The RD Sharma Solution books are the best form of tutors for the class 12 students at their home. Mathematics is a subject that gives chances for a lot of doubts to arise. Every mark that a student scores in the public exam is essential; students should not risk losing their marks. RD Sharma class 12th exercise 5.2, Determinants, is a part where students can score easily. Hence utmost care should be taken in solving the problems. With the help of the RD Sharma Class 12th exercise 5.2, students can do their homework effortlessly.
Also Read - RD Sharma Solution for Class 9 to 12 Maths
RD Sharma Class 12 Solutions Chapter 5 Determinants - Other Exercise determinants Exercise 5.2 Question 1
Answer:| 1 3 5 2 6 10 31 11 38 | = 0 Hint: We will solve the determinant with respect to row one.
Given:| 1 3 5 2 6 10 31 11 38 | Solution:| 1 3 5 2 6 10 31 11 38 | = 1 | 6 10 11 38 | − 3 | 2 10 31 38 | + 5 | 2 6 31 11 | = 1 ( 6 × 38 − 10 × 11 ) − 3 ( 2 × 38 − 10 × 31 ) + 5 ( 2 × 11 − 6 × 31 ) = ( 228 − 110 ) − 3 ( 76 − 310 ) + 5 ( 22 − 186 ) = 118 − 3 ( − 234 ) + 5 ( − 164 ) = 118 + 702 − 820 = 820 − 820 = 0 determinants Exercise 5.2 Question 1 Sub Question 2
Answer: Answer: | 67 19 21 39 13 14 81 24 26 | = − 43 Hint: First we will use column operation to get a simple determinant.
Given: | 67 19 21 39 13 14 81 24 26 |
Solution : | 67 19 21 39 13 14 81 24 26 |
Applying C 1 → C 1 − 4 C 3 = | − 17 19 21 − 17 13 14 − 23 24 26 | Applying R 1 → R 1 − R 2 = | 0 6 7 − 17 13 14 − 23 24 26 | Applying R 2 → R 2 − 2 R 1 | 0 6 7 − 17 1 0 − 23 24 26 |
on expanding w.r.t row
= 0 | 1 0 24 26 | − 6 | − 17 0 − 23 26 | + 7 | − 17 1 − 23 24 | = 0 − 6 ( − 17 × 26 − 0 × − 23 ) + 7 ( − 17 × 24 − 1 × − 23 ) = − 6 ( − 442 ) + 7 ( − 408 + 23 ) = 2652 + 7 ( − 385 ) = 2652 − 2695 = − 43
determinants Exercise 5.2 Question 1 Sub Question 3
Answer:| a h g h b f g f c | | a h g h b f g f c | = a b c − a f 2 − c h 2 + 2 f g h − b g 2 = a b c − a f 2 − c h 2 + 2 f g h − b g 2 | a h g h b f g f c | = a b c − a f 2 − c h 2 + 2 f g h − b g 2 | a h g h b f g f c | = a b c − a f 2 − c h 2 + 2 f g h − b g 2 | a h g h b f g f c | = a b c − a f 2 − c h 2 + 2 f g h − b g 2 Hint: We will expand it w.r.t
R 1 Given: | a h g h b f g f c | = a b c − a f 2 − c h 2 + 2 f g h − b g 2 | a h g h b f g f c | Solution: | a h g h b f g f c | Expanding w.r.t
R 1 = a | b f f c | − h | h f g c | + g | h b g f | = a ( b c − f 2 ) − h ( h c − g f ) + g ( h f − b g ) = a b c − a f 2 − h 2 c + f g h + f g h − b g 2 = a b c − a f 2 − c h 2 + 2 f g h − b g 2 determinants Exercise 5.2 Question 1 sub question 4
Answer:| 1 − 3 2 4 − 1 2 3 5 2 | = 40 Hint: First we will use row operation to make some element 0
Given: | 1 − 3 2 4 − 1 2 3 5 2 | Solution: Applying R 1 → R 1 − R 2 and R 2 → R 2 − R 3 = | − 3 − 2 0 1 − 6 0 3 5 2 | Now we will expand it w.r.t R 1 = − 3 | − 6 0 5 2 | − ( − 2 ) | 1 0 3 2 | + 0 | 1 − 6 3 5 | = − 3 ( − 12 − 5 × 0 ) + 2 ( 1 × 2 − 3 × 0 ) + 0 = − 3 ( − 12 ) + 2 ( 2 ) = 36 + 4 = 40 determinants Exercise 5.2 Question 1 sub question 5
