{"id":21869,"date":"2024-01-04T11:54:26","date_gmt":"2024-01-04T06:24:26","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=21869"},"modified":"2024-01-05T11:00:57","modified_gmt":"2024-01-05T05:30:57","slug":"inter-1st-year-maths-1a-matrices-solutions-ex-3i","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-1st-year-maths-1a-matrices-solutions-ex-3i\/","title":{"rendered":"Inter 1st Year Maths 1A Matrices Solutions Ex 3(i)"},"content":{"rendered":"

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

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

Solve the following systems of homogeneous equations.<\/span><\/p>\n

Question 1.
\n2x + 3y – z = 0, x – y – 2z = 0, 3x + y + 3z = 0
\nSolution:
\nThe Coefficient matrix is \\(\\left[\\begin{array}{ccc}
\n2 & 3 & -1 \\\\
\n1 & -1 & -2 \\\\
\n3 & 1 & 3
\n\\end{array}\\right]\\)
\ndet of \\(\\left[\\begin{array}{ccc}
\n2 & 3 & -1 \\\\
\n1 & -1 & -2 \\\\
\n3 & 1 & 3
\n\\end{array}\\right]=\\left[\\begin{array}{ccc}
\n2 & 3 & -1 \\\\
\n1 & -1 & -2 \\\\
\n3 & 1 & 3
\n\\end{array}\\right]\\)
\n= 2(-3 + 2) – 3(3 + 6) – 1(1 + 3)
\n= -2 – 27 – 4
\n= -33 \u2260 0, \u03c1(A) = 3
\nHence the system has the trivial solution x = y = z = 0 only.<\/p>\n

\"Inter<\/p>\n

Question 2.
\n3x + y – 2z = 0, x + y + z = 0, x – 2y + z = 0
\nHint: If the determinant of the coefficient matrix \u2260 0 then the system has a trivial solution (i.e.) \u03c1(A) = 3.
\nSolution:
\nThe coefficient matrix is \\(\\left[\\begin{array}{ccc}
\n3 & 1 & -2 \\\\
\n1 & 1 & 1 \\\\
\n1 & -2 & 1
\n\\end{array}\\right]\\)
\n\\(\\left|\\begin{array}{ccc}
\n3 & 1 & -2 \\\\
\n1 & 1 & 1 \\\\
\n1 & -2 & 1
\n\\end{array}\\right|\\)
\n= 3(1 + 2) – 1(1 – 1) – 2(-2 – 1)
\n= 9 + 6
\n= 15 \u2260 0, \u03c1(A) = 3
\nHence the system has the trivial solutions x = y = z = 0 only.<\/p>\n

\"Inter<\/p>\n

Question 3.
\nx + y – 2z = 0, 2x + y – 3z = 0, 5x + 4y – 9z = 0
\nSolution:
\nThe coefficient matrix is \\(\\left[\\begin{array}{rrr}
\n1 & 1 & -2 \\\\
\n2 & 1 & -3 \\\\
\n5 & 4 & -9
\n\\end{array}\\right]\\) = A (say)
\n|A| = \\(\\left|\\begin{array}{rrr}
\n1 & 1 & -2 \\\\
\n2 & 1 & -3 \\\\
\n5 & 4 & -9
\n\\end{array}\\right|\\)
\n= 1(-9 + 12) – 1(-18 + 15) – 2(8 – 5)
\n= 3 + 3 – 6
\n= 0
\n\u2234 Rank of A = 2 as the submatrix \\(\\left[\\begin{array}{ll}
\n1 & 1 \\\\
\n2 & 1
\n\\end{array}\\right]\\) is non-singular, \u03c1(A) < 3
\nHence the system has a non-trivial solution.
\nA = \\(\\left[\\begin{array}{lll}
\n1 & 1 & -2 \\\\
\n2 & 1 & -3 \\\\
\n5 & 4 & -9
\n\\end{array}\\right]\\)
\nR2<\/sub> \u2192 R2<\/sub> – 2R1<\/sub>, R3<\/sub> \u2192 R3<\/sub> – 3R1<\/sub>
\nA ~ \\(\\left[\\begin{array}{ccc}
\n1 & 1 & -2 \\\\
\n0 & -1 & 1 \\\\
\n0 & -1 & -1
\n\\end{array}\\right]\\)
\nThe system of equation is equivalent to the given system of equations are x + y – 2z = 0, -y + z = 0
\nLet z = k
\n\u21d2 y = k, x = k
\n\u2234 x = y = z = k for real number k.<\/p>\n

\"Inter<\/p>\n

Question 4.
\nx + y – z = 0, x – 2y + z = 0, 3x + 6y – 5z = 0
\nSolution:
\nCoefficient matrix A = \\(\\left[\\begin{array}{ccc}
\n1 & 1 & -1 \\\\
\n1 & -2 & 1 \\\\
\n3 & 6 & -5
\n\\end{array}\\right]\\)
\nR2<\/sub> \u2192 R2<\/sub> – R1<\/sub>, R3<\/sub> \u2192 R3<\/sub> – 3R1<\/sub>
\nA ~ \\(\\left[\\begin{array}{ccc}
\n1 & 1 & -1 \\\\
\n0 & -3 & 2 \\\\
\n0 & 3 & -2
\n\\end{array}\\right]\\)
\n\u21d2 det A = 0 as R2<\/sub>, R3<\/sub> are identical.
\nand rank (A) = 2 as the submatrix \\(\\left[\\begin{array}{cc}
\n1 & 1 \\\\
\n0 & -3
\n\\end{array}\\right]\\) is non-singular.
\nHence the system has a non-trivial solution, \u2235 \u03c1(A) < 3
\nThe system of equations equivalent to the given system of equations are
\nx + y – z = 0
\n3y – 2z = 0
\nLet z = k
\n\u21d2 y = \\(\\frac{2 k}{3}\\)
\nx = \\(\\frac{k}{3}\\)
\n\u2234 x = \\(\\frac{k}{3}\\), y = \\(\\frac{2 k}{3}\\), z = k for any real number of k.<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(i) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(i) Solve the following systems of homogeneous equations. Question 1. 2x + 3y – z = 0, x – y – … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21869"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=21869"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21869\/revisions"}],"predecessor-version":[{"id":21870,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/21869\/revisions\/21870"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=21869"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=21869"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=21869"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}