{"id":34978,"date":"2022-08-24T17:26:01","date_gmt":"2022-08-24T11:56:01","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=34978"},"modified":"2022-08-24T17:26:14","modified_gmt":"2022-08-24T11:56:14","slug":"inter-1st-year-maths-1a-matrices-solutions-ex-3e","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1a-matrices-solutions-ex-3e\/","title":{"rendered":"Inter 1st Year Maths 1A Matrices Solutions Ex 3(e)"},"content":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions<\/a> Inter 1st Year Maths 1A Matrices Solutions Exercise 3(e) will help students to clear their doubts quickly.<\/p>\n

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(e)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nFind the adjoint and inverse of the following matrices.
\n(i) \\(\\left[\\begin{array}{cc}
\n2 & -3 \\\\
\n4 & 6
\n\\end{array}\\right]\\)
\nSolution:
\n\"Inter<\/p>\n

(ii) \\(\\left[\\begin{array}{cc}
\n\\cos \\alpha & -\\sin \\alpha \\\\
\n\\sin \\alpha & \\cos \\alpha
\n\\end{array}\\right]\\)
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

(iii) \\(\\left[\\begin{array}{lll}
\n1 & 0 & 2 \\\\
\n2 & 1 & 0 \\\\
\n3 & 2 & 1
\n\\end{array}\\right]\\)
\nSolution:
\n\"Inter<\/p>\n

(iv) \\(\\left[\\begin{array}{lll}
\n2 & 1 & 2 \\\\
\n1 & 0 & 1 \\\\
\n2 & 2 & 1
\n\\end{array}\\right]\\)
\nSolution:
\n\"Inter<\/p>\n

Question 2.
\nIf A = \\(\\left[\\begin{array}{cc}
\na+i b & c+i d \\\\
\n-c+i d & a-i b
\n\\end{array}\\right]\\), a2<\/sup> + b2<\/sup> + c2<\/sup> + d2<\/sup> = 1, then find the inverse of A.
\nSolution:
\n\"Inter<\/p>\n

Question 3.
\nIf A = \\(\\left[\\begin{array}{ccc}
\n1 & -2 & 3 \\\\
\n0 & -1 & 4 \\\\
\n-2 & 2 & 1
\n\\end{array}\\right]\\), then find A-1<\/sup>
\nSolution:
\n\"Inter<\/p>\n

Question 4.
\nIf A = \\(\\left|\\begin{array}{ccc}
\n-1 & -2 & -2 \\\\
\n2 & 1 & -2 \\\\
\n2 & -2 & 1
\n\\end{array}\\right|\\), then show that the adjoint of A = 3A’ find A-1<\/sup>.
\nSolution:
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 5.
\nIf abc \u2260 0, find the inverse of \\(\\left[\\begin{array}{lll}
\na & 0 & 0 \\\\
\n0 & b & 0 \\\\
\n0 & 0 & c
\n\\end{array}\\right]\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nIf A = \\(\\left[\\begin{array}{lll}
\n0 & 1 & 1 \\\\
\n1 & 0 & 1 \\\\
\n1 & 1 & 0
\n\\end{array}\\right]\\) and B = \\(\\frac{1}{2}\\left[\\begin{array}{lll}
\nb+c & c-a & b-a \\\\
\nc-b & c+a & a-b \\\\
\nb-c & a-c & a+b
\n\\end{array}\\right]\\) then show that ABA-1<\/sup> is a diagonal matrix.
\nSolution:
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n

Question 2.
\nIf 3A = \\(\\left[\\begin{array}{ccc}
\n1 & 2 & 2 \\\\
\n2 & 1 & -2 \\\\
\n-2 & 2 & -1
\n\\end{array}\\right]\\) then show that A-1<\/sup> = A’
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 3.
\nIf A = \\(\\left[\\begin{array}{rrr}
\n3 & -3 & 4 \\\\
\n2 & -3 & 4 \\\\
\n0 & -1 & 1
\n\\end{array}\\right]\\), then show that A-1<\/sup> = A3<\/sup>
\nSolution:
\n\"Inter
\n\u2234 A4<\/sup> = I
\ndet A = 3(1) – 3(-2) + 4(-2) = 1
\n\u2235 A \u2260 0 \u21d2 A-1<\/sup> exists
\n\u2235 A4<\/sup> = I
\nMultiply with A-1<\/sup>
\nA4<\/sup> (A-1<\/sup>) = I (A-1<\/sup>)
\n\u21d2 A3<\/sup> (AA-1<\/sup>) = A-1<\/sup>
\n\u21d2 A3<\/sup> (I) = A-1<\/sup>
\n\u2234 A-1<\/sup> = A3<\/sup><\/p>\n

Question 4.
\nIf AB = I or BA = I, then prove that A is invertible and B = A-1<\/sup>
\nSolution:
\nGiven AB = I
\n\u21d2 AB| = |1|
\n\u21d2 |A| |B| = 1
\n\u21d2 |A| \u2260 0
\n\u2234 A is a non-singular matrix and BA = I
\n\u21d2 |BA| = |I|
\n\u21d2 |B| |A| = 1
\n\u21d2 |A| \u2260 0
\n\u2234 A is a non-singular matrix.
\nAB = I or BA = I, A is invertible.
\n\u2234 A-1<\/sup> exists.
\nAB = I
\n\u21d2 A-1<\/sup>AB = A-1<\/sup>I
\n\u21d2 IB = A-1<\/sup>
\n\u21d2 B = A-1<\/sup>
\n\u2234 B = A-1<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(e) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(e) I. Question 1. Find the adjoint and inverse of the following matrices. (i) Solution: (ii) Solution: (iii) Solution: (iv) Solution: …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/34978"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=34978"}],"version-history":[{"count":2,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/34978\/revisions"}],"predecessor-version":[{"id":34980,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/34978\/revisions\/34980"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=34978"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=34978"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=34978"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}