{"id":541031,"date":"2021-09-25T10:22:29","date_gmt":"2021-09-25T10:22:29","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=541031"},"modified":"2022-12-26T09:28:25","modified_gmt":"2022-12-26T09:28:25","slug":"rd-sharma-solutions-for-class-12-maths-chapter-8-solution-of-simultaneous-linear-equations","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-12-maths-chapter-8-solution-of-simultaneous-linear-equations\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Chapter 8\u2013Solution of Simultaneous Linear Equations"},"content":{"rendered":"\n

Class 12: Maths Chapter 8 solutions. Complete Class 12 Maths Chapter 8 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 8\u2013Solution of Simultaneous Linear Equations<\/h2>\n\n\n\n

RD Sharma 12th Maths Chapter 8, Class 12 Maths Chapter 8 solutions<\/p>\n\n\n\n

Exercise 8.1 Page No: 8.14<\/p>\n\n\n\n

1. Solve the following system of equations by matrix method:<\/strong><\/p>\n\n\n\n

(i) 5x + 7y + 2 = 0<\/strong><\/p>\n\n\n\n

4x + 6y + 3 = 0<\/strong><\/p>\n\n\n\n

(ii) 5x + 2y = 3<\/strong><\/p>\n\n\n\n

3x + 2y = 5<\/strong><\/p>\n\n\n\n

(iii) 3x + 4y \u2013 5 = 0<\/strong><\/p>\n\n\n\n

x \u2013 y + 3 = 0<\/strong><\/p>\n\n\n\n

(iv) 3x + y = 19<\/strong><\/p>\n\n\n\n

3x \u2013 y = 23<\/strong><\/p>\n\n\n\n

(v) 3x + 7y = 4<\/strong><\/p>\n\n\n\n

x + 2y = -1<\/strong><\/p>\n\n\n\n

(vi) 3x + y = 7<\/strong><\/p>\n\n\n\n

5x + 3y = 12<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given 5x + 7y + 2 = 0 and 4x + 6y + 3 = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Hence, x = 9\/2 and y = -7\/2<\/p>\n\n\n\n

(ii) Given 5x + 2y = 3<\/p>\n\n\n\n

3x + 2y = 5<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Hence, x = -1 and y = 4<\/p>\n\n\n\n

(iii) Given 3x + 4y \u2013 5 = 0<\/p>\n\n\n\n

x \u2013 y + 3 = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Hence, X = 1 Y = \u2013 2<\/p>\n\n\n\n

(iv) Given 3x + y = 19<\/p>\n\n\n\n

3x \u2013 y = 23<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(v) Given 3x + 7y = 4<\/p>\n\n\n\n

x + 2y = -1<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(vi) Given 3x + y = 7<\/p>\n\n\n\n

5x + 3y = 12<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

2. Solve the following system of equations by matrix method:<\/strong><\/p>\n\n\n\n

(i) x + y \u2013z = 3
2x + 3y + z = 10
3x \u2013 y \u2013 7z = 1<\/strong><\/p>\n\n\n\n

