{"id":162600,"date":"2023-10-27T13:04:14","date_gmt":"2023-10-27T07:34:14","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=162600"},"modified":"2023-10-27T17:05:55","modified_gmt":"2023-10-27T11:35:55","slug":"ml-aggarwal-class-12-maths-solutions-section-a-chapter-4-ex-4-6","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-12-maths-solutions-section-a-chapter-4-ex-4-6\/","title":{"rendered":"ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.6"},"content":{"rendered":"

Continuous practice using ML Aggarwal Class 12 Solutions ISC<\/a> Chapter 4 Determinants Ex 4.6 can lead to a stronger grasp of mathematical concepts.<\/p>\n

ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.6<\/h2>\n

Using cramer’s rule, solve the following (1 to 7) systems of linear equations:
\n1. (i) 3x + y = 5
\nx + 2y = 3
\n(ii) 2x – y = 17
\n3x + 5y = 6
\nSolution:
\n(i) The given system is equivalent to
\nAX = B
\nwhere A = \\(\\left[\\begin{array}{ll}
\n3 & 1 \\\\
\n1 & 2
\n\\end{array}\\right]\\) ;
\nX = \\(=\\left[\\begin{array}{l}
\nx \\\\
\ny
\n\\end{array}\\right]\\) ;
\nB = \\(\\left[\\begin{array}{l}
\n5 \\\\
\n3
\n\\end{array}\\right]\\)
\nHere |A| = \\(\\left|\\begin{array}{ll}
\n3 & 1 \\\\
\n1 & 2
\n\\end{array}\\right|\\)
\n= 6 – 1
\n= 5 \u2260 0
\n\u2234 given system has unique solution.
\n|A1<\/sub>| = \\(\\left|\\begin{array}{ll}
\n5 & 1 \\\\
\n3 & 2
\n\\end{array}\\right|\\)
\n= 10 – 3 = 7
\n|A2<\/sub>| = \\(\\left|\\begin{array}{ll}
\n3 & 5 \\\\
\n1 & 3
\n\\end{array}\\right|\\)
\n= 9 – 5 = 4
\nThen by cramer’s rule, we have
\nx = \\(\\frac{\\left|\\mathrm{A}_1\\right|}{|\\mathrm{A}|}\\)
\n= \\(\\frac{7}{5}\\)
\nand y = \\(\\frac{\\left|\\mathrm{A}_2\\right|}{|\\mathrm{A}|}\\)
\n= \\(\\frac{4}{5}\\)<\/p>\n

(ii) For given system of equations, we have
\nD = \\(\\left|\\begin{array}{rr}
\n2 & -1 \\\\
\n3 & 5
\n\\end{array}\\right|\\)
\n= 10 + 3
\n= 13 \u2260 0
\n\u2234 given system of eqns. have unique solution.
\nD1<\/sub> = \\(\\left|\\begin{array}{rr}
\n17 & -1 \\\\
\n6 & 5
\n\\end{array}\\right|\\)
\n= 85 +6
\n= 91
\nand D2<\/sub> = \\(\\left|\\begin{array}{rr}
\n2 & 17 \\\\
\n3 & 6
\n\\end{array}\\right|\\)
\n= 12 – 51
\n= – 39
\nThen by Cramer’s rule, we have
\nx = \\(\\frac{D_1}{D}\\)
\n= \\(\\frac{91}{13}\\)
\n= 7;
\ny = \\(\\frac{D_2}{D}\\)
\n= [\/latex]-\\frac{39}{13}[\/latex]
\n= – 3
\nHence x = 7, y = – 3 be the required solution.<\/p>\n

\"ML<\/p>\n

Question 2.
\n(i) 2x – 7y – 13 = 0 ; 5x + 6y – 9 = 0
\n(ii) \\(\\frac{2}{x}+\\frac{3}{y}\\) = 2 ; \\(\\frac{5}{x}+\\frac{8}{7}=\\frac{31}{6}\\)
\nSolution:
\n(i) The given system of equations be:
\n2x – 7y = 13
\n5x + 6y = 9
\nHere D = \\(\\left|\\begin{array}{rr}
\n2 & -7 \\\\
\n5 & 6
\n\\end{array}\\right|\\)
\n= 12 + 35
\n= 47 \u2260 0
\n\u2234 Thus given system has unique solution.
\nD1<\/sub> = \\(\\left|\\begin{array}{rr}
\n13 & -7 \\\\
\n9 & 6
\n\\end{array}\\right|\\)
\n= 78 +63
\n= 141
\nD2<\/sub> = \\(\\left|\\begin{array}{rr}
\n2 & 13 \\\\
\n5 & 9
\n\\end{array}\\right|\\)
\n= 18 – 65
\n= – 47
\n\u2234 x = \\(\\frac{D_1}{D}=\\frac{141}{47}\\)
\n= 3 ;
\ny = \\(\\frac{D_2}{D}=-\\frac{47}{47}\\)
\n= – 1
\nHence the required solution of given system be x = 3 and y = – 1.<\/p>\n