Answer:| 1 4 9 4 9 16 9 16 25 | = − 8 Hint: We will expand it w.r.t
R 1 Given:| 1 4 9 4 9 16 9 16 25 | Solution:| 1 4 9 4 9 16 9 16 25 | Expanding w.r.t
R 1 = 1 | 9 16 16 25 | − 4 | 4 16 9 25 | + 9 | 4 9 9 16 | = 1 ( 9 × 25 − 16 × 16 ) − 4 ( 4 × 25 − 9 × 16 ) + 9 ( 4 × 16 − 9 × 9 ) = ( 225 − 256 ) − 4 ( 100 − 144 ) + 9 ( 64 − 81 ) = − 31 + 176 − 153 = 176 − 184 = − 8 determinants Exercise 5.2 Question 1 sub question 6
Answer:| 6 − 3 2 2 − 1 2 − 10 5 2 | = 0 Hint: In this question we will convert some elements into 0
Given:| 6 − 3 2 2 − 1 2 − 10 5 2 | Solution: | 6 − 3 2 2 − 1 2 − 10 5 2 | Applying R 1 → R 1 − R 2 and R 2 → R 2 − R 3 = | 4 − 2 0 12 − 6 0 − 10 5 2 | On expanding it w.r.t
R 1 = 4 | − 6 0 5 2 | − ( − 2 ) | 12 0 − 10 2 | + 0 | 12 − 6 − 10 5 | = 4 ( − 6 × 2 − 5 × 0 ) + 2 ( 12 × 2 − ( − 10 ) × 0 ) + 0 = 4 ( − 12 ) + 2 ( 24 ) = − 48 + 48 = 0 Hence
| 6 − 3 2 2 − 1 2 − 10 5 2 | = 0 determinants Exercise 5.2 Question 1 sub question 7
Answer: | 1 3 9 27 3 9 27 1 9 27 1 3 27 1 3 9 | = 512000 Hint: First we will make a column with same elements
Given: | 1 3 9 27 3 9 27 1 9 27 1 3 27 1 3 9 |
Solution: | 1 3 9 27 3 9 27 1 9 27 1 3 27 1 3 9 |
Applying C 1 → C 1 + C 2 + C 3 + C 4 = | 40 3 9 27 40 9 27 1 40 27 1 3 40 1 3 9 | Applying R 1 → R 1 − R 2 and R 2 → R 2 − R 3 and R 3 → R 3 − R 4 = | 0 − 6 − 18 26 0 − 18 26 − 2 0 26 − 2 − 6 40 1 3 9 | = | 0 0 0 40 − 6 − 18 26 1 − 18 26 − 2 3 26 − 2 − 6 9 | R 1
= 0 | − 18 26 1 26 − 2 3 − 2 − 6 9 | − 0 | − 6 26 1 − 18 − 2 3 26 − 6 9 | + 0 | − 6 − 18 1 − 18 26 3 26 − 2 9 | − 40 | − 6 − 18 26 − 18 26 − 2 26 − 2 − 6 | = 0 − 0 + 0 − 40 { − 6 | 26 − 2 − 2 − 6 | − ( − 18 ) | − 18 − 2 26 − 6 | + 26 | − 18 26 26 − 2 | } = − 40 [ − 6 { 26 ( − 6 ) − ( − 2 ) ( − 2 ) } + 18 { ( − 18 ) ( − 6 ) − ( − 2 ) ( 26 ) } + 26 { ( − 18 ) ( − 2 ) − 26 × 26 } ] = − 40 [ − 6 ( − 156 − 4 ) + 18 ( 108 + 52 ) + 26 ( 36 − 676 ) ] = − 40 [ − 6 ( − 160 ) + 18 ( 160 ) + 26 ( − 640 ) ] = − 40 ( 960 + 2880 − 16640 ) = − 40 ( − 12800 ) = 512000
Hence,| 1 3 9 27 3 9 27 1 9 27 1 3 27 1 3 9 | = 512000
determinants Exercise 5.2 Question 1 Sub Question 8
Answer:| 102 18 36 1 3 4 17 3 6 | = 0 Hint: We will solve the determinant w.r.t
R 1 Given: | 102 18 36 1 3 4 17 3 6 | Solution: | 102 18 36 1 3 4 17 3 6 | = 102 | 3 4 3 6 | − 18 | 1 4 17 6 | + 36 | 1 3 17 3 | = 102 ( 6 × 3 − 3 × 4 ) − 18 ( 6 × 1 − 4 × 17 ) + 36 ( 1 × 3 − 17 × 3 ) = 102 ( 18 − 12 ) − 18 ( 6 − 68 ) + 36 ( 3 − 51 ) = 102 ( 6 ) − 18 ( − 62 ) + 36 ( − 48 ) = 612 + 1116 − 1728 = 1728 − 1728 = 0 Hence
| 102 18 36 1 3 4 17 3 6 | = 0 determinants Exercise 5.2 Question 2
Answer:| 8 2 7 12 3 5 16 4 3 | = 0 Hint: We will try to do any two column or row equal
Given:| 8 2 7 12 3 5 16 4 3 | Solution:| 8 2 7 12 3 5 16 4 3 | On taking common 4 from C 1 = 4 | 2 2 7 3 3 5 4 4 3 | If any two rows or columns of a determinant are identical.