(ii) x + y + z = 3<\/strong><\/p>\n\n\n\n

2x \u2013 y + z = -1<\/strong><\/p>\n\n\n\n

2x + y \u2013 3z = -9<\/strong><\/p>\n\n\n\n

(iii) 6x \u2013 12y + 25z = 4<\/strong><\/p>\n\n\n\n

4x + 15y \u2013 20z = 3<\/strong><\/p>\n\n\n\n

2x + 18y + 15z = 10<\/strong><\/p>\n\n\n\n

(iv) 3x + 4y + 7z = 14<\/strong><\/p>\n\n\n\n

2x \u2013 y + 3z = 4<\/strong><\/p>\n\n\n\n

x + 2y \u2013 3z = 0<\/strong><\/p>\n\n\n\n

(v) (2\/x) \u2013 (3\/y) + (3\/z) = 10<\/strong><\/p>\n\n\n\n

(1\/x) + (1\/y) + (1\/z) = 10<\/strong><\/p>\n\n\n\n

(3\/x) \u2013 (1\/y) + (2\/z) = 13<\/strong><\/p>\n\n\n\n

(vi) 5x + 3y + z = 16<\/strong><\/p>\n\n\n\n

2x + y + 3z = 19<\/strong><\/p>\n\n\n\n

x + 2y + 4z = 25<\/strong><\/p>\n\n\n\n

(vii) 3x + 4y + 2z = 8<\/strong><\/p>\n\n\n\n

2y \u2013 3z = 3<\/strong><\/p>\n\n\n\n

x \u2013 2y + 6z = -2<\/strong><\/p>\n\n\n\n

(viii) 2x + y + z = 2<\/strong><\/p>\n\n\n\n

x + 3y \u2013 z = 5<\/strong><\/p>\n\n\n\n

3x + y \u2013 2z = 6<\/strong><\/p>\n\n\n\n

(ix) 2x + 6y = 2<\/strong><\/p>\n\n\n\n

3x \u2013 z = -8<\/strong><\/p>\n\n\n\n

2x \u2013 y + z = -3<\/strong><\/p>\n\n\n\n

(x) 2y \u2013 z = 1<\/strong><\/p>\n\n\n\n

x \u2013 y + z = 2<\/strong><\/p>\n\n\n\n

2x \u2013 y = 0<\/strong><\/p>\n\n\n\n

(xi) 8x + 4y + 3z = 18<\/strong><\/p>\n\n\n\n

2x + y + z = 5<\/strong><\/p>\n\n\n\n

x + 2y + z = 5<\/strong><\/p>\n\n\n\n

(xii) x + y + z = 6<\/strong><\/p>\n\n\n\n

x + 2z = 7<\/strong><\/p>\n\n\n\n

3x + y + z = 12<\/strong><\/p>\n\n\n\n

(xiii) (2\/x) + (3\/y) + (10\/z) = 4,<\/strong><\/p>\n\n\n\n

(4\/x) \u2013 (6\/y) + (5\/z) = 1,<\/strong><\/p>\n\n\n\n

(6\/x) + (9\/y) \u2013 (20\/z) = 2, x, y, z \u2260 0<\/strong><\/p>\n\n\n\n

(xiv) x \u2013 y + 2z = 7<\/strong><\/p>\n\n\n\n

3x + 4y \u2013 5z = -5<\/strong><\/p>\n\n\n\n

2x \u2013 y + 3z = 12<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given x + y \u2013z = 3<\/p>\n\n\n\n

2x + 3y + z = 10<\/p>\n\n\n\n

3x \u2013 y \u2013 7z = 1<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= (\u2013 20) \u2013 1(\u2013 17) \u2013 1(11)<\/p>\n\n\n\n

= \u2013 20 + 17 + 11 = 8<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub><\/strong> = (\u2013 1)1 + 1<\/sup> \u2013 21 + 1 = \u2013 20<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 7 \u2013 1 = 8<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> 1 + 3 = 4<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 14 \u2013 3 = 17<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 7 + 3 = \u2013 4<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 1 + 2 = \u2013 3<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 2 \u2013 9 = \u2013 11<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 1 \u2013 3 = 4<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 3 \u2013 2 = 1<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(ii) Given x + y + z = 3<\/p>\n\n\n\n

2x \u2013 y + z = -1<\/p>\n\n\n\n

2x + y \u2013 3z = -9<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= (3 \u2013 1) \u2013 1(\u2013 6 \u2013 2) + 1(2 + 2)<\/p>\n\n\n\n

= 2 + 8 + 4<\/p>\n\n\n\n

= 14<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 3 \u2013 1 = 2<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 3 \u2013 1 = 4<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> 1 + 1 = 2<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 6 \u2013 2 = 8<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 3 \u2013 2 = \u2013 5<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 1 \u2013 2 = 1<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 2 + 2 = 4<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 1 \u2013 2 = 1<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 1 \u2013 2 = \u2013 3<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iii) Given 6x \u2013 12y + 25z = 4<\/p>\n\n\n\n

4x + 15y \u2013 20z = 3<\/p>\n\n\n\n

2x + 18y + 15z = 10<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 6(225 + 360) + 12(60 + 40) + 25(72 \u2013 30)<\/p>\n\n\n\n

= 3510 + 1200 + 1050<\/p>\n\n\n\n

= 5760<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> (225 + 360) = 585<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> (\u2013 180 \u2013 450) = 630<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> (240 \u2013 375) = \u2013 135<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> (60 + 40) = \u2013 100<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> (90 \u2013 50) = 40<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> (\u2013 120 \u2013 100) = 220<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> (72 \u2013 30) = 42<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup>(108 + 24) = \u2013 132<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> (90 + 48) = 138<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iv) Given 3x + 4y + 7z = 14<\/p>\n\n\n\n

2x \u2013 y + 3z = 4<\/p>\n\n\n\n

x + 2y \u2013 3z = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 3(3 \u2013 6) \u2013 4(\u2013 6 \u2013 3) + 7(4 + 1)<\/p>\n\n\n\n

= \u2013 9 + 36 + 35<\/p>\n\n\n\n

= 62<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 3 \u2013 6 = \u2013 3<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 12 \u2013 14 = 26<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup>12 + 7 = 19<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 6 \u2013 3 = 9<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 3 \u2013 7 = \u2013 10<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 9 \u2013 14 = 5<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 4 + 1 = 5<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 6 \u2013 4 = \u2013 2<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 3 \u2013 8 = \u2013 11<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(v) Given (2\/x) \u2013 (3\/y) + (3\/z) = 10<\/p>\n\n\n\n