(ii) putting \\(\\frac{1}{x}\\) = u and \\(\\frac{1}{y}\\) = v in given system of eqn’s we have,
\n2u + 3v = 2
\n5u + 8v = \\(\\frac{31}{6}\\)
\nHere, D = \\(\\left|\\begin{array}{ll}
\n2 & 3 \\\\
\n5 & 8
\n\\end{array}\\right|\\)
\n= 16 – 15
\n= 1 \u2260 0
\n\u2234 given system of equations has unique solution.
\nD1<\/sub> = \\(\\left|\\begin{array}{rr}
\n2 & 3 \\\\
\n\\frac{31}{6} & 8
\n\\end{array}\\right|\\)
\n= 16 – \\(\\frac{31}{2}\\)
\n= \\(\\frac{1}{2}\\)
\nD2<\/sub> = \\(\\left|\\begin{array}{rr}
\n2 & 2 \\\\
\n5 & \\frac{31}{6}
\n\\end{array}\\right|\\)
\n= \\(\\frac{31}{3}\\) – 10
\n= \\(\\frac{1}{3}\\)
\nThus by cramer’s rule, we have
\nu = \\(\\frac{D_1}{D_2}=\\frac{\\frac{1}{2}}{1}=\\frac{1}{2}\\)
\nand v = \\(\\frac{D_2}{D}=\\frac{1}{3}\\)
\n\u21d2 \\(\\frac{1}{x}=\\frac{1}{2}\\)
\n\u21d2 x = 2
\nand \\(\\frac{1}{y}=\\frac{1}{3}\\)
\n\u21d2 y = 3
\nHence the required solution of given system be x = 2 and y = 3.<\/p>\n

\"ML<\/p>\n

Question 3.
\n3x + ay = 4
\n2x + ay = 2, a \u2260 0.
\nSolution:
\nThe given system of equations is,
\n3x + ay = 4 ………….(1)
\nand 2x + ay = 2 ………..(2)
\nHere |A| = \\(\\left|\\begin{array}{ll}
\n3 & a \\\\
\n2 & a
\n\\end{array}\\right|\\)
\n= 3a – 2a = a
\n\u2234 The given system of equations has unique solution and it is given by using cramer’s rule.
\nx = \\(\\frac{|\\mathrm{A}|}{|\\mathrm{A}|}\\),
\ny = \\(\\frac{\\left|\\mathrm{A}_2\\right|}{|\\mathrm{A}|}\\) …………..(3)
\nWhere, |A1<\/sub>| = \\(\\left|\\begin{array}{ll}
\n4 & a \\\\
\n2 & a
\n\\end{array}\\right|\\)
\n= 4a – 2a = 2a
\nand |A2<\/sub>| = \\(\\left|\\begin{array}{ll}
\n3 & 4 \\\\
\n2 & 2
\n\\end{array}\\right|\\)
\n= 6 – 8 = – 2
\n\u2234 From (3) ;
\nx = \\(\\frac{2 a}{a}\\) = 2 ;
\ny = \\(\\frac{-2}{a}\\)<\/p>\n

\"ML<\/p>\n

Question 4.
\n(i) 5x – 7y + z = 11
\n6x – 8y -z = 11
\n3x + 2y – 6z = 7
\n(ii) x + y + z + 1 = 0
\nx + 2y + 3z + 4 = 0
\nx + 3y + 4z + 6 = 0
\nSolution:
\n(i) The given system of eqns. is,
\n5x – 7y + z = 11 ………….(1)
\n6x – 8y – z = 15 ………….(2)
\nand 3x + 2y – 6z = 7 ………….(3)
\nThe given system of eqns is equivalent to
\nAX = B
\nWhere A = \\(\\left[\\begin{array}{rrr}
\n5 & -7 & 1 \\\\
\n6 & -8 & -1 \\\\
\n3 & 2 & -6
\n\\end{array}\\right]\\) ;
\nX = \\(\\left[\\begin{array}{l}
\nx \\\\
\ny \\\\
\nz
\n\\end{array}\\right]\\) ;
\nB = \\(\\left[\\begin{array}{r}
\n11 \\\\
\n15 \\\\
\n7
\n\\end{array}\\right]\\)
\n\u2234 |A| = \\(\\left|\\begin{array}{rrr}
\n5 & -7 & 1 \\\\
\n6 & -8 & -1 \\\\
\n3 & 2 & -6
\n\\end{array}\\right|\\)
\nExpanding along R1<\/sub> ; we get
\n= 5 (48 + 2) + 7 (- 36 + 3) + (12 + 24)
\n= 250 – 231 + 36
\n= 55 \u2260 0
\n\u2234 given system of eqns. has unique solution and by Cramer’s rule
\nx = \\(\\frac{\\left|\\mathrm{A}_1\\right|}{|\\mathrm{A}|}\\) ;
\ny = \\(\\frac{\\left|\\mathrm{A}_2\\right|}{|\\mathrm{A}|}\\) ;
\nz = \\(\\frac{\\left|\\mathrm{A}_3\\right|}{|\\mathrm{A}|}\\) …………..(2)
\nHence, |A1<\/sub>| = \\(\\left|\\begin{array}{rrr}
\n11 & -7 & 1 \\\\
\n15 & -8 & -1 \\\\
\n7 & 2 & -6
\n\\end{array}\\right|\\)
\n= 11 (48 + 2) + 7 (- 90 + 7) + 1 (30 + 56)
\n= 550 – 581 + 86
\n= 55<\/p>\n

|A2<\/sub>| = \\(\\left|\\begin{array}{rrr}
\n5 & 11 & 1 \\\\
\n6 & 15 & -1 \\\\
\n3 & 7 & -6
\n\\end{array}\\right|\\)
\n= 5 (- 90 + 7) – 11 (- 36 + 3) + 1 (42 – 45)
\n= – 415 + 363 – 3
\n= – 418 + 363
\n= – 55<\/p>\n

|A3<\/sub>| = \\(\\left|\\begin{array}{rrr}
\n5 & -7 & 11 \\\\
\n6 & -8 & 15 \\\\
\n3 & 2 & 7
\n\\end{array}\\right|\\)
\n= 5 (- 56 – 30) + 7 (42 – 45) + 11 (12 + 24)
\n= – 430 – 21 + 396
\n= – 55<\/p>\n

