{"id":6260,"date":"2024-01-11T01:24:47","date_gmt":"2024-01-10T19:54:47","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=6260"},"modified":"2024-01-12T17:48:14","modified_gmt":"2024-01-12T12:18:14","slug":"ts-inter-1st-year-maths-1a-solutions-chapter-3-ex-3g","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1a-solutions-chapter-3-ex-3g\/","title":{"rendered":"TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(g)"},"content":{"rendered":"

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

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

I.
\nExamine whether the following systems of equations are consistent or inconsistent and if consistent find the complete solutions,
\nQuestion 1.
\nx + y + z = 4
\n2x + 5y – 2z = 3
\nx + 7y – 7z = 5
\nAnswer:
\nAugmented matrix of the above system is
\n\"TS
\nRank of the matrix \u03c1(A) = 2 and \u03c1(AB) = 3.
\nSince \u03c1(A) \u2260 \u03c1(AB), the given system of equa\u00actions are inconsistent.<\/p>\n

Question 2.
\nx + y + z = 6
\nx – y + z = 2
\n2x – y + 3z = 9
\nAnswer:
\nAugmented matrix [AB] = \\(\\left[\\begin{array}{rrrr}
\n1 & 1 & 1 & 6 \\\\
\n1 & -1 & 1 & 2 \\\\
\n2 & -1 & 3 & 9
\n\\end{array}\\right]\\)
\nApply operations R2<\/sub> – R1<\/sub>, R3<\/sub> – 2R1<\/sub>, we get
\n\"TS
\nHere \u03c1(A) = 3 and \u03c1(AB) = 3
\nSince \u03c1(A) = \u03c1(AB), the given system is consistent and has a unique solution.
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\nx + y + z = 1
\n2x + y + z = 2
\nx + 2y + 2z = 1 (March 2015-T.S)
\nAnswer:
\nAugmented matrix of the system is
\n\"TS
\n\u03c1(AB) = 2 and \u03c1(A) = 2 and \u03c1(A) = \u03c1(AB) < 3
\nThe given system of equations is consis\u00actent and has infinitely many solutions.
\nThe given system is equivalent to x + y + z = 1 and y + z = 0.
\nSolution set is
\n\"TS<\/p>\n

Question 4.
\nx + y + z = 9
\n2x + 5y + 7z = 52
\n2x + y – z = 0
\nAnswer:
\nAugmented matrix of the system
\n\"TS
\nHere \u03c1(A) = \u03c1(AB) = 3; and the system of given equations is consistent; and has a unique solution.
\nAlso x = 1, y = 3, z = 5 form the solution.<\/p>\n

Question 5.
\nx + y + z = 6
\nx + 2y + 3z = 10
\nx + 2y + 4z = 1
\nAnswer:
\nAugmented matrix of the system is
\n\"TS<\/p>\n

Question 6.
\nx – 3y – 8z = – 10
\n3x + y – 4z = 0
\n2x + 5y + 6z = 13
\nAnswer:
\nThe augmented matrix of the above system is
\n\"TS
\nSince \u03c1(A) = 2 = \u03c1(AB) < 3, given system of equations is consistent with infinitely many solutions.
\nThe given system is equivalent to
\nx – 3y – 8z = – 10,
\ny + 2z = 3
\nPut z = t then y = 3 – 2t
\n\u2234 x = – 10 + 3(3 – 2t) + 8t
\n= -10 + 9 – 6t + 8t
\n= 2t – 1
\nHence the solutions are given by
\nx = 2t – 1, y = 3 – 2t and z = t
\nWhere t is any scalar.<\/p>\n

Question 7.
\n2x + 3y + z = 9
\nx + 2y + 3z = 6
\n3x + y + 2z = 8
\nAnswer:
\nAugmented matrix of the above system is
\n\"TS
\n\u03c1(A) = \u03c1(AB) = 3; system is consistent and has a unique solution given by
\nx = \\(\\frac{35}{18}\\), y = \\(\\frac{29}{18}\\), z = \\(\\frac{5}{18}\\)<\/p>\n

\"TS<\/p>\n

Question 8.
\nx + y + 4z = 6
\n3x + 2y – 2z = 9
\n5x + y + 2z = 13
\nAnswer:
\nAugmented matrix of the system
\n\"TS
\n\u03c1(A) = \u03c1(AB) = 3;
\nHence the system is consistent and has a unique solution given by
\nx = 2, y = 2, z = \\(\\frac{1}{2}\\).<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these\u00a0TS Intermediate Maths 1A Solutions Chapter 3 Matrices Ex 3(g) to find a better approach to solving the problems. TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(g) I. Examine whether the following systems of equations are consistent or inconsistent and if consistent find the complete solutions, Question 1. x + … 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\/6260"}],"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=6260"}],"version-history":[{"count":3,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/6260\/revisions"}],"predecessor-version":[{"id":6273,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/6260\/revisions\/6273"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=6260"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=6260"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=6260"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}