id":20722,"date":"2020-01-22T15:40:43","date_gmt":"2020-01-22T10:10:43","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20722"},"modified":"2021-07-02T15:27:54","modified_gmt":"2021-07-02T09:57:54","slug":"2nd-puc-basic-maths-question-bank-chapter-1-ex-1-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-1-ex-1-1\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 1 Matrices and Determinants Ex 1.1"},"content":{"rendered":"

Students can Download Basic Maths Exercise 1.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Basic Maths Question Bank Chapter 1 Matrices and Determinants Ex 1.1<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Matrices and Determinants Ex 1.1 One Mark Questions and Answers<\/strong><\/p>\n

Question 1.
\nIf A = \\(\\left[ \\begin{matrix} 1 & -2 \\\\ 3 & 4 \\end{matrix} \\right]\\) . Find 2A and 3A
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\nIf A = \\(\\left[ \\begin{matrix} 1 & -3 & 5 \\\\ 6 & 2 & 4 \\end{matrix} \\right]\\) find 5A’.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nIf A = \\(\\left[ \\begin{matrix} 1 & 2 & 4 \\\\ -1 & 3 & -2 \\end{matrix} \\right]\\) and B = \\(\\left[ \\begin{matrix} 3 & -4 & -1 \\\\ 1 & 5 & -2 \\end{matrix} \\right]\\) .Find A+ B and A – B
\nAnswer:
\n\"2nd<\/p>\n

Question 4.
\nIf\u00a0 \\(\\left[ \\begin{matrix} x+y & 3 \\\\ 2 & -y+x \\end{matrix} \\right] +\\left[ \\begin{matrix} 2 & 3 \\\\ 4 & 1 \\end{matrix} \\right] =\\left[ \\begin{matrix} 0 & 6 \\\\ 6 & 0 \\end{matrix} \\right]\\) . Find x and y.
\nAnswer:
\n\"2nd<\/p>\n

Question 5.
\nIf\u00a0 \\(\\left[ \\begin{matrix} { x }^{ 2 } & 1 \\\\ 2 & -1 \\end{matrix} \\right] +\\left[ \\begin{matrix} 2x & 2 \\\\ -1 & 2 \\end{matrix} \\right] =\\left[ \\begin{matrix} -1 & 3 \\\\ 1 & 1 \\end{matrix} \\right]\\) Find x.
\nAnswer:
\n\"2nd
\n\u21d2 x2<\/sup> + 2x = -1
\nx2<\/sup> + 2x + 1 = 0
\n(x + 1)2<\/sup> = 0 \u21d2 x + 1 = 0
\n\u21d2 x = -1<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nIf A = \\(\\left[ \\begin{matrix} 4 & 0 & 0 \\\\ 0 & 4 & 0 \\\\ 0 & 0 & 4 \\end{matrix} \\right]\\) . Find A – A’
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 7.
\nIf A = \\(\\left[ \\begin{matrix} 2 & -x \\\\ x & -7 \\end{matrix} \\right]\\) . Find A + A’.
\nAnswer:
\n\"2nd<\/p>\n

Question 8.
\nIf A = \\(\\left[ \\begin{matrix} 1 & 2 \\\\ 3 & -1 \\end{matrix} \\right]\\) B = \\(\\left[ \\begin{matrix} 2 & 5 \\\\ -3 & -1 \\end{matrix} \\right]\\) C = \\(\\left[ \\begin{matrix} 1 & -1 \\\\ 4 & -3 \\end{matrix} \\right]\\) Find 2A – 3B – C
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nFind the matrix A. if 2A + B = \\(\\left[ \\begin{matrix} 2 & 0 \\\\ 1 & -3 \\end{matrix} \\right]\\) , where u = \\(\\left[ \\begin{matrix} 1 & -1 \\\\ 3 & 0 \\end{matrix} \\right]\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 10.
\nIf A = \\(\\left[ \\begin{matrix} 3 & -1 \\\\ 4 & 5 \\end{matrix} \\right]\\) , Find X such that A – 2x = \\(\\left[ \\begin{matrix} 1 & 4 \\\\ 2 & -3 \\end{matrix} \\right]\\) .
\nAnswer:
\n\"2nd<\/p>\n

Question 11.
\nIf A + B + C = 0 where A = \\(\\left[ \\begin{matrix} 3 & -2 \\\\ 2 & 0 \\\\ 1 & 4 \\end{matrix} \\right]\\) , B = \\(\\left[ \\begin{matrix} 2 & -1 \\\\ -4 & 2 \\\\ -3 & 3 \\end{matrix} \\right]\\) . Find C
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nIf A = B = \\(\\left[ \\begin{matrix} 3 & 1 \\\\ -3 & 4 \\end{matrix} \\right]\\) verify (A – B)’ = A’ – B’
\nAnswer:
\n\"2nd<\/p>\n