The value of the determinant is zero
= 4 × 0 ( ∵ C 1 = C 2 ) = 0 Hence
| 8 2 7 12 3 5 16 4 3 | = 0 determinants Exercise 5.2 Question 2 Sub Question 2
Answer: | 6 − 3 2 2 − 1 2 − 10 5 2 | = 0 Hint : We will try to do any two column or row equal
Given: | 6 − 3 2 2 − 1 2 − 10 5 2 | Solution :
| 6 − 3 2 2 − 1 2 − 10 5 2 | On taking ( − 2 ) common from C 1 = ( − 2 ) | − 3 − 3 2 − 1 − 1 2 5 5 2 | If any two rows or columns of a determinant are identical.
The value of the determinant is zero
= ( − 2 ) × 0 ( ∵ C 1 = C 2 ) = 0 Hence
| 6 − 3 2 2 − 1 2 − 10 5 2 | = 0 determinants Exercise 5.2 Question 2 Sub Question 3
Answer: | 2 3 7 13 17 5 15 20 12 | = 0 Hint: We will try to do any two column or row equal
Given: | 2 3 7 13 17 5 15 20 12 | Solution: | 2 3 7 13 17 5 15 20 12 | Applying R 1 → R 1 + R 2 = | 15 20 12 13 17 5 15 20 12 | If any two rows or columns of a determinant are identical.
The value of the determinant is zero
= 0 ( ∵ R 1 = R 3 ) Hence,
| 2 3 7 13 17 5 15 20 12 | = 0 determinants Exercise 5.2 Question 2 Sub Question 5
Answer:| a + b 2 a + b 3 a + b 2 a + b 3 a + b 4 a + b 4 a + b 5 a + b 6 a + b | = 0 Hint: We will try to do any two column or row equal
Given:| a + b 2 a + b 3 a + b 2 a + b 3 a + b 4 a + b 4 a + b 5 a + b 6 a + b | Solution:| a + b 2 a + b 3 a + b 2 a + b 3 a + b 4 a + b 4 a + b 5 a + b 6 a + b | Applying C 2 → C 2 − C 1 and C 3 → C 3 − C 2 = | a + b ( 2 a + b ) − ( a + b ) ( 3 a + b ) − ( 2 a + b ) 2 a + b ( 3 a + b ) − ( 2 a + b ) ( 4 a + b ) − ( 3 a + b ) 4 a + b ( 5 a + b ) − ( 4 a + b ) ( 6 a + b ) − ( 5 a + b ) | = | a + b 2 a + b − a − b 3 a + b − 2 a − b 2 a + b 3 a + b − 2 a − b 4 a + b − 3 a − b 4 a + b 5 a + b − 4 a − b 6 a + b − 5 a − b | = | a + b a a 2 a + b a a 4 a + b a a | If any two rows or columns of a determinant are identical.