\u2234 From (2) ;
\nx = \\(\\frac{55}{55}\\) = 1 ;
\ny = \\(\\frac{-55}{55}\\) = – 1
\nand z = \\(\\frac{-55}{55}\\) = – 1
\nThus x = 1 ; y = – 1 and z = – 1<\/p>\n

(ii) The given system of eqns is equivalent to AX = B
\nwhere A = \\(\\left[\\begin{array}{lll}
\n1 & 1 & 1 \\\\
\n1 & 2 & 3 \\\\
\n1 & 3 & 4
\n\\end{array}\\right]\\) ;
\nX = \\(\\left[\\begin{array}{l}
\nx \\\\
\ny \\\\
\nz
\n\\end{array}\\right]\\)
\nand B = \\(\\left[\\begin{array}{l}
\n-1 \\\\
\n-4 \\\\
\n-6
\n\\end{array}\\right]\\)
\nHere, |A| = \\(\\left|\\begin{array}{lll}
\n1 & 1 & 1 \\\\
\n1 & 2 & 3 \\\\
\n1 & 3 & 4
\n\\end{array}\\right|\\)
\n= 1 (- 1) – 1 (1) + 1 (1)
\n= – 1 – 1 + 1
\n= – 1 \u2260 0
\n\u2234 The given system of equations has unique solution and it is given by using cramer’s rule.
\nx = \\(\\frac{\\left|A_1\\right|}{|A|}\\) ;
\ny = \\(\\frac{\\left|A_2\\right|}{|A|}\\)
\nand z = \\(\\frac{\\left|A_3\\right|}{|A|}\\) ………..(1)
\nwhere, |A1<\/sub>| = \\(\\left|\\begin{array}{lll}
\n-1 & 1 & 1 \\\\
\n-4 & 2 & 3 \\\\
\n-6 & 3 & 4
\n\\end{array}\\right|\\)
\n= – 1 (- 1) – 1 (2) + 1 (0)
\n= 1 – 2
\n= – 1
\n|A2<\/sub>| = \\(\\left|\\begin{array}{lll}
\n1 & -1 & 1 \\\\
\n1 & -4 & 3 \\\\
\n1 & -6 & 4
\n\\end{array}\\right|\\)
\n= 1 (2) + 1 (1) + 1 (- 2)
\n= 2 + 1 – 2 = 1
\n|A3<\/sub>| = \\(\\left|\\begin{array}{lll}
\n1 & 1 & -1 \\\\
\n1 & 2 & -4 \\\\
\n1 & 3 & -6
\n\\end{array}\\right|\\)
\n= 1 (0) – 1 (- 2) – 1 (1)
\n= 2 – 1 = 1
\n\u2234 From (1) ; we have
\nx = \\(\\frac{-1}{-1}\\) = 1 ;
\ny = \\(\\frac{1}{-1}\\) = – 1;
\nz = \\(\\frac{1}{-1}\\) = – 1.<\/p>\n

\"ML<\/p>\n

Question 5.
\n(i) 4x – 2y + 9z + 2 = 0
\n3x + 4y + z – 5 = 0
\nx – 3y + 2z – 8 = 0<\/p>\n

(ii) x + y = 2
\n2x – z = 1
\n2y – 3z = 1
\nSolution:
\nGiven system of eqn’s can be written as :
\n4x – 2y + 9z = – 2
\n3x + 4y + z = 5
\nx – 3y + 2z = 8
\nHere D = \\(\\left|\\begin{array}{rrr}
\n4 & -2 & 9 \\\\
\n3 & 4 & 1 \\\\
\n1 & -3 & 2
\n\\end{array}\\right|\\) ;
\nExpanding along R1<\/sub>
\n= 4 (8 + 3) + 2 (6 – 1) + 9 (- 9 – 4)
\n= 44 + 10 – 117
\n= – 63 \u2260 0
\nThus given system of eqns has unique solution.
\nD1<\/sub> = \\(\\left|\\begin{array}{rrr}
\n-2 & -2 & 9 \\\\
\n5 & 4 & 1 \\\\
\n8 & -3 & 2
\n\\end{array}\\right|\\) ;
\nexpanding along R1<\/sub>
\n= – 2 (8 + 3) + 2 (10 – 8) + 9 (- 15 – 32)
\n= – 22 + 4 – 423
\n= – 441
\nD2<\/sub> = \\(\\left|\\begin{array}{rrr}
\n4 & -2 & 9 \\\\
\n3 & 5 & 1 \\\\
\n1 & 8 & 2
\n\\end{array}\\right|\\) ;
\nExpanding along R1<\/sub>
\n= 4 (10 – 8) + 2 (6 – 1) + 9 (24 – 5)
\n= 8 + 10 + 171
\n= 189
\nand D3<\/sub> = \\(\\left|\\begin{array}{rrr}
\n4 & -2 & -2 \\\\
\n3 & 4 & 5 \\\\
\n1 & -3 & 8
\n\\end{array}\\right|\\) ;
\nExpanding along R1<\/sub>
\n= 4 (32 + 15) + 2 (24 – 5) – 2 (- 9 – 4)
\n= 188 + 38 + 262
\n= 252
\nThus the Cramer’s rule, we have
\nx = \\(\\frac{D_1}{D}\\)
\n= \\(\\frac{-441}{-63}\\)
\n= 7 ;
\ny = \\(\\frac{D_2}{D}\\)
\n= \\(\\frac{189}{-63}\\)
\n= – 3 ;
\nz = \\(\\frac{D_3}{D}\\)
\n= \\(\\frac{252}{-63}\\)
\n= – 4
\nHence the required solution of given system be
\nx = 7 ;
\ny = – 3
\nand z = – 4<\/p>\n