Part – B<\/span><\/p>\n

2nd PUC Basic Maths Matrices and Determinants Ex 1.1 Two Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nIf A = \\(\\left[ \\begin{matrix} 4 & -1 \\\\ 0 & 3 \\\\ 2 & -3 \\end{matrix} \\right]\\) . Find (i) \\(\\frac{5 A}{2} (ii) \\frac{-2 A}{3}\\) .
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nIf A = \\(\\left[ \\begin{matrix} 1 & 3 & -1 \\\\ -1 & 0 & 2 \\end{matrix} \\right]\\) and B = \\(\\left[ \\begin{matrix} 4 & -1 & 2 \\\\ 1 & 3 & -2 \\end{matrix} \\right]\\) . Find (i) 2A + 3B (ii) A – 3B (iii) A + \\(\\frac { 1 }{ 2 }\\)B.
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 3.
\nIf A = \\(\\left[ \\begin{matrix} 3 & 2 \\\\ 1 & 4 \\end{matrix} \\right]\\) B = \\(\\left[ \\begin{matrix} 1 & -1 \\\\ -2 & 3 \\end{matrix} \\right]\\) and C = \\(\\left[ \\begin{matrix} -3 & 4 \\\\ 2 & -1 \\end{matrix} \\right]\\) . Find 34A – 2B – 4C
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nIf A = \\(\\left[ \\begin{matrix} 1 & -3 \\\\ -4 & -1 \\end{matrix} \\right]\\) and B = \\(\\left[ \\begin{matrix} 3 & 4 \\\\ -5 & 1 \\end{matrix} \\right]\\) and 0 is a null matrix of order 2 \u00d7 2. Find the matrix C such that
\n(i) 2C = A + B
\n(ii) A + C = 0
\n(iii) B + 5c = A
\n(iv) 3A + 5B + 2C = 0
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 5.
\n\"2nd
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nFind x and y given that \\(\\left[ \\begin{matrix} -9 \\\\ 2 \\end{matrix} \\right] \\quad -\\quad \\left[ \\begin{matrix} 5 \\\\ -1 \\end{matrix} \\right] =\\quad \\left[ \\begin{matrix} x \\\\ y \\end{matrix} \\right]\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 7.
\nIf A = \\(\\left[ \\begin{matrix} 3 & 1 & 4 \\\\ 5 & 6 & 3x+1 \\end{matrix} \\right]\\) and B = \\(\\left[ \\begin{matrix} 3 & 5 \\\\ 1 & 6 \\\\ 4 & 3 \\end{matrix} \\right]\\) .
\nFind x given that A = B’
\nAnswer:
\n\"2nd<\/p>\n

Question 8.
\nSolve for x and y \\(x\\left[ \\begin{matrix} 2 \\\\ 1 \\end{matrix} \\right] +y\\left[ \\begin{matrix} 3 \\\\ 5 \\end{matrix} \\right] +\\left[ \\begin{matrix} 4 \\\\ 6 \\end{matrix} \\right] =\\left[ \\begin{matrix} 12 \\\\ 17 \\end{matrix} \\right]\\) .
\nAnswer:
\n\"2nd
\n\u21d2 2x + 3y + 4 = 12
\nx + 5y + 6 = 7
\n\u21d2 2x + 3y = 8 …. (1) solving these equations
\nx + 5y = 11 ….. (2) \u00d7 2
\n\"2nd
\nand x = 11 – 5y = 11 – 10 = 1
\n\u2234 x = 1, y = 2<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nFind x and y given that \\(\\left[ \\begin{matrix} x+y & 3 \\\\ -1 & x-y \\end{matrix} \\right] =\\left[ \\begin{matrix} 4 & 3 \\\\ -1 & 8 \\end{matrix} \\right]\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 10.
\nIf \\(\\left[ \\begin{matrix} 2 & 3 \\\\ 7 & 5 \\end{matrix} \\right] +\\left[ \\begin{matrix} 2 & x-2 \\\\ y-1 & 5 \\end{matrix} \\right] =\\left[ \\begin{matrix} 4 & 1 \\\\ 7 & 10 \\end{matrix} \\right]\\) Find x and y-1
\nAnswer:
\n\"2nd
\n\u21d2x + 1 = 1 \u21d2 x = 0
\ny + 6 = 7 \u21d2 y = 1
\n\u2234 x = 0, y = 1<\/p>\n