The value of the determinant is zero
= 0 ( ∵ C 2 = C 3 ) Hence
| a + b 2 a + b 3 a + b 2 a + b 3 a + b 4 a + b 4 a + b 5 a + b 6 a + b | = 0 determinants Exercise 5.2 Question 2 Sub Question 6
Answer:| 1 a a 2 − b c 1 b b 2 − a c 1 c c 2 − a b | = 0 Hint: First we will split the determinant into two determinant
Given: | 1 a a 2 − b c 1 b b 2 − a c 1 c c 2 − a b | Solution:| 1 a a 2 − b c 1 b b 2 − a c 1 c c 2 − a b | = | 1 a a 2 1 b b 2 1 c c 2 | − | 1 a b c 1 b a c 1 c a b | ∵ Using | x 1 x 2 x 3 + a y 1 y 2 y 3 + b z 1 z 2 z 3 + c | = | x 1 x 2 x 3 y 1 y 2 y 3 z 1 z 2 z 3 | + | x 1 x 2 a y 1 y 2 b z 1 z 2 c | == | 1 a a 2 1 b b 2 1 c c 2 | − | 1 a b c 1 b a c 1 c a b | We will multiply and divide by abc in second determinent
= | 1 a a 2 1 b b 2 1 c c 2 | − a b c a b c | 1 a b c 1 b a c 1 c a b | On multiplying R 1 by a , R 2 by b and R 3 by c in determinant ( 2 ) = | 1 a a 2 1 b b 2 1 c c 2 | − 1 a b c | a a 2 a b c b b 2 a b c c c 2 a b c | = | 1 a a 2 1 b b 2 1 c c 2 | − a b c a b c | a a 2 1 b b 2 1 c c 2 1 | = | 1 a a 2 1 b b 2 1 c c 2 | − | a a 2 1 b b 2 1 c c 2 1 | C 1 ↔ C 2 in determinant ( 1 ) = − | a 1 a 2 b 1 b 2 c 1 c 2 | − | a a 2 1 b b 2 1 c c 2 1 | and If any two rows or columns of a determinant are interchanged,
then sign of the determinant is changed.
Again C 2 ↔ C 3 in determinant ( 1 ) = ( − ) ( − ) | a a 2 1 b b 2 1 c c 2 1 | − | a a 2 1 b b 2 1 c c 2 1 | = | a a 2 1 b b 2 1 c c 2 1 | − | a a 2 1 b b 2 1 c c 2 1 | = 0 Hence
| 1 a a 2 − b c 1 b b 2 − a c 1 c c 2 − a b | = 0 determinants Exercise 5.2 Question 2 Sub Question 7
Answer: | 49 1 6 39 7 4 26 2 3 | = 0 Hint: We will try to do any two column or row equal
Given: | 49 1 6 39 7 4 26 2 3 | Solution: | 49 1 6 39 7 4 26 2 3 | Applying C 3 → 8 × C 3 + C 2 = | 49 1 6 × 8 + 1 39 7 4 × 8 + 7 26 2 3 × 8 + 2 | = | 49 1 48 + 1 39 7 32 + 7 26 2 24 + 2 | = | 49 1 49 39 7 39 26 2 26 | If any two rows or columns of a determinant are identical.
The value of the determinant is zero
= 0 ( ∵ C 1 = C 3 ) Hence
| 49 1 6 39 7 4 26 2 3 | = 0 determinants Exercise 5.2 Question 2 Sub Question 8
Answer:
| 0 x y − x 0 z − y − z 0 | = 0 Hint: We will try to do any two column or row equal
Given:
| 0 x y − x 0 z − y − z 0 | Solution:
| 0 x y − x 0 z − y − z 0 | On multiplying C 1 , C 2 and C 3 by z , y and x respectively = 1 x y z | 0 x y y x − x z 0 z x − y z − z y 0 | Take y , x and z common from R 1 , R 2 and R 3 respectively = 1 x y z ( y z x ) | 0 x x − z 0 z − y − y 0 | = | 0 x x − z 0 z − y − y 0 | On applying C 2 → C 2 − C 3 = | 0 0 x − z − z z − y − y 0 | If any two rows or columns of a determinant are identical.
The value of the determinant is zero
= 0 ( ∵ C 1 = C 2 ) Hence
| 0 x y − x 0 z − y − z 0 | = 0 determinants Exercise 5.2 Question 2 Sub Question 9
Answer:| 1 43 6 7 35 4 3 17 2 | = 0 Hint: We will try to do any two column or row equal
Given:| 1 43 6 7 35 4 3 17 2 | Solution:| 1 43 6 7 35 4 3 17 2 | Applying C 3 → 7 × C 3 + C 1 = | 1 43 6 × 7 + 1 7 35 4 × 7 + 7 3 17 2 × 7 + 3 | = | 1 43 42 + 1 7 35 28 + 7 3 17 14 + 3 | = | 1 43 43 7 35 35 3 17 17 | If any two rows or columns of a determinant are identical.