(1\/x) + (1\/y) + (1\/z) = 10<\/p>\n\n\n\n

(3\/x) \u2013 (1\/y) + (2\/z) = 13<\/p>\n\n\n\n\n\n

= 5(4 \u2013 6) \u2013 3(8 \u2013 3) + 1(4 \u2013 2)<\/p>\n\n\n\n

= \u2013 10 \u2013 15 + 3<\/p>\n\n\n\n

= \u2013 22<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> (4 \u2013 6) = \u2013 2<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup>(12 \u2013 2) = \u2013 10<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup>(9 \u2013 1) = 8<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> (8 \u2013 3) = \u2013 5<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 20 \u2013 1 = 19<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 15 \u2013 2 = \u2013 13<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> (4 \u2013 2) = 2<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 10 \u2013 3 = \u2013 7<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 5 \u2013 6 = \u2013 1<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(vi) Given 5x + 3y + z = 16<\/p>\n\n\n\n

2x + y + 3z = 19<\/p>\n\n\n\n

x + 2y + 4z = 25<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 3(12 \u2013 6) \u2013 4(0 + 3) + 2(0 \u2013 2)<\/p>\n\n\n\n

= 18 \u2013 12 \u2013 4<\/p>\n\n\n\n

= 2<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> (12 \u2013 6) = 6<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup>(24 + 4) = \u2013 28<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup>(\u2013 12 \u2013 4) = \u2013 16<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> (0 + 3) = \u2013 3<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 18 \u2013 2 = 16<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 9 \u2013 0 = 9<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> (0 \u2013 2) = \u2013 2<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> (\u2013 6 \u2013 4) = 10<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 6 \u2013 0 = 6<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(vii) Given 3x + 4y + 2z = 8<\/p>\n\n\n\n

2y \u2013 3z = 3<\/p>\n\n\n\n

x \u2013 2y + 6z = -2<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 2(\u2013 6 + 1) \u2013 1(\u2013 2 + 3) + 1(1 \u2013 9)<\/p>\n\n\n\n

= \u2013 10 \u2013 1 \u2013 8<\/p>\n\n\n\n

= \u2013 19<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> \u2013 6 + 1 = \u2013 5<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup>(24 + 4) = \u2013 28<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 1 \u2013 3 = \u2013 4<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 2 + 3 = \u2013 1<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 4 \u2013 3 = \u2013 7<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 2 \u2013 1 = 3<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup>1 \u2013 9 = \u2013 8<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup>2 \u2013 3 = \u2013 1<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 6 \u2013 1 = 5<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(viii) Given 2x + y + z = 2<\/p>\n\n\n\n

x + 3y \u2013 z = 5<\/p>\n\n\n\n

3x + y \u2013 2z = 6<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 2(\u2013 6 + 1) \u2013 1(\u2013 2 + 3) + 1(1 \u2013 9)<\/p>\n\n\n\n

= \u2013 10 \u2013 1 \u2013 8<\/p>\n\n\n\n

= \u2013 19<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> \u2013 6 + 1 = \u2013 5<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup>(24 + 4) = \u2013 28<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 1 \u2013 3 = \u2013 4<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 2 + 3 = \u2013 1<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 4 \u2013 3 = \u2013 7<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 2 \u2013 1 = 3<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup>1 \u2013 9 = \u2013 8<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup>2 \u2013 3 = \u2013 1<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 6 \u2013 1 = 5<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(ix) Given 2x + 6y = 2<\/p>\n\n\n\n

3x \u2013 z = -8<\/p>\n\n\n\n

2x \u2013 y + z = -3<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 2(0 \u2013 1) \u2013 6(3 + 2)<\/p>\n\n\n\n

= \u2013 2 \u2013 30<\/p>\n\n\n\n

= \u2013 32<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 0 \u2013 1 = \u2013 1<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup>6 + 0 = \u2013 6<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 6 = \u2013 6<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 3 + 2 = 5<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 2 \u2013 0 = 2<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 2 \u2013 0 = 2<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 3 \u2013 0 = \u2013 3<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 2 \u2013 12 = 14<\/p>\n\n\n\n

C3<\/sub>3<\/sub><\/strong> = (\u2013 1)3 + 1<\/sup> 0 \u2013 18 = \u2013 18<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(x) Given 2y \u2013 z = 1<\/p>\n\n\n\n

x \u2013 y + z = 2<\/p>\n\n\n\n

2x \u2013 y = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 0 + 4 \u2013 1<\/p>\n\n\n\n

= 3<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 1 \u2013 0 = 1<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup>1 \u2013 2 = 1<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup>0 + 1 = 1<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 2 \u2013 0 = 2<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 1 \u2013 0 = \u2013 1<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 0 \u2013 2 = 2<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 4 \u2013 0 = 4<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 2 \u2013 0 = \u2013 2<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 1 + 2 = 1<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(xi) Given 8x + 4y + 3z = 18<\/p>\n\n\n\n

2x + y + z = 5<\/p>\n\n\n\n

x + 2y + z = 5<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 8(\u2013 1) \u2013 4(1) + 3(3)<\/p>\n\n\n\n

