{"id":34986,"date":"2022-08-24T17:30:21","date_gmt":"2022-08-24T12:00:21","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=34986"},"modified":"2022-08-24T17:30:36","modified_gmt":"2022-08-24T12:00:36","slug":"inter-1st-year-maths-1a-matrices-solutions-ex-3h","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1a-matrices-solutions-ex-3h\/","title":{"rendered":"Inter 1st Year Maths 1A Matrices Solutions Ex 3(h)"},"content":{"rendered":"

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

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

Solve the following systems of equations.<\/span>
\n(i) by using Cramer’s rule and matrix inversion method, when the coefficient matrix is non-singular.<\/span>
\n(ii) by using the Gauss-Jordan method. Also, determine whether the system has a unique solution or an infinite number of solutions, or no solution, and find the solutions if exist.<\/span><\/p>\n

Question 1.
\n5x – 6y + 4z = 15
\n7x + 4y – 3z = 19
\n2x + y + 6z = 46
\nSolution:
\n(i) Cramer’s rule:
\n\u0394 = \\(\\left|\\begin{array}{ccc}
\n5 & -6 & 4 \\\\
\n7 & 4 & -3 \\\\
\n2 & 1 & 6
\n\\end{array}\\right|\\)
\n= 5(24 + 3) + 6(42 + 6) + 4(7 – 8)
\n= 135 + 288 – 4
\n= 419
\n\u03941<\/sub> = \\(\\left|\\begin{array}{ccc}
\n15 & -6 & 4 \\\\
\n19 & 4 & -3 \\\\
\n46 & 1 & 6
\n\\end{array}\\right|\\)
\n= 15(24 + 3) + 6(114 + 138) + 4(19 – 184)
\n= 405 + 1512 – 660
\n= 1917 – 660
\n= 1257
\n\u03942<\/sub> = \\(\\left|\\begin{array}{ccc}
\n5 & 15 & 4 \\\\
\n7 & 19 & -3 \\\\
\n2 & 46 & 6
\n\\end{array}\\right|\\)
\n= 5(114 + 138) – 15(42 + 6) + 4(322 – 38)
\n= 1260 – 720 + 1136
\n= 1676
\n\u03943<\/sub> = \\(\\left|\\begin{array}{ccc}
\n5 & -6 & 15 \\\\
\n7 & 4 & 19 \\\\
\n2 & 1 & 46
\n\\end{array}\\right|\\)
\n= 5(184 – 19) + 6(322 – 38) + 15(7 – 8)
\n= 825 + 1704 – 15
\n= 2529 – 15
\n= 2514
\n\"Inter
\nSolution is x = 3, y = 4, z = 6.<\/p>\n

(ii) Matrix inversion method:
\n\"Inter
\n\"Inter
\n\"Inter
\nSolution is x = 3, y = 4, z = 6<\/p>\n

(iii) Gauss-Jordan method:
\n\"Inter
\n\u2234 Unique solution exists.
\nSolution is x = 3, y = 4, z = 6.<\/p>\n

\"Inter<\/p>\n

Question 2.
\nx + y + z = 1
\n2x + 2y + 3z = 6
\nx + 4y + 9z = 3
\nSolution:
\n(i) Cramer’s rule:
\n\u0394 = \\(\\left|\\begin{array}{lll}
\n1 & 1 & 1 \\\\
\n2 & 2 & 3 \\\\
\n1 & 4 & 9
\n\\end{array}\\right|\\)
\n= 1(18 – 12) – 1(18 – 3) + 1(8 – 2)
\n= 6 – 15 + 6
\n= -3
\n\u03941<\/sub> = \\(\\left|\\begin{array}{lll}
\n1 & 1 & 1 \\\\
\n6 & 2 & 3 \\\\
\n3 & 4 & 9
\n\\end{array}\\right|\\)
\n= 1(18 – 12) – 1(54 – 9) + 1(24 – 6)
\n= 6 – 45 + 18
\n= -21
\n\u03942<\/sub> = \\(\\left|\\begin{array}{lll}
\n1 & 1 & 1 \\\\
\n2 & 6 & 3 \\\\
\n1 & 3 & 9
\n\\end{array}\\right|\\)
\n= 1(54 – 9) – 1(18 – 3) + 1(6 – 6)
\n= 45 – 15
\n= 30
\n\u03943<\/sub> = \\(\\left|\\begin{array}{lll}
\n1 & 1 & 1 \\\\
\n2 & 2 & 6 \\\\
\n1 & 4 & 3
\n\\end{array}\\right|\\)
\n= 1(6 – 24) – 1(6 – 6) + 1(8 – 2)
\n= -18 – 0 + 6
\n= -12
\n\"Inter
\nSolution is x = 7, y = -10, z = 4<\/p>\n