The value of the determinant is zero
= 0 ( ∵ C 2 = C 3 ) Hence
| 1 43 6 7 35 4 3 17 2 | = 0 determinants Exercise 5.2 Question 10
Answer:| 1 2 2 2 3 2 4 2 2 2 3 2 4 2 5 2 3 2 4 2 5 2 6 2 4 2 5 2 6 2 7 2 | = 0 Hint: We will try to do any two column or row equal
Given:| 1 2 2 2 3 2 4 2 2 2 3 2 4 2 5 2 3 2 4 2 5 2 6 2 4 2 5 2 6 2 7 2 | Solution:| 1 2 2 2 3 2 4 2 2 2 3 2 4 2 5 2 3 2 4 2 5 2 6 2 4 2 5 2 6 2 7 2 | On applying C 3 → C 3 − C 2 and C 4 → C 4 − C 1 = | 1 2 2 2 3 2 − 2 2 4 2 − 1 2 2 2 3 2 4 2 − 3 2 5 2 − 2 2 3 2 4 2 5 2 − 4 2 6 2 − 3 2 4 2 5 2 6 2 − 5 2 7 2 − 4 2 | = | 1 2 2 2 9 − 4 16 − 1 2 2 3 2 16 − 9 25 − 4 3 2 4 2 25 − 16 36 − 9 4 2 5 2 36 − 25 49 − 16 | = | 1 2 2 2 5 15 2 2 3 2 7 21 3 2 4 2 9 27 4 2 5 2 11 33 | On taking common 3 from C 4 = 3 | 1 2 2 2 5 5 2 2 3 2 7 7 3 2 4 2 9 9 4 2 5 2 11 11 | If any two rows or columns of a determinant are identical.
The value of the determinant is zero
= 3 × 0 ( ∵ C 3 = C 4 ) = 0 Hence
| 1 2 2 2 3 2 4 2 2 2 3 2 4 2 5 2 3 2 4 2 5 2 6 2 4 2 5 2 6 2 7 2 | = 0 determinants Exercise 5.2 Question 12
Answer:| ( 2 x + 2 − x ) 2 ( 2 x − 2 − x ) 2 1 ( 3 x + 3 − x ) 2 ( 3 x − 3 − x ) 2 1 ( 4 x + 4 − x ) 2 ( 4 x − 4 − x ) 2 1 | = 0 Hint: We will try to do any two column or row equal
Given:| ( 2 x + 2 − x ) 2 ( 2 x − 2 − x ) 2 1 ( 3 x + 3 − x ) 2 ( 3 x − 3 − x ) 2 1 ( 4 x + 4 − x ) 2 ( 4 x − 4 − x ) 2 1 | Solution: First ( a + b ) 2 − ( a − b ) 2 = ( a 2 + 2 a b + b 2 ) − ( a 2 − 2 a b + b 2 ) = a 2 + 2 a b + b 2 − a 2 + 2 a b − b 2 = 4 a b | ( 2 x + 2 − x ) 2 ( 2 x − 2 − x ) 2 1 ( 3 x + 3 − x ) 2 ( 3 x − 3 − x ) 2 1 ( 4 x + 4 − x ) 2 ( 4 x − 4 − x ) 2 1 | On applying C 1 → C 1 − C 2 ( 2 x + 2 − x ) 2 − ( 2 x − 2 − x ) 2 ( 2 x − 2 − x ) 2 1 ( 3 x + 3 − x ) 2 − ( 3 x − 3 − x ) 2 ( 3 x − 3 − x ) 2 1 ( 4 x + 4 − x ) 2 − ( 4 x − 4 − x ) 2 ( 4 x − 4 − x ) 2 1 ∣ = | 4 × 2 x × 2 − x ( 2 x − 2 − x ) 2 1 4 × 3 x × 3 − x ( 3 x − 3 − x ) 2 1 4 × 4 x × 4 − x ( 4 x − 4 − x ) 2 1 | { ( a + b ) 2 − ( a − b ) 2 = 4 a b a m × a n = a m + n } = | 4 × 2 x + ( − x ) ( 2 x − 2 − x ) 2 1 4 × 3 x + ( − x ) ( 3 x − 3 − x ) 2 1 4 × 4 x + ( − x ) ( 4 x − 4 − x ) 2 1 | = | 4 × 2 0 ( 2 x − 2 − x ) 2 1 4 × 3 0 ( 3 x − 3 − x ) 2 1 4 × 4 0 ( 4 x − 4 − x ) 2 1 | = | 4 ( 2 x − 2 − x ) 2 1 4 ( 3 x − 3 − x ) 2 1 4 ( 4 x − 4 − x ) 2 1 | Taking 4 common from C 1 = 4 | 1 ( 2 x − 2 − x ) 2 1 1 ( 3 x − 3 − x ) 2 1 1 ( 4 x − 4 − x ) 2 1 | If any two rows or columns of a determinant are identical.