= \u2013 8 \u2013 4 + 9<\/p>\n\n\n\n

= \u2013 3<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 1 \u2013 2 = \u2013 1<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 4 \u2013 6 = 2<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> 4 \u2013 3 = 1<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 2 \u2013 1 = \u2013 1<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 8 \u2013 3 = 5<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 8 \u2013 6 = \u2013 2<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 4 \u2013 1 = 3<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 16 \u2013 4 = \u2013 12<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 8 \u2013 8 = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(xii) Given x + y + z = 6<\/p>\n\n\n\n

x + 2z = 7<\/p>\n\n\n\n

3x + y + z = 12<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

= 1(\u2013 2) \u2013 1(1 \u2013 6) + 1(1)<\/p>\n\n\n\n

= \u2013 2 + 5 + 1<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 0 \u2013 2 = \u2013 2<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 1 \u2013 1 = 0<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> 2 \u2013 0 = 2<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 1 \u2013 6 = 5<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 1 \u2013 3 = \u2013 2<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 2 \u2013 1 = \u2013 1<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 1 \u2013 0 = 1<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 1 \u2013 3 = 2<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 0 \u2013 1 = \u2013 1<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(xiii) Given (2\/x) + (3\/y) + (10\/z) = 4,<\/p>\n\n\n\n

(4\/x) \u2013 (6\/y) + (5\/z) = 1,<\/p>\n\n\n\n

(6\/x) + (9\/y) \u2013 (20\/z) = 2, x, y, z \u2260 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

AX = B<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

|A| = 2(75) \u2013 3(\u2013 110) + 10(72)<\/p>\n\n\n\n

= 150 + 330 + 720<\/p>\n\n\n\n

= 1200<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 120 \u2013 45 = 75<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 60 \u2013 90 = 150<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> 15 + 60 = 75<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 80 \u2013 30 = 110<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 40 \u2013 60 = \u2013 100<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 10 \u2013 40 = 30<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 36 + 36 = 72<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 18 \u2013 18 = 0<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 12 \u2013 12 = \u2013 24<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(xiv) Given x \u2013 y + 2z = 7<\/p>\n\n\n\n

3x + 4y \u2013 5z = -5<\/p>\n\n\n\n

2x \u2013 y + 3z = 12<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

A X = B<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

|A| = 1(12 \u2013 5) + 1(9 + 10) + 2(\u2013 3 \u2013 8)<\/p>\n\n\n\n

= 7 + 19 \u2013 22<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

So, the above system has a unique solution, given by<\/p>\n\n\n\n

X = A \u2013 1<\/sup>B<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 12 \u2013 5 = 7<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 3 + 2 = 1<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> 5 \u2013 8 = \u2013 3<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 9 + 10 = \u2013 19<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 3 \u2013 4 = \u2013 1<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 5 \u2013 6 = 11<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 3 \u2013 8 = \u2013 11<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 1 + 2 = \u2013 1<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 4 + 3 = 7<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

3. Show that each one of the following systems of linear equations is consistent and also find their solutions:<\/strong><\/p>\n\n\n\n

(i) 6x + 4y = 2<\/strong><\/p>\n\n\n\n

9x + 6y = 3<\/strong><\/p>\n\n\n\n

(ii) 2x + 3y = 5<\/strong><\/p>\n\n\n\n

6x + 9y = 15<\/strong><\/p>\n\n\n\n

(iii) 5x + 3y + 7z = 4<\/strong><\/p>\n\n\n\n

3x + 26y + 2z = 9<\/strong><\/p>\n\n\n\n

7x + 2y + 10z = 5<\/strong><\/p>\n\n\n\n

(v) x + y + z = 6<\/strong><\/p>\n\n\n\n

x + 2y + 3z = 14<\/strong><\/p>\n\n\n\n

x + 4y + 7z = 30<\/strong><\/p>\n\n\n\n

(vi) 2x + 2y \u2013 2z = 1<\/strong><\/p>\n\n\n\n

4x + 4y \u2013 z = 2<\/strong><\/p>\n\n\n\n

6x + 6y + 2z = 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given 6x + 4y = 2<\/p>\n\n\n\n

9x + 6y = 3<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 36 \u2013 36 = 0<\/p>\n\n\n\n

So, A is singular, Now X will be consistence if (Adj A) x B = 0<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 6 = 6<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 9 = \u2013 9<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 4 = \u2013 4<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 2<\/sup> 6 = 6<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(ii) Given 2x + 3y = 5<\/p>\n\n\n\n

6x + 9y = 15<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 18 \u2013 18 = 0<\/p>\n\n\n\n

So, A is singular,<\/p>\n\n\n\n

Now X will be consistence if (Adj A) x B = 0<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 9 = 9<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 6 = \u2013 6<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 3 = \u2013 3<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 2<\/sup> 2 = 2<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(iii) Given 5x + 3y + 7z = 4<\/p>\n\n\n\n