(ii) The given system of equations is:
\nx + y = 2
\n2x – z = 1
\n2y – 3z = 1
\nHere D = \\(\\left|\\begin{array}{rrr}
\n1 & 1 & 0 \\\\
\n2 & 0 & -1 \\\\
\n0 & 2 & -3
\n\\end{array}\\right|\\) ;
\nExpanding along R1<\/sub>
\n= 1 (0 + 2) – 1 (- 6 – 0)
\n= 2 + 6
\n= 8 \u2260 0
\nThus given system of eqns. has unique solution.
\nD1<\/sub> = \\(\\left|\\begin{array}{rrr}
\n2 & 1 & 0 \\\\
\n1 & 0 & -1 \\\\
\n1 & 2 & -3
\n\\end{array}\\right|\\)
\n= 2 (0 + 2) – 1 (- 3 + 1)
\n= 4 + 2
\n= 6
\nD2<\/sub> = \\(\\left|\\begin{array}{rrr}
\n1 & 2 & 0 \\\\
\n2 & 1 & -1 \\\\
\n0 & 1 & -3
\n\\end{array}\\right|\\)
\n= 1 (- 3 + 1) – 2 (- 6 – 0)
\n= – 2 + 12
\n= 0
\nD3<\/sub> = \\(\\left|\\begin{array}{lll}
\n1 & 1 & 2 \\\\
\n2 & 0 & 1 \\\\
\n0 & 2 & 1
\n\\end{array}\\right|\\)
\n= 1 (0 – 2) – 1 (2 – 0) 2(4 – 0)
\n= – 2 – 2 + 8
\n= 4
\nThus, by cramer’s rule, we have
\nx = \\(\\frac{D_1}{D}\\)
\n= \\(\\frac{6}{8}=\\frac{3}{4}\\) ;<\/p>\n

y = \\(\\frac{\\mathrm{D}_2}{\\mathrm{D}}\\)
\n= \\(\\frac{10}{8}=\\frac{5}{4}\\) ;<\/p>\n

and z = \\(\\frac{D_3}{D}\\)
\n= \\(\\frac{4}{8}=\\frac{1}{2}\\)
\nHence the required solution of given system be
\nx = \\(\\frac{3}{4}\\) ;
\ny = \\(\\frac{5}{4}\\) and
\nz = \\(\\frac{1}{2}\\).<\/p>\n

\"ML<\/p>\n

Question 6.
\n(i) 2x – 3y = 1
\nx + 3z = 11
\nx + 2y + z = 7<\/p>\n

(ii) \\(\\frac{2}{x}+\\frac{3}{y}+\\frac{10}{z}\\) = 4
\n\\(\\frac{4}{x}-\\frac{6}{y}+\\frac{5}{z}\\) = 1
\n\\(\\frac{6}{x}+\\frac{9}{y}-\\frac{20}{z}\\) = 2
\nSolution:
\n(i) The given system of eqns. are
\n2x – 3y = 1
\nx + 3z = 11
\nx + 2y + z = 7
\nHere, D = \\(\\left|\\begin{array}{rrr}
\n2 & -3 & 0 \\\\
\n1 & 0 & 3 \\\\
\n1 & 2 & 1
\n\\end{array}\\right|\\) ;
\nExpanding along R1<\/sub> ;
\n= 2 (0 – 6) + 3 (1 – 3) + 0
\n= – 12 – 6
\n= – 18 \u2260 0
\nThus, given system of equations has unique solution.
\nD1<\/sub> = \\(\\left|\\begin{array}{rrr}
\n1 & -3 & 0 \\\\
\n11 & 0 & 3 \\\\
\n7 & 2 & 1
\n\\end{array}\\right|\\)
\n= 1 (0 – 6) + 3 (11 – 21) + 0
\n= – 6 – 30
\n= – 36
\nD2<\/sub> = \\(\\left|\\begin{array}{rrr}
\n2 & 1 & 0 \\\\
\n1 & 11 & 3 \\\\
\n1 & 7 & 1
\n\\end{array}\\right|\\)
\n= 2 (11 – 21) + 1 (1 – 3) + 0
\n= – 20 + 2
\n= – 18
\nD3<\/sub> = \\(\\left|\\begin{array}{rrr}
\n2 & -3 & 1 \\\\
\n1 & 0 & 11 \\\\
\n1 & 2 & 7
\n\\end{array}\\right|\\)
\n= 2 (0 – 22) + 3 (7 – 11) + 1 (2 – 0)
\n= – 44 – 12 + 2
\n= – 54
\nThen by Cramer’s rule, we have
\nx = \\(\\frac{D_1}{D}\\)
\n= \\( \\frac{-36}{-18}\\)
\n= 2;
\ny = \\(\\frac{D_2}{D}\\)
\n= \\(\\frac{-18}{-18}\\)
\n= 1;
\nand z = \\(\\frac{D_3}{D}\\)
\n= \\( \\frac{-54}{-18}\\)
\n= 3
\nthus, the required solution of given system be
\nx = 2 ; y = 1 and z = 3.<\/p>\n

(ii) putting \\(\\frac{1}{x}\\) = u ;
\n\\(\\frac{1}{y}\\) = v ;
\n\\(\\frac{1}{z}\\) = w
\nin given system of equations ; we get
\nHere D = \\(\\left|\\begin{array}{rrr}
\n2 & 3 & 10 \\\\
\n4 & -6 & 5 \\\\
\n6 & 9 & -20
\n\\end{array}\\right|\\) ;
\nexpanding along R1<\/sub>
\n= 2 (120 – 45) – 3 (- 80 – 30) + 10 (36 +36)
\n= + 150 + 330 + 720
\n= 1200 \u2260 0
\nThus, given system has unique solution.<\/p>\n