Part – C<\/span><\/p>\n

2nd PUC Basic Maths Matrices and Determinants Ex 1.1 Three Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nFind A and B if
\n(a) 2A + B = \\(\\left[ \\begin{matrix} 1 & -1 \\\\ 0 & 1 \\end{matrix} \\right]\\) and A – 3B = \\(\\left[ \\begin{matrix} 0 & 1 \\\\ 1 & 0 \\end{matrix} \\right]\\) .
\n(b) 2A + B = \\(\\left[ \\begin{matrix} 2 & 3 & 1 \\\\ 1 & 4 & 0 \\end{matrix} \\right]\\) , 3A + B = \\(\\left[ \\begin{matrix} 4 & 6 & 1 \\\\ 2 & 3 & 5 \\end{matrix} \\right]\\)
\n(c) 2A – 3B = \\(\\left[ \\begin{matrix} 2 & -4 \\\\ -12 & 1 \\end{matrix} \\right]\\) , A + 5B = \\(\\left[ \\begin{matrix} 1 & 24 \\\\ 33 & 7 \\end{matrix} \\right]\\)
\n(d) 2A + B = \\(\\left[ \\begin{matrix} 3 & -1 \\\\ -2 & 5 \\end{matrix} \\right]\\) and A – 2B = \\(\\left[ \\begin{matrix} 4 & 2 \\\\ -1 & 5 \\end{matrix} \\right]\\) .
\nAnswer:
\n\"2nd
\n\"2nd
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nIf A = \\(\\left[ \\begin{matrix} 3 & 8 & 1 \\\\ 2 & -6 & 3 \\\\ 7 & 4 & -5 \\end{matrix} \\right]\\) , B = \\(\\left[ \\begin{matrix} 4 & 0 & 2 \\\\ 6 & 2 & 3 \\\\ 1 & 3 & 2 \\end{matrix} \\right]\\) . verify 3(A + B) = 3A + 3B.
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 3.
\nFind x, y, a, b if = \\(\\left[ \\begin{matrix} 3x+4y & 2 & x-2y \\\\ a+b & 2a-b & -1 \\end{matrix} \\right] =\\left[ \\begin{matrix} 2 & 2 & 4 \\\\ 5 & -5 & -1 \\end{matrix} \\right]\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 4.
\nFind x if \\(\\left[ \\begin{matrix} { x }^{ 3 } & 1 \\\\ 3 & 3 \\end{matrix} \\right] +\\left[ \\begin{matrix} { -2x }^{ 2 } & 3 \\\\ 1 & 4 \\end{matrix} \\right] =\\left[ \\begin{matrix} -x & 4 \\\\ 4 & 7 \\end{matrix} \\right]\\)
\nAnswer:
\n\"2nd
\n\u21d2 x3<\/sup> – 2x2<\/sup> = x; x3<\/sup> – 2x2<\/sup> + x = 0 ; x(x2<\/sup> – 2x + 1) = 0
\nx = 0 or (x – 1)2<\/sup> = 0 \u21d2 x = 0 or x = 1<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nIf A = \\(\\left[ \\begin{matrix} 2 & -1 & 3 \\\\ -2 & 1 & 0 \\end{matrix} \\right]\\) , B = \\(\\left[ \\begin{matrix} -1 & 0 & 4 \\\\ 5 & 1 & -3 \\end{matrix} \\right]\\) C = \\(\\left[ \\begin{matrix} 4 & -1 & 1 \\\\ 6 & 2 & -1 \\end{matrix} \\right]\\) . Find the matrix X such that 2A – 3B – x = C
\nAnswer:
\n\"2nd<\/p>\n

Question 6.
\nFind a,b,c given that \\(4\\left[ \\begin{matrix} a & 2 & 3 \\\\ 4 & 5 & -6 \\\\ 7 & -8 & 9 \\end{matrix} \\right] -5\\left[ \\begin{matrix} 3 & 2 & 1 \\\\ b & 6 & -4 \\\\ 9 & -8 & 7 \\end{matrix} \\right] =\\left[ \\begin{matrix} 13 & -2 & 7 \\\\ 6 & -5 & -4 \\\\ -17 & 2c & 1 \\end{matrix} \\right]\\)
\nAnswer:
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 1.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Basic Maths Question Bank Chapter 1 Matrices and Determinants Ex 1.1 Part – A 2nd …<\/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\/20722"}],"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=20722"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20722\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20722"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20722"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20722"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}