{"id":20831,"date":"2020-01-23T09:38:35","date_gmt":"2020-01-23T04:08:35","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20831"},"modified":"2021-07-02T15:27:19","modified_gmt":"2021-07-02T09:57:19","slug":"2nd-puc-basic-maths-question-bank-chapter-1-ex-1-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-1-ex-1-2\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 1 Matrices and Determinants Ex 1.2"},"content":{"rendered":"

Students can Download Basic Maths Exercise 1.2 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 1.2<\/h2>\n

Part – A<\/span><\/p>\n

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

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

Question 2.
\nIf A = \\(\\left[ \\begin{matrix} 1 & 2 \\\\ 0 & -3 \\end{matrix} \\right]\\) . Find A2<\/sup>
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nIf A = \\(\\left[ \\begin{matrix} 3 & -1 \\\\ 2 & -4 \\end{matrix} \\right]\\) . Find A2<\/sup>
\nAnswer:
\n\"2nd<\/p>\n

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

Question 5.
\nIf \\(\\left[ \\begin{matrix} x \\\\ y \\end{matrix} \\right] =\\left[ \\begin{matrix} 2 & 3 \\\\ 1 & 0 \\end{matrix} \\right] \\left[ \\begin{matrix} 1 \\\\ 0 \\end{matrix} \\right]\\) . Find x and y.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

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

Question 7.
\nIf \\(\\left[ \\begin{matrix} 4 \\\\ 1 \\\\ 3 \\end{matrix} \\right] \\left[ \\begin{matrix} x & y & z \\end{matrix} \\right] =\\left[ \\begin{matrix} 4 & 8 & 4 \\\\ 1 & 2 & 1 \\\\ 3 & 6 & 2 \\end{matrix} \\right]\\) . Find x,y,z.
\nAnswer:
\n\"2nd<\/p>\n

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

\"KSEEB<\/p>\n

Question 9.
\nFind A.B, if A = \\(\\left[ \\begin{matrix} 1 & 3 & 2 \\end{matrix} \\right]\\) and B = \\(\\left[ \\begin{matrix} 4 \\\\ 2 \\\\ 1 \\end{matrix} \\right]\\) .
\nAnswer:
\n\"2nd<\/p>\n

Question 10.
\nFind x, if \\(\\left[ \\begin{matrix} 2 & \\times & 2 \\end{matrix} \\right] \\left[ \\begin{matrix} 1 \\\\ 4 \\\\ 2 \\end{matrix} \\right] =\\left[ 3 \\right] \\) .
\nAnswer:
\n\"2nd
\n[2 + 4x + 4] = [3]
\n\u21d2 4x = 3 – 6 = -3 ; x = \\(\\frac { -3 }{ 4 }\\)<\/p>\n

Part – B<\/span><\/p>\n

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

Question 1.
\nIf A = \\(\\left[ \\begin{matrix} 2 & -2 \\\\ 3 & -1 \\end{matrix} \\right]\\) and B = \\(\\left[ \\begin{matrix} 3 & -1 \\\\ -2 & 0 \\end{matrix} \\right]\\) . Find AB and BA
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

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

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

Question 4.
\nIf A = \\(\\left[ \\begin{matrix} 3 & -1 & 2 \\\\ 1 & 2 & 2 \\\\ 2 & 0 & 5 \\end{matrix} \\right]\\) , Find A2
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

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

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

Question 7.
\nIf A = \\(\\left[ \\begin{matrix} 1 & 2 \\\\ 1 & 4 \\end{matrix} \\right]\\) and B = \\(\\left[ \\begin{matrix} 4 & -3 \\\\ 2 & 1 \\end{matrix} \\right]\\) and C = \\(\\left[ \\begin{matrix} 1 & 0 \\\\ -2 & 4 \\end{matrix} \\right]\\) .
\nverify that (i) A(BC) = (AB)A (ii) A(B+C) = AB + AC
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nIf A = \\(\\left[ \\begin{matrix} 1 & 3 \\\\ 1 & 0 \\end{matrix} \\right]\\) . prove that A2<\/sup> – A – 31 = 0
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 9.
\nIf A = \\(\\left[ \\begin{matrix} 2 & -1 \\\\ -1 & 2 \\end{matrix} \\right]\\) . Show that A2<\/sup> – 4A + 31 = 0
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nIf A = \\(\\left[ \\begin{matrix} 1 & 2 \\\\ 3 & 4 \\end{matrix} \\right]\\) Show that A2<\/sup> – 5A = 21.
\nAnswer:
\n\"2nd<\/p>\n