D1<\/sub> = \\(\\left|\\begin{array}{rrr}
\n4 & 3 & 10 \\\\
\n1 & -6 & 5 \\\\
\n2 & 9 & -20
\n\\end{array}\\right|\\)
\n= 4 (120 – 45) – 3 (- 20 – 10) + 10 (9 +12)
\n= 300 + 90 + 210
\n= 600
\nD2<\/sub> = \\(\\left|\\begin{array}{rrr}
\n2 & 4 & 10 \\\\
\n4 & 1 & 5 \\\\
\n6 & 2 & -20
\n\\end{array}\\right|\\)
\n= 2 (- 20 – 10) – 4 (- 80 – 30) + 10 (8 – 6)
\n= – 60 + 440 + 20
\n= 400
\nD3<\/sub> = \\(\\left|\\begin{array}{rrr}
\n2 & 3 & 4 \\\\
\n4 & -6 & 1 \\\\
\n6 & 9 & 2
\n\\end{array}\\right|\\)
\n= 2 (- 12 – 9) – 3 (8 – 6) + 4 (36 + 36)
\n= – 42 – 6 + 288
\n= 240
\nThen by cramer’s rule, we have
\nu = \\(\\frac{D_1}{D}=\\frac{600}{1200}=\\frac{1}{2}\\)
\n\u21d2 \\(\\frac{1}{x}=\\frac{1}{2}\\)
\n\u21d2 x = 2
\nv = \\(\\frac{D_2}{D}=\\frac{400}{1200}=\\frac{1}{3}\\)
\n\u21d2 \\(\\frac{1}{y}=\\frac{1}{3}\\)
\n\u21d2 y = 3
\nand w = \\(\\frac{D_3}{D}=\\frac{240}{1200}=\\frac{1}{5}\\)
\n\u21d2 \\(\\frac{1}{z}=\\frac{1}{5}\\)
\n\u21d2 z = 5.
\nHence cthe required solution of given system be
\nx = 2 ; y = 3 and z = 5.<\/p>\n

\"ML<\/p>\n

Question 7.
\nx + y + z = 1
\nax + by + cz = k
\na2<\/sup>x + b2<\/sup>y + c2<\/sup>z = k where a, b, c are all different.
\nSolution:
\nThe given system of eqns. is equivalent to AX = B
\nwhere A = \\(\\left|\\begin{array}{ccc}
\n1 & 1 & 1 \\\\
\na & b & c \\\\
\na^2 & b^2 & c^2
\n\\end{array}\\right|\\);
\nX = \\(\\left[\\begin{array}{l}
\nx \\\\
\ny \\\\
\nz
\n\\end{array}\\right]\\)
\nand B = \\(\\left[\\begin{array}{c}
\n1 \\\\
\nk \\\\
\nk^2
\n\\end{array}\\right]\\)
\nHere |A| = \\(\\left|\\begin{array}{ccc}
\n1 & 1 & 1 \\\\
\na & b & c \\\\
\na^2 & b^2 & c^2
\n\\end{array}\\right|\\);
\noperate C2<\/sub> \u2192 C2<\/sub> – C1<\/sub>
\nC3<\/sub> \u2192 C3<\/sub> – C1<\/sub>
\n= \\(\\left|\\begin{array}{ccc}
\n1 & 0 & 0 \\\\
\na & b-a & c-a \\\\
\na^2 & b^2-a^2 & c^2-a^2
\n\\end{array}\\right|\\) ;
\nTaking (b – a) common from C2<\/sub> and (c – a) common from C3<\/sub><\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 8.
\nThe sum of three numbers is 20. If we multiply the first number by 2 and add the second number to the result and subtract the third number, we get 23. By adding second and third numbers to three times the first number, we get 46. Represent the above problem algebraically and use Cramer\u2019s rule to find the numbers from these equations.
\nSolution:
\nThe given problem can be formulated mathematically as under:
\nx + y + z = 20
\n2x + y – z = 23
\n3x + y + z = 46
\nHere D = \\(\\left|\\begin{array}{rrr}
\n1 & 1 & 1 \\\\
\n2 & 1 & -1 \\\\
\n3 & 1 & 1
\n\\end{array}\\right|\\);
\nexpanding along R1<\/sub>
\n= 1(1 + 1) – 1 (2 + 3) + 1 (2 – 3)
\n= 2 – 5 – 1
\n= – 4 \u2260 0
\nThus given system of eqn\u2019s has unique solution.
\nD1<\/sub> = \\(\\left|\\begin{array}{rrr}
\n20 & 1 & 1 \\\\
\n23 & 1 & -1 \\\\
\n46 & 1 & 1
\n\\end{array}\\right|\\)
\n= 20 (1 + 1) – 1 (23 + 46) + 1 (23 – 46)
\n= 40 – 69 – 23
\n= – 52
\nD2<\/sub> = \\(\\left|\\begin{array}{rrr}
\n1 & 20 & 1 \\\\
\n2 & 23 & -1 \\\\
\n3 & 46 & 1
\n\\end{array}\\right|\\)
\n= 1 (23 + 46) – 20 (2 + 3) + 1 (92 – 69)
\n= 69 – 100 + 23
\n= – 8
\nD3<\/sub> = \\(\\left|\\begin{array}{lll}
\n1 & 1 & 20 \\\\
\n2 & 1 & 23 \\\\
\n3 & 1 & 46
\n\\end{array}\\right|\\)
\n= 1 (46 – 23) – 1 (92 – 69) + 20 (2 – 3)
\n= 23 – 23 – 20
\n= – 20
\nThus, by cramer’s rule, we get
\nx = \\(\\frac{D_1}{D}\\)
\n=\\frac{-52}{-4}[\/latex]
\n= 13 ;
\ny = \\(\\frac{D_2}{D}\\)
\n= \\(\\frac{-8}{-4}\\)
\n= 2 ;
\nz = \\(\\frac{D_3}{D}\\)
\n= \\(\\frac{-20}{-4}\\)
\n= 5
\nThus the required three numbers are 13, 2 and 5 respectively.<\/p>\n