3x + 26y + 2z = 9<\/p>\n\n\n\n

7x + 2y + 10z = 5<\/p>\n\n\n\n\n\n

|A| = 5(260 \u2013 4) \u2013 3(30 \u2013 14) + 7(6 \u2013 182)<\/p>\n\n\n\n

= 5(256) \u2013 3(16) + 7(176)<\/p>\n\n\n\n

|A| = 0<\/p>\n\n\n\n

So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solution according to as:<\/p>\n\n\n\n

(Adj A) x B\u22600 or (Adj A) x B = 0<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 260 \u2013 4 = 256<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 30 \u2013 14 = \u2013 16<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> 6 \u2013 182 = \u2013 176<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 30 \u2013 14 = \u2013 16<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 50 \u2013 49 = 1<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 10 \u2013 21 = 11<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 6 \u2013 182 = \u2013 176<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 10 \u2013 21 = 11<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 130 \u2013 9 = 121<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now, AX = B has infinite many solution<\/p>\n\n\n\n

Let z = k<\/p>\n\n\n\n

Then, 5x + 3y = 4 \u2013 7k<\/p>\n\n\n\n

3x + 26y = 9 \u2013 2k<\/p>\n\n\n\n

This can be written as<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(v) Given x + y + z = 6<\/p>\n\n\n\n

x + 2y + 3z = 14<\/p>\n\n\n\n

x + 4y + 7z = 30<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 1(2) \u2013 1(4) + 1(2)<\/p>\n\n\n\n

= 2 \u2013 4 + 2<\/p>\n\n\n\n

|A| = 0<\/p>\n\n\n\n

So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solution according to as:<\/p>\n\n\n\n

(Adj A) x B\u22600 or (Adj A) x B = 0<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 14 \u2013 12 = 2<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 7 \u2013 4 = \u2013 3<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> 3 \u2013 2 = 1<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 7 \u2013 3 = \u2013 4<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 7 \u2013 1 = 6<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 3 \u2013 1 = 2<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 4 \u2013 2 = 2<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 4 \u2013 1 = \u2013 3<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 2 \u2013 1 = 1<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now, AX = B has infinite many solution<\/p>\n\n\n\n

Let z = k<\/p>\n\n\n\n

Then, x + y = 6 \u2013 k<\/p>\n\n\n\n

x + 2y = 14 \u2013 3k<\/p>\n\n\n\n

This can be written as:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(vi) Given x + y + z = 6<\/p>\n\n\n\n

x + 2y + 3z = 14<\/p>\n\n\n\n

x + 4y + 7z = 30<\/p>\n\n\n\n

This can be written as<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 2(14) \u2013 2(14) \u2013 2(0)<\/p>\n\n\n\n

|A| = 0<\/p>\n\n\n\n

So, A is singular. Thus, the given system is either inconsistent or it is consistent<\/p>\n\n\n\n

with infinitely many solution according to as:<\/p>\n\n\n\n

(Adj A) x B\u22600 or (Adj A) x B = 0<\/p>\n\n\n\n

Cofactors of A are:<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup>8 + 6 = 14<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 4 + 12 = \u2013 16<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 2 + 8 = 6<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 8 + 6 = \u2013 14<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 4 + 12 = 16<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 2 + 8 = \u2013 6<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 24 \u2013 24 = 0<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 12 \u2013 12 = 0<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 8 \u2013 8 = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

4. Show that each one of the following systems of linear equations is consistent:<\/strong><\/p>\n\n\n\n

(i) 2x + 5y = 7<\/strong><\/p>\n\n\n\n

6x + 15y = 13<\/strong><\/p>\n\n\n\n

(ii) 2x + 3y = 5<\/strong><\/p>\n\n\n\n

6x + 9y = 10<\/strong><\/p>\n\n\n\n

(iii) 4x \u2013 2y = 3<\/strong><\/p>\n\n\n\n

6x \u2013 3y = 5<\/strong><\/p>\n\n\n\n

(iv) 4x \u2013 5y \u2013 2z = 2<\/strong><\/p>\n\n\n\n

5x \u2013 4y + 2z = -2<\/strong><\/p>\n\n\n\n

2x + 2y + 8z = -1<\/strong><\/p>\n\n\n\n

(v) 3x \u2013 y \u2013 2z = 2<\/strong><\/p>\n\n\n\n

2y \u2013 z = -1<\/strong><\/p>\n\n\n\n

3x \u2013 5y = 3<\/strong><\/p>\n\n\n\n

(vi) x + y \u2013 2z = 5<\/strong><\/p>\n\n\n\n

x \u2013 2y + z = -2<\/strong><\/p>\n\n\n\n

-2x + y + z = 4<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given 2x + 5y = 7<\/p>\n\n\n\n

6x + 15y = 13<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 30 \u2013 30 = 0<\/p>\n\n\n\n