(ii) Matrix inversion method:
\n\"Inter
\n\"Inter
\n\"Inter
\n\u2234 Solution is x = 7, y = -10, z = 4<\/p>\n

(iii) Gauss-Jordan method:
\nAugmented matrix is A = \\(\\left[\\begin{array}{llll}
\n1 & 1 & 1 & 1 \\\\
\n2 & 2 & 3 & 6 \\\\
\n1 & 4 & 9 & 3
\n\\end{array}\\right]\\)
\nR2<\/sub> \u2192 R2<\/sub> – 2R1<\/sub>, R3<\/sub> \u2192 R3<\/sub> – R1<\/sub>
\n\"Inter
\nUnique solution exists.
\n\u2234 Solution is x = 7, y = -10, z = 4<\/p>\n

\"Inter<\/p>\n

Question 3.
\nx – y + 3z = 5
\n4x + 2y – z = 0
\n-x + 3y + z = 5
\nSolution:
\n(i) Cramer’s rule:
\n\u0394 = \\(\\left|\\begin{array}{ccc}
\n1 & -1 & 3 \\\\
\n4 & 2 & -1 \\\\
\n-1 & 3 & 1
\n\\end{array}\\right|\\)
\n= 1(2 + 3) + 1(4 – 1) + 3(12 + 2)
\n= 5 + 3 + 42
\n= 50
\n\u03941<\/sub> = \\(\\left|\\begin{array}{ccc}
\n5 & -1 & 3 \\\\
\n0 & 2 & -1 \\\\
\n5 & 3 & 1
\n\\end{array}\\right|\\)
\n= 5(2 + 3) + 1(0 + 5) + 3(0 – 10)
\n= 25 + 5 – 30
\n= 0
\n\u03942<\/sub> = \\(\\left|\\begin{array}{ccc}
\n1 & 5 & 3 \\\\
\n4 & 0 & -1 \\\\
\n-1 & 5 & 1
\n\\end{array}\\right|\\)
\n= 1(0 + 5) – 5(4 – 1) + 3(20 – 0)
\n= 5 – 15 + 60
\n= 50
\n\u03943<\/sub> = \\(\\left|\\begin{array}{ccc}
\n1 & -1 & 5 \\\\
\n4 & 2 & 0 \\\\
\n-1 & 3 & 5
\n\\end{array}\\right|\\)
\n= 1(10 – 0) + 1(20 – 0) + 5(12 + 2)
\n= 10 + 20 + 70
\n= 100
\n\"Inter
\n\u2234 Solution is x = 0, y = 1, z = 2.<\/p>\n

(ii) Matrix inversion method:
\n\"Inter
\n\"Inter
\n\"Inter
\nSolution is x = 0, y = 1, z = 2<\/p>\n

(iii) Gauss Jordan method:
\n\"Inter
\nUnique solution exists.
\n\u2234 Solution is x = 0, y = 1, z = 2<\/p>\n

\"Inter<\/p>\n

Question 4.
\n2x + 6y + 11 = 0
\n6x + 20y – 6z + 3 = 0
\n6y – 18z + 1 = 0
\nSolution:
\n(i) Cramer’s rule:
\n\u0394 = \\(\\left|\\begin{array}{ccc}
\n2 & 6 & 0 \\\\
\n6 & 20 & -6 \\\\
\n0 & 6 & -18
\n\\end{array}\\right|\\)
\n= 2(-360 + 36) – 6(-108 – 0)
\n= -648 + 648
\n= 0
\n\u2234 Cramer’s rule and matrix inversion method cannot be used.
\n\u2235 \u0394 = 0<\/p>\n

(ii) Gauss Jordan method:
\n\"Inter
\n\u03c1(A) = 2, \u03c1(AB) = 3
\n\u03c1(A) \u2260 \u03c1(AB)
\n\u2234 The given system of equations does not have a solution.<\/p>\n