\"ML<\/p>\n

Question 9.
\nWhich of the following equations are consistent? If consistent, solve them.
\n(i) 3x + y = 5
\n6x + 2y = 11<\/p>\n

(ii) x + 2y = 3
\n4x + 8y = 12
\nSolution:
\n(i) The given system of eqns is;
\n3x + y = 5
\n6x + 2y = 11
\nHere, D = \\(\\left|\\begin{array}{ll}
\n3 & 1 \\\\
\n6 & 2
\n\\end{array}\\right|\\)
\n= 6 – 6 = 0
\nD = \\(\\left|\\begin{array}{ll}
\n3 & 1 \\\\
\n6 & 2
\n\\end{array}\\right|\\)
\n= 10 – 11
\n= – 1
\nThus by Cramer\u2019s rule given system has no solution and given system is inconsistent.<\/p>\n

(ii) Given system of eqn’s is
\nx + 2y = 3
\n4x + 8y = 12
\nHere, D = \\(\\left|\\begin{array}{ll}
\n1 & 2 \\\\
\n4 & 8
\n\\end{array}\\right|\\)
\n= 8 – 8 = 0
\nD1<\/sub> = \\(\\left|\\begin{array}{rr}
\n3 & 2 \\\\
\n12 & 8
\n\\end{array}\\right|\\)
\n= 24 – 24 = 0
\nD2<\/sub> = \\(\\left|\\begin{array}{rr}
\n1 & 3 \\\\
\n4 & 12
\n\\end{array}\\right|\\)
\n= 12 – 12 = 0
\nThus D = D1<\/sub> = D2<\/sub> = 0
\nThen by Cramer\u2019s rule. the given system is consistent and has infinitely many solutions.<\/p>\n

For Solution :
\nputting y = k in first equation of given system, we have
\nx = 3 – 2k. where k be any number
\nClearly x = 3 – 2k and y = k satisfies the 2nd eqn. of given system.
\nHence, the given system is consistent and has infinitely many solutions given by
\nx = 3 – 2k ; y = k; where k be any number.<\/p>\n

\"ML<\/p>\n

Question 10.
\nWhich of the following equations are consistent ? If consistent, solve them.
\n(i) x – y + 3z = 6
\nx + 3y – 3z = – 4
\n5x + 3y + 3z = 14<\/p>\n

(ii) 4x – 2y + 6z = 8
\n2x – y + z = 4
\n2x – y + z = 13.<\/p>\n

(iii) 2x + y – 2z = 4
\nx – 2y + z = – 2
\n5x – 5y + z = – 2
\nSolution:
\n(i) The given system of eqns is
\nx – y + 3z = 6 ………….(1)
\nx + 3y – 3z = – 4 …………(2)
\nand 5x + 3y – 3z = 14 ………….(30
\nHere |A| = \\(\\left|\\begin{array}{rrr}
\n1 & -1 & 3 \\\\
\n1 & 3 & -3 \\\\
\n5 & 3 & 3
\n\\end{array}\\right|\\)
\n= 1 (9 + 9) + 1 (3 + 15) – 3 (3 – 15)
\n= 18 + 18 – 36 = 0
\nAlso |A1<\/sub>| = \\(\\)
\n= 6 (18) + 1 (30) + 3 (- 54)
\n= 108 + 30 – 162
\n= – 24 \u2260 0
\n\u2234 The given system of eqns has no solution and they are inconsistent.<\/p>\n

(ii) Given system of eqn’s is
\n4x – 2y + 6z = 8
\n2x – y + z = 4
\n2x – y + z = 13
\nHere D = \\(\\left|\\begin{array}{rrr}
\n4 & -2 & 6 \\\\
\n2 & -1 & 3 \\\\
\n2 & -1 & 3
\n\\end{array}\\right|\\) = 0
\n[\u2235 R2<\/sub> and R3<\/sub> are identical]<\/p>\n

D1<\/sub> = \\(\\left|\\begin{array}{rrr}
\n8 & -2 & 6 \\\\
\n4 & -1 & 3 \\\\
\n13 & -1 & 3
\n\\end{array}\\right|\\) ;
\nExpanding along R1<\/sub>
\n= 8 (- 3 + 3) + 2 (12 – 39) + 6 (- 4 + 13)
\n= 0 – 54 + 54
\n= 0<\/p>\n

D2<\/sub> = \\(\\left|\\begin{array}{rrr}
\n4 & 8 & 6 \\\\
\n2 & 4 & 3 \\\\
\n2 & 13 & 3
\n\\end{array}\\right|\\) ;
\nExpanding along R1<\/sub>
\n= 4 (12 – 39) – 8 (6 – 6) + 6 (26 – 8)
\n= – 108 + 0 + 108
\n= 0<\/p>\n

D3<\/sub> = \\(\\left|\\begin{array}{rrr}
\n4 & -2 & 8 \\\\
\n2 & -1 & 4 \\\\
\n2 & -1 & 13
\n\\end{array}\\right|\\) ;
\nExpanding along R1<\/sub>
\n= 4 (- 13 + 4) + 2 (26 – 8) + 8 (- 2 + 2)
\n= – 36 + 36 + 0
\n= 0
\nHere, D = D1<\/sub> = D2<\/sub> = D3<\/sub> = 0
\nThus, by Cramer\u2019s rule, given system of eqn\u2019s has infinitely many solutions and system of eqn\u2019s is consistent.<\/p>\n

