{"id":6212,"date":"2024-01-12T09:29:33","date_gmt":"2024-01-12T03:59:33","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=6212"},"modified":"2024-01-16T17:46:38","modified_gmt":"2024-01-16T12:16:38","slug":"ts-inter-1st-year-maths-1a-solutions-chapter-3-ex-3e","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1a-solutions-chapter-3-ex-3e\/","title":{"rendered":"TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e)"},"content":{"rendered":"

Students must practice these\u00a0TS Intermediate Maths 1A Solutions<\/a> Chapter 3 Matrices Ex 3(e) to find a better approach to solving the problems.<\/p>\n

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

I.
\nQuestion 1.
\nFind the adjoint and inverse of the following matrices. (March 2002)
\ni) \\(\\left[\\begin{array}{rr}
\n2 & -3 \\\\
\n4 & 6
\n\\end{array}\\right]\\)
\nAnswer:
\nIf A = \\(\\left[\\begin{array}{ll}
\na & b \\\\
\nc & d
\n\\end{array}\\right]\\) then adj A = \\(\\left[\\begin{array}{rr}
\n\\mathrm{d} & -\\mathrm{b} \\\\
\n-\\mathrm{c} & \\mathrm{a}
\n\\end{array}\\right]\\)
\n\"TS<\/p>\n

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

iii) Find the adjoint and inverse of the matrix \\(\\left[\\begin{array}{lll}
\n1 & 0 & 2 \\\\
\n2 & 1 & 0 \\\\
\n3 & 2 & 1
\n\\end{array}\\right]\\).
\nAnswer:
\nFind cofactors of elements in the matrix as
\n\"TS<\/p>\n

\"TS<\/p>\n

iv) \\(\\left|\\begin{array}{lll}
\n2 & 1 & 2 \\\\
\n1 & 0 & 1 \\\\
\n2 & 2 & 1
\n\\end{array}\\right|\\)
\nAnswer:
\n\"TS<\/p>\n

Question 2.
\nIf A = \\(\\left[\\begin{array}{cc}
\n\\mathrm{a}+\\mathrm{i b} & \\mathrm{c}+\\mathrm{i d} \\\\
\n-\\mathrm{c}+\\mathrm{i d} & \\mathrm{a}-\\mathrm{i b}
\n\\end{array}\\right]\\), a2<\/sup> + b2<\/sup> + c2<\/sup> + d2<\/sup> = 1
\nAnswer:
\ndet A = (a + ib) (a – ib) – (c + id) (- c + id)
\n= (a2<\/sup> – i2<\/sup> b2<\/sup>) – (- c2<\/sup> + i2<\/sup>d2<\/sup>)
\n= a2<\/sup> + b2<\/sup> + c2<\/sup> + d2<\/sup> (\u2235 i2<\/sup> = -1)
\n= 1
\nAdj A = \\(\\left[\\begin{array}{cc}
\na-i b & -c-i d \\\\
\nc-i d & a+i b
\n\\end{array}\\right]\\)
\nA-1<\/sup> = \\(\\frac{{Adj} A}{{det} A}=\\left[\\begin{array}{cc}
\na-i b & -c-i d \\\\
\nc-i d & a+i b
\n\\end{array}\\right]\\)<\/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>. (Board Model Paper)
\nAnswer:
\n\"TS<\/p>\n

Question 4.
\nIf A = \\(\\left[\\begin{array}{rrr}
\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>
\nAnswer:
\n\"TS<\/p>\n

Question 5.
\nIf abc \u2260 0; find the inverse of \\(\\left[\\begin{array}{lll}
\n\\mathrm{a} & 0 & 0 \\\\
\n0 & b & 0 \\\\
\n0 & 0 & c
\n\\end{array}\\right]\\) (May 2006)
\nAnswer:
\n\"TS<\/p>\n

\"TS<\/p>\n

II.
\nQuestion 1.
\nIf A = \\(\\left[\\begin{array}{lll}
\n\\mathrm{b}+\\mathrm{c} & \\mathrm{c}-\\mathrm{a} & \\mathrm{b}-\\mathrm{a} \\\\
\n\\mathrm{c}-\\mathrm{b} & \\mathrm{c}+\\mathrm{a} & \\mathrm{a}-\\mathrm{b} \\\\
\n\\mathrm{b}-\\mathrm{c} & \\mathrm{a}-\\mathrm{c} & \\mathrm{a}+\\mathrm{b}
\n\\end{array}\\right]\\) and B = \\(\\frac{1}{2}\\left[\\begin{array}{lll}
\n\\mathrm{b}+\\mathrm{c} & \\mathrm{c}-\\mathrm{a} & \\mathrm{b}-\\mathrm{a} \\\\
\n\\mathrm{c}-\\mathrm{b} & \\mathrm{c}+\\mathrm{a} & \\mathrm{a}-\\mathrm{b} \\\\
\n\\mathrm{b}-\\mathrm{c} & \\mathrm{a}-\\mathrm{c} & \\mathrm{a}+\\mathrm{b}
\n\\end{array}\\right]\\), then show that ABA-1<\/sup> is a diagonal matrix.
\nAnswer:
\n\"TS<\/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-I<\/sup> = A’.
\nAnswer:
\n\"TS
\n\u2234 A.A’ = I and by definition A’ = A-1<\/sup>
\nsimilarly A’.A = I<\/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>
\nAnswer:
\n\"TS
\nSo, the multiplicative inverse of A exists and it is A3<\/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>.
\nAnswer:
\nGiven AB = I
\n\u21d2 |AB| = |I|
\n\u21d2 |A| |B| = 1
\n\u21d2 |A| \u2260 0
\n\u2234 A is a non-singular matrix.
\nAlso BA = I
\n\u21d2 |B| |A| = |I|
\n\u21d2 |A| |B| = 1
\n\u21d2 |A| *0
\n\u2234 A is a non-singular matrix.
\n\u21d2 A is invertible
\n\u21d2 A-1<\/sup> exists AB = I
\n\u21d2 A-1<\/sup> AB = A-1<\/sup>I
\n\u21d2 (A-1<\/sup> A) B = A-1<\/sup>I
\n\u21d2 IB = A-1<\/sup>I
\n\u21d2 B = A-1<\/sup>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these\u00a0TS Intermediate Maths 1A Solutions Chapter 3 Matrices Ex 3(e) to find a better approach to solving the problems. TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(e) I. Question 1. Find the adjoint and inverse of the following matrices. (March 2002) i) Answer: If A = then adj A = … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/6212"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=6212"}],"version-history":[{"count":3,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/6212\/revisions"}],"predecessor-version":[{"id":6225,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/6212\/revisions\/6225"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=6212"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=6212"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=6212"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}