So, A is singular,<\/p>\n\n\n\n

Now X will be consistence if (Adj A) x B = 0<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 15 = 15<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 6 = \u2013 6<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 5 = \u2013 5<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 2<\/sup> 2 = 2<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(ii) Given 2x + 3y = 5<\/p>\n\n\n\n

6x + 9y = 10<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 18 \u2013 18 = 0<\/p>\n\n\n\n

So, A is singular,<\/p>\n\n\n\n

Now X will be consistence if (Adj A) x B = 0<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 9 = 9<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 6 = \u2013 6<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 3 = \u2013 3<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 2<\/sup> 2 = 2<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iii) Given 4x \u2013 2y = 3<\/p>\n\n\n\n

6x \u2013 3y = 5<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = \u2013 12 + 12 = 0<\/p>\n\n\n\n

So, A is singular,<\/p>\n\n\n\n

Now X will be consistence if (Adj A) x B = 0<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> \u2013 3 = \u2013 3<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 6 = \u2013 6<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 2 = 2<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 2<\/sup> 4 = 4<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iv) Given 4x \u2013 5y \u2013 2z = 2<\/p>\n\n\n\n

5x \u2013 4y + 2z = -2<\/p>\n\n\n\n

2x + 2y + 8z = -1<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 4(\u2013 36) + 5(36) \u2013 2(18)<\/p>\n\n\n\n

|A| = 0<\/p>\n\n\n\n

Cofactors of A are:<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> \u2013 32 \u2013 4 = \u2013 36<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 40 + 4 = \u2013 36<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 10 \u2013 8 = \u2013 18<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 40 \u2013 4 = \u2013 36<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 32 + 4 = 36<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 8 + 10 = \u2013 18<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 10 + 8 = 18<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 8 + 10 = \u2013 18<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 16 + 25 = 9<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(v) Given 3x \u2013 y \u2013 2z = 2<\/p>\n\n\n\n

2y \u2013 z = -1<\/p>\n\n\n\n

3x \u2013 5y = 3<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 3(\u2013 5) + 1(3) \u2013 2(\u2013 6)<\/p>\n\n\n\n

|A| = 0<\/p>\n\n\n\n

Cofactors of A are<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 0 \u2013 5 = \u2013 5<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 0 \u2013 10 = 10<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> 1 + 4 = 5<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 0 + 3 = \u2013 3<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 0 + 6 = 6<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 3 \u2013 0 = 3<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 0 \u2013 6 = \u2013 6<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 15 + 3 = 12<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 6 \u2013 0 = 6<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(vi) Given x + y \u2013 2z = 5<\/p>\n\n\n\n

x \u2013 2y + z = -2<\/p>\n\n\n\n

-2x + y + z = 4<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 1(\u2013 3) \u2013 1(3) \u2013 2(\u2013 3) = \u2013 3 \u2013 3 + 6<\/p>\n\n\n\n

|A| = 0<\/p>\n\n\n\n

Cofactors of A are:<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> \u2013 2 \u2013 1 = \u2013 3<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 1 + 2 = \u2013 3<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> 1 \u2013 4 = \u2013 3<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 1 + 2 = \u2013 3<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 1 \u2013 4 = \u2013 3<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 1 + 2 = \u2013 3<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 1 \u2013 4 = \u2013 3<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 1 + 2 = \u2013 3<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 2 \u2013 1 = \u2013 3<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

x \u2013 y = 3, 2x + 3y + 4z = 17, y + 2z = 7<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

2x \u2013 3y + 5z = 11, 3x + 2y \u2013 4z = -5, x + y \u2013 2z = -3.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 2(0) + 3(\u2013 2) + 5(1)<\/p>\n\n\n\n

= \u2013 1<\/p>\n\n\n\n

Now, the cofactors of A<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> \u2013 4 + 4 = 0<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 6 \u2013 5 = \u2013 1<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> 12 \u2013 10 = 2<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 6 + 4 = 2<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 4 \u2013 5 = \u2013 9<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 8 \u2013 15 = 23<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 3 \u2013 2 = 1<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 2 + 3 = \u2013 5<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 4 + 9 = 13<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

x + 2y + 5z = 10, x \u2013 y \u2013 z = -2, 2x + 3y \u2013 z = -11.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 1(1 + 3) + 2(\u2013 1 + 2) + 5(3 + 2)<\/p>\n\n\n\n

= 4 + 2 + 25<\/p>\n\n\n\n

= 27<\/p>\n\n\n\n

Now, the cofactors of A<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 1 + 3 = 4<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 2 \u2013 15 = 17<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 2 + 5 = 3<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 1 + 2 = \u2013 1<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 1 \u2013 10 = \u2013 11<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 1 \u2013 5 = 6<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 3 + 2 = 5<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 3 \u2013 4 = 1<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 1 \u2013 2 = \u2013 3<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 1(1 + 6) + 2(2 \u2013 0) + 0<\/p>\n\n\n\n