For solution :
\nputting z = k in eqn. (1) and eqn. (3) of given system, we have
\n4x – 2y = 8 – 6k
\n\u21d2 2x – y = 4 – 3k
\nand 2x – y + 3k = 13
\n\u21d2 2x – y = 13 – 3k
\n\u2234 4 – 3k = 13 – 3k
\n\u21d2 4=13
\nwhich is false.
\nThus given system is inconsistent and has no solution.<\/p>\n

(iii) The given system of eqns is ;
\n2x + y – 2z = 4
\nx – 2y + z = – 2
\n5x – 5y + z = – 2
\nHere D = \\(\\left|\\begin{array}{rrr}
\n2 & 1 & -2 \\\\
\n1 & -2 & 1 \\\\
\n5 & -5 & 1
\n\\end{array}\\right|\\)
\n= 2 (- 2 + 5) – 1 (1 – 5) – 2 (- 5 + 10)
\n= 6 + 4 – 10 = 0
\nD1<\/sub> = \\(\\left|\\begin{array}{rrr}
\n4 & 1 & -2 \\\\
\n-2 & -2 & 1 \\\\
\n-2 & -5 & 1
\n\\end{array}\\right|\\)
\n= 4 (- 2 + 5) – 1 (- 2 + 2) – 2 (- 5 + 10)
\n= 12 + 0 – 12 = 0<\/p>\n

D2<\/sub> = \\(\\left|\\begin{array}{rrr}
\n2 & 4 & -2 \\\\
\n1 & -2 & 1 \\\\
\n5 & -2 & 1
\n\\end{array}\\right|\\)
\n= 2 (- 2 + 2) – 4 (1 – 5) – 2 (- 2 + 10)
\n= 0 + 16 – 16 = 0<\/p>\n

D3<\/sub> = \\(\\left|\\begin{array}{rrr}
\n2 & 1 & 4 \\\\
\n1 & -2 & -2 \\\\
\n5 & -5 & -2
\n\\end{array}\\right|\\)
\n= 2 (4 – 10) – 1 (- 2 + 10) + 4 (- 5 + 10)
\n= – 12 – 8 + 20 = 0
\nThus D = D1<\/sub> = D2<\/sub> = D3<\/sub> = 0
\nThen by Cramers rule, the given system is consistent and has infinitely many solutions.<\/p>\n

For solution:
\nputting z = k in first two equations of given system, we have
\n2x + y = 4 + 2k
\nx – 2y = – 2 – k
\nHere |A| = \\(\\left|\\begin{array}{rr}
\n2 & 1 \\\\
\n1 & -2
\n\\end{array}\\right|\\)
\n= – 4 – 1
\n= – 5 \u2260 0
\n\u2234 given system has unique solution.
\nHere A1<\/sub> = \\(\\left|\\begin{array}{rr}
\n4+2 k & 1 \\\\
\n-2-k & -2
\n\\end{array}\\right|\\)
\n= – 2 (4 + 2k) – (- 2 – k)
\n= – 8 – 4k + 2 + k
\n= – 6 – 3k
\nand A2<\/sub> = \\(\\left|\\begin{array}{cc}
\n2 & 4+2 k \\\\
\n1 & -2-k
\n\\end{array}\\right|\\)
\n= 2 (- 2 – k) – (4 + 2k)
\n= – 4 – 2k – 4 – 2k
\n= – 8 – 4k
\nThus, by Cramer\u2019s rule, we have
\nx = \\(\\frac{\\mathrm{A}_1}{\\mathrm{~A}}\\)
\n= \\(\\frac{-6-3 k}{-5}\\)
\n= \\(\\frac{3(k+2)}{5}\\)
\nand y = \\(\\frac{A_2}{A}\\)
\n= \\(\\frac{-8-4 k}{-5}\\)
\n= \\(\\frac{4(k+2)}{5}\\)
\nand z = k
\nThese values of x, y and z satisfies the 3rd eqn. of given system.
\nHence the required solution of given system be
\nx = \\(\\frac{3}{5}\\) (k + 2) ;
\ny = \\(\\frac{4}{5}\\) (k + 2); = k,
\nwhere k be any number.<\/p>\n

\"ML<\/p>\n

Question 11.
\nWhich of the following systems has non trivial solutions ? If so, find these solutions.
\n(i) 5x + 5y + 2z = 0
\n2x + 5y + 4z = 0
\n4x + 5y + 2z = 0
\n(ii) 2x – 3y – z = 0
\nx + 3y – 2z =0
\nx – 3y = 0
\nSolution:
\n(i) The given homogeneous system of eqns. is
\n5x + 5y + 2z = 0
\n2x + 5y + 4z = 0
\n4x + 5y + 2z = 0
\nHere, D = \\(\\left|\\begin{array}{lll}
\n5 & 5 & 2 \\\\
\n2 & 5 & 4 \\\\
\n4 & 5 & 2
\n\\end{array}\\right|\\)
\n= 5 (10 – 20) – 5 (4 – 16) + 2 (10 – 20)
\n= – 50 + 60 – 20
\n= – 10 \u2260 0
\nThus, given system has trivial solution.
\ni.e., x = 0 = y = z.<\/p>\n