\"Inter<\/p>\n

Question 5.
\n2x – y + 3z = 9
\nx + y + z = 6
\nx – y + z = 2
\nSolution:
\n(i) Cramer’s rule:
\n\u0394 = \\(\\left|\\begin{array}{ccc}
\n2 & -1 & 3 \\\\
\n1 & 1 & 1 \\\\
\n1 & -1 & 1
\n\\end{array}\\right|\\)
\n= 2(1 + 1) + 1(1 – 1) + 3(-1 – 1)
\n= 4 + 0 – 6
\n= -2
\n\u03941<\/sub> = \\(\\left|\\begin{array}{ccc}
\n9 & -1 & 3 \\\\
\n6 & 1 & 1 \\\\
\n2 & -1 & 1
\n\\end{array}\\right|\\)
\n= 9(1 + 1) + 1(6 – 2) + 3(-6 – 2)
\n= 18 + 4 – 24
\n= -2
\n\u03942<\/sub> = \\(\\left|\\begin{array}{lll}
\n2 & 9 & 3 \\\\
\n1 & 6 & 1 \\\\
\n1 & 2 & 1
\n\\end{array}\\right|\\)
\n= 2(6 – 2) – 9(1 – 1) + 3(2 – 6)
\n= 8 – 0 – 12
\n= -4
\n\u03943<\/sub> = \\(\\left|\\begin{array}{ccc}
\n2 & -1 & 9 \\\\
\n1 & 1 & 6 \\\\
\n1 & -1 & 2
\n\\end{array}\\right|\\)
\n= 2(2 + 6) + 1(2 – 6) + 9(-1 – 1)
\n= 16 – 4 – 18
\n= -6
\n\"Inter
\nSolution is x = 1, y = 2, z = 3.<\/p>\n

(ii) Matrix inversion method:
\n\"Inter
\n\"Inter
\nSolution is x = 1, y = 2, z = 3.<\/p>\n

(iii) Gauss-Jordan method:
\n\"Inter
\n\u2234 The given equations have a unique solution.
\nSolution is x = 1, y = 2, z = 3<\/p>\n

\"Inter<\/p>\n

Question 6.
\n2x – y + 8z = 13
\n3x + 4y + 5z = 18
\n5x – 2y + 7z = 20
\nSolution:
\n(i) Cramer’s rule:
\n\u0394 = \\(\\left|\\begin{array}{ccc}
\n2 & -1 & 8 \\\\
\n3 & 4 & 5 \\\\
\n5 & -2 & 7
\n\\end{array}\\right|\\)
\n= 2(28 + 10) + 1(21 – 25) + 8(-6 – 20)
\n= 76 – 4 – 208
\n= -136
\n\u03941<\/sub> = \\(\\left|\\begin{array}{ccc}
\n13 & -1 & 8 \\\\
\n18 & 4 & 5 \\\\
\n20 & -2 & 7
\n\\end{array}\\right|\\)
\n= 13(28 + 10) + 1(126 – 100) + 8(-36 – 80)
\n= 494 + 26 – 928
\n= -408
\n\u03942<\/sub> = \\(\\left|\\begin{array}{lll}
\n2 & 13 & 8 \\\\
\n3 & 18 & 5 \\\\
\n5 & 20 & 7
\n\\end{array}\\right|\\)
\n= 2(126 – 100) – 13(21 – 25) + 8(60 – 90)
\n= 52 + 52 – 240
\n= -136
\n\u03943<\/sub> = \\(\\left|\\begin{array}{ccc}
\n2 & -1 & 13 \\\\
\n3 & 4 & 18 \\\\
\n5 & -2 & 20
\n\\end{array}\\right|\\)
\n= 2(80 + 36) + 1(60 – 90) + 13(-6 – 20)
\n= 232 – 30 – 338
\n= -136
\n\"Inter
\n\u2234 Solution is x = 3, y = 1, z = 1<\/p>\n

(ii) Matrix inversion method:
\n\"Inter
\n\"Inter
\n\"Inter
\n\u2234 Solution is x = 3, y = 1, z = 1<\/p>\n

(iii) Gauss Jordan method:
\n\"Inter
\n\u2234 The given equations have a unique solution and Solution is x = 3, y = 1, z = 1.<\/p>\n