Question 11.
\nSolve for x, y:
\n(i) \\(\\left[ \\begin{matrix} 2 & -1 \\\\ 3 & 1 \\end{matrix} \\right] \\left[ \\begin{matrix} x \\\\ y \\end{matrix} \\right] =\\left[ \\begin{matrix} 10 \\\\ 2 \\end{matrix} \\right]\\)
\n(ii) \\(\\left[ \\begin{matrix} 1 & 3 \\\\ -2 & 4 \\end{matrix} \\right] \\left[ \\begin{matrix} x \\\\ y \\end{matrix} \\right] =\\left[ \\begin{matrix} -1 \\\\ 0 \\end{matrix} \\right]\\)
\n(iii) \\(\\left[ \\begin{matrix} x & y \\end{matrix} \\right] \\left[ \\begin{matrix} 2 & -1 \\\\ 3 & 1 \\end{matrix} \\right] =\\left[ \\begin{matrix} -1 & 4 \\end{matrix} \\right]\\) .
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nIf x = \\(\\left[ \\begin{matrix} 2 & 1 & 3 \\\\ 0 & 1 & 4 \\end{matrix} \\right]\\) , y = \\(\\left[ \\begin{matrix} 1 \\\\ 0 \\\\ 3 \\end{matrix} \\right]\\) and z = [2 1]. verify that x(yz) = (xy)z.
\nAnswer:
\n\"2nd
\nFrom 1 & 2 we get (xy)z = x(yz).<\/p>\n

Part – C<\/span><\/p>\n

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

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

\"KSEEB<\/p>\n

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

\"KSEEB<\/p>\n

Question 3.
\nIf A = \\(\\left[ \\begin{matrix} i & 0 \\\\ 0 & -1 \\end{matrix} \\right]\\) B = \\(\\left[ \\begin{matrix} 0 & -1 \\\\ 1 & 0 \\end{matrix} \\right]\\) and C = \\(\\left[ \\begin{matrix} 0 & i \\\\ i & 0 \\end{matrix} \\right]\\) where i2 = -1, show that
\n(i) A2 = B2 = C2 = -1
\n(ii) AB = -BA = -C
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 4.
\nIf A = \\(\\left[ \\begin{matrix} -1 & 1 & 0 \\\\ 3 & -3 & 3 \\\\ 5 & -5 & 5 \\end{matrix} \\right]\\) , B = \\(\\left[ \\begin{matrix} 0 & 4 & 3 \\\\ 1 & -3 & -3 \\\\ -1 & 4 & 4 \\end{matrix} \\right]\\) . Show that A2<\/sup> . B2<\/sup> = A2<\/sup>.
\nAnswer:
\n\"2nd
\nHence A2<\/sup>B2<\/sup> = A2<\/sup> is proved.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nIf f(x) = x2<\/sup> – 5x + 7, find f(A) where A = \\(\\left[ \\begin{matrix} 3 & 1 \\\\ -1 & 2 \\end{matrix} \\right]\\) .
\nAnswer:
\n\"2nd<\/p>\n

Question 6.
\nIf A = \\(\\left[ \\begin{matrix} 1 & -1 & 0 \\\\ 2 & 3 & 4 \\\\ 0 & 1 & 2 \\end{matrix} \\right]\\) B = \\(\\left[ \\begin{matrix} 2 & 2 & -4 \\\\ -4 & 2 & 4 \\\\ 2 & -1 & 5 \\end{matrix} \\right]\\) Show that BA = 61
\nAnswer:
\n\"2nd
\nHence BA = 61 is proved<\/p>\n

Question 7.
\nIf A = \\(\\left[ \\begin{matrix} 1 & 2 \\\\ 3 & 4 \\end{matrix} \\right]\\) and A + 2B = A2<\/sup>, find B
\nAnswer:
\nGiven A + 2B = A2<\/sup>
\n2B = A2<\/sup> – A
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nIf \\(\\left[ \\begin{matrix} 1 & 2 \\\\ 4 & 7 \\end{matrix} \\right] \\left[ \\begin{matrix} a & b \\\\ c & d \\end{matrix} \\right] =\\left[ \\begin{matrix} 1 & 0 \\\\ 0 & 1 \\end{matrix} \\right]\\) . Find a, b, c, d.
\nAnswer:
\n\"2nd
\n\u2234 a = -7, b = 2, c = 4, d = -1.<\/p>\n

Question 9.
\nIf A = \\(\\left[ \\begin{matrix} a & b \\\\ c & d \\end{matrix} \\right]\\) and I = \\(\\left[ \\begin{matrix} 1 & 0 \\\\ 0 & 1 \\end{matrix} \\right]\\) . Show that A2<\/sup> – (a + d)A – (bc – ad) I is a null matrix.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nIf A = \\(\\left[ \\begin{matrix} 1 & 2 & 2 \\\\ 2 & 1 & 2 \\\\ 2 & 2 & 1 \\end{matrix} \\right]\\) . Prove that A2<\/sup> – 4A – 51 is a null matrix of order 3 \u00d7 3.
\nAnswer:
\n\"2nd
\n\u2234 A2<\/sup> – 4A – 51 is a null matrix.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 1.2 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 1.2 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\/20831"}],"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=20831"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20831\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20831"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20831"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20831"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}