= 7 + 4<\/p>\n\n\n\n

= 11<\/p>\n\n\n\n

Now, the cofactors of A<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 1 + 6 = 7<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 2 \u2013 0 = 2<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 6 \u2013 0 = \u2013 6<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 2 \u2013 0 = \u2013 2<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 1 \u2013 0 = 1<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 3 \u2013 0 = \u2013 3<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 4 \u2013 0 = \u2013 4<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 2 \u2013 0 = 2<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 1 + 4 = 5<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 3(3 \u2013 0) + 4(2 \u2013 5) + 2(0 \u2013 3)<\/p>\n\n\n\n

= 9 \u2013 12 \u2013 6<\/p>\n\n\n\n

= \u2013 9<\/p>\n\n\n\n

Now, the cofactors of A<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> 3 \u2013 0 = 3<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 4 \u2013 0 = 4<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 20 \u2013 6 = \u2013 26<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 2 \u2013 5 = 3<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 3 \u2013 2 = 1<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> 15 \u2013 4 = \u2013 11<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> 0 \u2013 3 = \u2013 3<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> 0 + 4 = \u2013 4<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> 9 + 8 = 17<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

|A| = 1(\u2013 1 \u2013 1) \u2013 2(\u2013 2 \u2013 0) + 0<\/p>\n\n\n\n

= \u2013 2 + 4<\/p>\n\n\n\n

= 2<\/p>\n\n\n\n

Now, the cofactors of A<\/p>\n\n\n\n

C11<\/sub> = (\u2013 1)1 + 1<\/sup> \u2013 1 \u2013 1 = \u2013 2<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> 2 \u2013 0 = 2<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 2 \u2013 0 = \u2013 2<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> 2 \u2013 0 = \u2013 2<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 1<\/sup> 1 \u2013 0 = 1<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 1 \u2013 0 = 1<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 2<\/sup> \u2013 2 \u2013 0 = \u2013 2<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 1<\/sup> \u2013 1 \u2013 0 = 1<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 1<\/sup> \u2013 1 + 4 = 3<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

9. The sum of three numbers is 2. If twice the second number is added to the sum of first and third, the sum is 1. By adding second and third number to five times the first number, we get 6. Find the three numbers by using matrices.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the numbers are x, y, z<\/p>\n\n\n\n

x + y + z = 2
<\/a>\u2026\u2026 (i)<\/p>\n\n\n\n

Also, 2y + (x + z) + 1<\/p>\n\n\n\n

x + 2y + z = 1 \u2026\u2026 (ii)<\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

x + z + 5(x) = 6<\/p>\n\n\n\n

5x + y + z = 6 \u2026\u2026 (iii)<\/p>\n\n\n\n\n\n

A X = B<\/p>\n\n\n\n

|A| = 1(1) \u2013 1(\u2013 4) + 1(\u2013 9)<\/p>\n\n\n\n

= 1 + 4 \u2013 9<\/p>\n\n\n\n

= \u2013 4<\/p>\n\n\n\n

Hence, the unique solution given by x = A \u2013 1<\/sup>B<\/p>\n\n\n\n

C11 =<\/sub> (\u2013 1)1 + 1<\/sup> (2 \u2013 1) = 1<\/p>\n\n\n\n

C12<\/sub> = (\u2013 1)1 + 2<\/sup> (1 \u2013 5) = 4<\/p>\n\n\n\n

C13<\/sub> = (\u2013 1)1 + 3<\/sup> (1 \u2013 10) = \u2013 9<\/p>\n\n\n\n

C21<\/sub> = (\u2013 1)2 + 1<\/sup> (1 \u2013 1) = 0<\/p>\n\n\n\n

C22<\/sub> = (\u2013 1)2 + 2<\/sup> (1 \u2013 5) = \u2013 4<\/p>\n\n\n\n

C23<\/sub> = (\u2013 1)2 + 3<\/sup> (1 \u2013 5) = 4<\/p>\n\n\n\n

C31<\/sub> = (\u2013 1)3 + 1<\/sup> (1 \u2013 2) = \u2013 1<\/p>\n\n\n\n

C32<\/sub> = (\u2013 1)3 + 2<\/sup> (1 \u2013 1) = 0<\/p>\n\n\n\n

C33<\/sub> = (\u2013 1)3 + 3<\/sup> (2 \u2013 1) = 1<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

10. An amount of \u20b910,000 is put into three investments at the rate of 10, 12 and 15% per annum. The combined incomes are \u20b91310 and the combined income of first and second investment is \u20b9 190 short of the income from the third. Find the investment in each using matrix method.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the numbers are x, y, and z<\/p>\n\n\n\n

x + y + z = 10,000 \u2026\u2026 (i)<\/p>\n\n\n\n

Also,<\/p>\n\n\n\n

0.1x + 0.12y + 0.15z = 1310 \u2026\u2026 (ii)<\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