\"Inter<\/p>\n

Question 7.
\n2x – y + 3z = 8
\n-x + 2y + z = 4
\n3x + y – 4z = 0
\nSolution:
\n(i) Cramer’s rule:
\n\u0394 = \\(\\left|\\begin{array}{ccc}
\n2 & -1 & 3 \\\\
\n-1 & 2 & 1 \\\\
\n3 & 1 & -4
\n\\end{array}\\right|\\)
\n= 2(-8 – 1) + 1(4 – 3) + 3(-1 – 6)
\n= -18 + 1 – 21
\n= -38
\n\u03941<\/sub> = \\(\\left|\\begin{array}{ccc}
\n8 & -1 & 3 \\\\
\n4 & 2 & 1 \\\\
\n0 & 1 & -4
\n\\end{array}\\right|\\)
\n= 8(-8 – 1) + 1(-16 – 0) + 3(4 – 0)
\n= -72 – 16 + 12
\n= -76
\n\u03942<\/sub> = \\(\\left|\\begin{array}{ccc}
\n2 & 8 & 3 \\\\
\n-1 & 4 & 1 \\\\
\n3 & 0 & -4
\n\\end{array}\\right|\\)
\n= 2(-16 – 0) – 8(4 – 3) + 3(-0 – 12)
\n= -32 – 8 – 36
\n= -76
\n\u03943<\/sub> = \\(\\left|\\begin{array}{ccc}
\n2 & -1 & 8 \\\\
\n-1 & 2 & 4 \\\\
\n3 & 1 & 0
\n\\end{array}\\right|\\)
\n= 2(0 – 4) + 1(0 – 12) + 8(-1 – 6)
\n= -8 – 12 – 56
\n= -76
\n\"Inter
\n\u2234 Solution is x = 2, y = 2, z = 2.<\/p>\n

(ii) Matrix inversion method:
\n\"Inter
\n\"Inter
\nSolution is x = 2, y = 2, z = 2<\/p>\n

(iii) Gauss Jordan method:
\n\"Inter
\n\"Inter
\n\u2234 The given equations have a unique solution and solution is x = 2, y = 2, z = 2.<\/p>\n

\"Inter<\/p>\n

Question 8.
\nx + y + z = 9
\n2x + 5y + 7z = 52
\n2x + y – z = 0
\nSolution:
\n(i) Cramer’s rule:
\n\u0394 = \\(\\left|\\begin{array}{ccc}
\n1 & 1 & 1 \\\\
\n2 & 5 & 7 \\\\
\n2 & 1 & -1
\n\\end{array}\\right|\\)
\n= 1(-5 – 7) – 1(-2 – 14) + 1(2 – 10)
\n= -12 + 16 – 8
\n= -4
\n\u03941<\/sub> = \\(\\left|\\begin{array}{ccc}
\n9 & 1 & 1 \\\\
\n52 & 5 & 7 \\\\
\n0 & 1 & -1
\n\\end{array}\\right|\\)
\n= 9(-5 – 7) – 1(-52 – 0) + 1(52 – 0)
\n= -108 + 52 + 52
\n= -4
\n\u03942<\/sub> = \\(\\left|\\begin{array}{ccc}
\n1 & 9 & 1 \\\\
\n2 & 52 & 7 \\\\
\n2 & 0 & -1
\n\\end{array}\\right|\\)
\n= 1(-52 – 0) – 9(-2 – 14) + 1(0 – 104)
\n= -52 + 144 – 104
\n= -12
\n\u03943<\/sub> = \\(\\left|\\begin{array}{ccc}
\n1 & 1 & 9 \\\\
\n2 & 5 & 52 \\\\
\n2 & 1 & 0
\n\\end{array}\\right|\\)
\n= 1(0 – 52) – 1(0 – 104) + 9(2 – 10)
\n= -52 + 104 – 72
\n= -20
\n\"Inter<\/p>\n

(ii) Matrix inversion method:
\n\"Inter
\n\"Inter
\nSolution is x = 1, y = 3, z = 5<\/p>\n

(iii) Gauss Jordan method:
\n\"Inter
\n\u2234 The given equations have a unique solution and solution is x = 1, y = 3, z = 5.<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(h) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(h) Solve the following systems of equations. (i) by using Cramer’s rule and matrix inversion method, when the coefficient matrix is …<\/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\/34986"}],"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=34986"}],"version-history":[{"count":2,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/34986\/revisions"}],"predecessor-version":[{"id":34988,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/34986\/revisions\/34988"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=34986"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=34986"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=34986"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}