(ii) The given system of eqns is,
\n2x – 3y – z = 0 …………..(1)
\nx + 3y – 2z =0 …………..(2)
\nx – 3y = 0 …………..(3)
\nHere, |A| = \\(\\left|\\begin{array}{ccc}
\n2 & -3 & -1 \\\\
\n1 & 3 & -2 \\\\
\n1 & -3 & 0
\n\\end{array}\\right|\\)
\n= 2 (- 6) + 3 (0 + 2) – 1 (- 3 – 3)
\n= – 12 + 6 + 6
\n= 0
\n\u2234The given system of eqns. has infinite number of solutions.
\nLet z = k
\nFrom (1),
\n2x – 3y = k …………..(3)
\nand From (2),
\nx + 3y = 2k …………(4)
\nHere, |D| = \\(\\left|\\begin{array}{rr}
\n2 & -3 \\\\
\n1 & 3
\n\\end{array}\\right|\\)
\n= 6 + 3
\n= 9 \u2260 0
\nAlso, |D1<\/sub>| = \\(\\left|\\begin{array}{rr}
\nk & -3 \\\\
\n2 k & 3
\n\\end{array}\\right|\\)
\n= 3k + 6k
\n= 9k
\nand |D2<\/sub>| = \\(\\left|\\begin{array}{rr}
\n2 & k \\\\
\n1 & 2 k
\n\\end{array}\\right|\\)
\n= 4k – k
\n= 3k
\n\u2234 by Cramer\u2019s rule,
\nx = \\(\\frac{\\left|D_1\\right|}{|D|}\\)
\n= \\(\\frac{9 k}{9}\\) = k ;
\nand y = \\(\\frac{\\left|\\mathrm{D}_2\\right|}{|\\mathrm{D}|}\\)
\n= \\(\\frac{3 k}{9}\\)
\n= \\(\\frac{k}{3}\\)
\nPutting in eqn. (3) ; we get
\nL.H.S. = k – 3 \\(\\left(\\frac{k}{3}\\right)\\)
\n= k – k = 0
\n= R.H.S.
\nHence x = k, y = \\(\\frac{k}{3}\\) and z = k is a solution of given system for all values of k \u2208 R.<\/p>\n

\"ML<\/p>\n

Question 12.
\nIf the system of equations x – ky – z = 0, kx – y – z = 0, x + y – z = 0 has a non-zero solution, then find the possible values of k.
\nSolution:
\nGiven system of equations is;
\nx – ky – z = 0
\nkx – y – z = 0
\nx + y – z = 0
\nThe given homogeneous system of equations has non-trivial solution then
\nD = \\(\\left|\\begin{array}{rrr}
\n1 & -k & -1 \\\\
\nk & -1 & -1 \\\\
\n1 & 1 & -1
\n\\end{array}\\right|\\) = 0
\n\u21d2 1 (1 + 1) + k (- k + 1) – 1 (k + 1) = 0
\n\u21d2 2 – k2<\/sup> + k – k – 1 = O
\n\u21d2 k2<\/sup> = 1
\n\u21d2 k = \u00b1 1.<\/p>\n

Question 13.
\nFind the real value(s) of a for which the system of equations x + ay = 0, y + az = 0, z + ax = 0 has infinitely many solutions.
\nSolution:
\nThe homogeneous system of equations is;
\nx + ay = 0
\ny + az = 0
\nz + ax = 0
\nThe given homogeneous system of eqn’s has infinitely many solutions.
\n\u2234 |D| = 0
\n\u21d2 \\(\\left|\\begin{array}{lll}
\n1 & a & 0 \\\\
\n0 & 1 & a \\\\
\na & 0 & 1
\n\\end{array}\\right|\\) = 0
\nExpanding along r1<\/sub> ; we have
\n1 (1 – 0) – a (0 – a2<\/sup>) = 0
\n\u21d2 1 + a3<\/sup> = 0
\n\u21d2 (a + 1) (a2<\/sup> – a + 1) = 0
\na + 1 = 0 or
\na2<\/sup> – a + 1 = 0
\nit does not give real values of a.
\ni.e., a = – 1.<\/p>\n

\"ML<\/p>\n

Question 14.
\nIf the equations x = cy + bz, y = az + cx, z = bx + ay are consistent, prove that a2<\/sup> + b2<\/sup> + c2<\/sup> + 2abc = 1.
\nSolution:
\nGiven homogeneous system of eqn’s is equivalent to AX = O
\nwhere A = \\(\\left[\\begin{array}{rrr}
\n-1 & c & b \\\\
\nc & -1 & a \\\\
\nb & a & -1
\n\\end{array}\\right]\\) ;
\nX = \\(\\left[\\begin{array}{l}
\nx \\\\
\ny \\\\
\nz
\n\\end{array}\\right]\\) ;
\nO = \\(\\left[\\begin{array}{l}
\n0 \\\\
\n0 \\\\
\n0
\n\\end{array}\\right]\\)
\nThe given system of equations have non-trivial or infinitely many solution or consistent.
\nThen |A| = 0
\n\u21d2 \\(\\left|\\begin{array}{rrr}
\n-1 & c & b \\\\
\nc & -1 & a \\\\
\nb & a & -1
\n\\end{array}\\right|\\) = 0
\nexpanding along R1<\/sub> ; we have
\n– 1 (1 – a2<\/sup>) – c (- c – ab) + b (ac + b) = 0
\n\u21d2 – 1 + a2<\/sup> + c2<\/sup> + b2<\/sup> + 2abc = 0
\n\u21d2 a2<\/sup> + b 2<\/sup> + c2<\/sup> + 2abc = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

Continuous practice using ML Aggarwal Class 12 Solutions ISC Chapter 4 Determinants Ex 4.6 can lead to a stronger grasp of mathematical concepts. ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.6 Using cramer’s rule, solve the following (1 to 7) systems of linear equations: 1. (i) 3x + y = …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162600"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=162600"}],"version-history":[{"count":2,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162600\/revisions"}],"predecessor-version":[{"id":162605,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162600\/revisions\/162605"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=162600"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=162600"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=162600"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}