0.1x + 0.12y \u2013 0.15z = \u2013 190 \u2026\u2026 (iii)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

A X = B<\/p>\n\n\n\n

|A| = 1(\u2013 0.036) \u2013 1(\u2013 0.03) + 1(0)<\/p>\n\n\n\n

= \u2013 0.006<\/p>\n\n\n\n

Hence, the unique solution given by x = A \u2013 1<\/sup>B<\/p>\n\n\n\n

C11 =<\/sub> \u2013 0.036<\/p>\n\n\n\n

C12<\/sub> = 0.27<\/p>\n\n\n\n

C13<\/sub> = 0<\/p>\n\n\n\n

C21<\/sub> = 0.27<\/p>\n\n\n\n

C22<\/sub> = \u2013 0.25<\/p>\n\n\n\n

C23<\/sub> = \u2013 0.02<\/p>\n\n\n\n

C31<\/sub> = 0.03<\/p>\n\n\n\n

C32<\/sub> = \u2013 0.05<\/p>\n\n\n\n

C33<\/sub> = 0.02<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Exercise 8.2 Page No: 8.20<\/p>\n\n\n\n

Solve the following systems of homogeneous linear equations by matrix method:<\/p>\n\n\n\n

1. 2x \u2013 y + z = 0<\/strong><\/p>\n\n\n\n

3x + 2y \u2013 z = 0<\/strong><\/p>\n\n\n\n

x + 4y + 3z = 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

2x \u2013 y + z = 0<\/p>\n\n\n\n

3x + 2y \u2013 z = 0<\/p>\n\n\n\n

X + 4y + 3z = 0<\/p>\n\n\n\n

The system can be written as<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

A X = 0<\/p>\n\n\n\n

Now, |A| = 2(6 + 4) + 1(9 + 1) + 1(12 \u2013 2)<\/p>\n\n\n\n

|A| = 2(10) + 10 + 10<\/p>\n\n\n\n

|A| = 40 \u2260 0<\/p>\n\n\n\n

Since, |A|\u2260 0, hence x = y = z = 0 is the only solution of this homogeneous equation.<\/p>\n\n\n\n

2. 2x \u2013 y + 2z = 0
5x + 3y \u2013 z = 0
X + 5y \u2013 5z = 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given 2x \u2013 y + 2z = 0<\/p>\n\n\n\n

5x + 3y \u2013 z = 0<\/p>\n\n\n\n

X + 5y \u2013 5z = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

A X = 0<\/p>\n\n\n\n

Now, |A| = 2(\u2013 15 + 5) + 1(\u2013 25 + 1) + 2(25 \u2013 3)<\/p>\n\n\n\n

|A| = \u2013 20 \u2013 24 + 44<\/p>\n\n\n\n

|A| = 0<\/p>\n\n\n\n

Hence, the system has infinite solutions<\/p>\n\n\n\n

Let z = k<\/p>\n\n\n\n

2x \u2013 y = \u2013 2k<\/p>\n\n\n\n

5x + 3y = k<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

3. 3x \u2013 y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given 3x \u2013 y + 2z = 0<\/p>\n\n\n\n

4x + 3y + 3z = 0<\/p>\n\n\n\n

5x + 7y + 4z = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

A X = 0<\/p>\n\n\n\n

Now, |A| = 3(12 \u2013 21) + 1(16 \u2013 15) + 2(28 \u2013 15)<\/p>\n\n\n\n

|A| = \u2013 27 + 1 + 26<\/p>\n\n\n\n

|A| = 0<\/p>\n\n\n\n

Hence, the system has infinite solutions<\/p>\n\n\n\n

Let z = k<\/p>\n\n\n\n

3x \u2013 y = \u2013 2k<\/p>\n\n\n\n

4x + 3y = \u2013 3k<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

4. x + y \u2013 6z = 0
x \u2013 y + 2z = 0
\u2013 3x + y + 2z = 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given x + y \u2013 6z = 0<\/p>\n\n\n\n

x \u2013 y + 2z = 0<\/p>\n\n\n\n

\u2013 3x + y + 2z = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

A X = 0<\/p>\n\n\n\n

Now, |A| = 1(\u2013 2 \u2013 2) \u2013 1(2 + 6) \u2013 6(1 \u2013 3)<\/p>\n\n\n\n

|A| = \u2013 4 \u2013 8 + 12<\/p>\n\n\n\n

|A| = 0<\/p>\n\n\n\n

Hence, the system has infinite solutions<\/p>\n\n\n\n

Let z = k<\/p>\n\n\n\n

x + y = 6k<\/p>\n\n\n\n

x \u2013 y = \u2013 2k<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 8: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 8\u2013Solution of Simultaneous Linear Equations<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 12 Maths Chapter 8\u2013Solution of Simultaneous Linear Equations PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 12 Maths :<\/strong><\/h2>\n\n\n\n