The value of the determinant is zero
= 4 × 0 ( ∵ C 1 = C 3 ) = 0 Hence
| ( 2 x + 2 − x ) 2 ( 2 x − 2 − x ) 2 1 ( 3 x + 3 − x ) 2 ( 3 x − 3 − x ) 2 1 ( 4 x + 4 − x ) 2 ( 4 x − 4 − x ) 2 1 | = 0 determinants Exercise 5.2 Question 13
Answer:| sin α cos α cos ( α + δ ) sin β cos β cos ( β + δ ) sin γ cos γ cos ( γ + δ ) | = 0 Hint : We will try to do any two column or row equal
Given:| sin α cos α cos ( α + δ ) sin β cos β cos ( β + δ ) sin γ cos γ cos ( γ + δ ) | Solution:| sin α cos α cos ( α + δ ) sin β cos β cos ( β + δ ) sin γ cos γ cos ( γ + δ ) | On applying C 2 → cos δ ⋅ C 2 − sin δ . C 1 | sin α cos α ⋅ cos δ − sin α ⋅ sin δ cos ( α + δ ) sin β cos β ⋅ cos δ − sin β ⋅ sin δ cos ( β + δ ) sin γ cos γ ⋅ cos δ − sin γ ⋅ sin δ cos ( γ + δ ) | { ∵ cos ( α + δ ) = cos α cos δ − sin α sin δ } = | sin α cos ( α + δ ) cos ( α + δ ) sin β cos ( β + δ ) cos ( β + δ ) sin γ cos ( γ + δ ) cos ( γ + δ ) | If any two rows or columns of a determinant are identical.
The value of the determinant is zero
= 0 ( ∵ C 2 = C 3 ) Hence
| sin α cos α cos ( α + δ ) sin β cos β cos ( β + δ ) sin γ cos γ cos ( γ + δ ) | = 0 deteminants Exercise 5.2 Question 15
Answer:| cos ( x + y ) − sin ( x + y ) cos 2 y sin x cos x sin y − cos x sin x − cos y | = 0 Hint: We will try to do any two column or row identical
Given: | cos ( x + y ) − sin ( x + y ) cos 2 y sin x cos x sin y − cos x sin x − cos y | Solution:| cos ( x + y ) − sin ( x + y ) cos 2 y sin x cos x sin y − cos x sin x − cos y | On applying R 3 → − cos y × R 3 − sin y × R 2 = | cos ( x + y ) − sin ( x + y ) cos 2 y sin x cos x sin y ( − cos y ) ( − cos x ) − sin y sin x ( − cos y ) ( sin x ) − sin y cos x ( − cos y ) ( − cos y ) − sin y sin y | = | cos ( x + y ) − sin ( x + y ) cos 2 y sin x cos x sin y cos x cos y − sin y sin x − [ sin x cos y + sin y cos x ] ( cos 2 y − sin 2 y ) | ∵ cos ( x + y ) = cos x cos y − sin y sin x sin ( x + y ) = sin x cos y + sin y cos x cos 2 y − sin 2 y = cos 2 y = | cos ( x + y ) − sin ( x + y ) cos 2 y sin x cos x sin y cos ( x + y ) − sin ( x + y ) cos 2 y | If any two rows or columns of a determinant are identical.
The value of the determinant is zero
= 0 ( ∵ R 1 = R 3 ) Hence
| cos ( x + y ) − sin ( x + y ) cos 2 y sin x cos x sin y − cos x sin x − cos y | = 0 deteminants Exercise 5.2 Question 16
Answer:| 23 + 3 5 5 15 + 46 5 10 3 + 115 15 5 | = 0 Hint: We will try to make all the elements of two row or column zero
Given: | 23 + 3 5 5 15 + 46 5 10 3 + 115 15 5 | Solution:| 23 + 3 5 5 15 + 46 5 10 3 + 115 15 5 | On taking 5 common from C 2 and C 3 = 5 × 5 | 23 + 3 1 1 15 + 46 5 2 3 + 115 3 5 | On applying C 1 → C 1 − 3 C 2 − 23 C 3 = 5 | 23 + 3 − 3 − 23 1 1 15 + 46 − 15 − 46 5 2 3 + 115 − 3 − 115 3 5 | = 5 | 0 1 1 0 5 2 0 3 5 | If all the elements of a row or column are zero
The value of the determinant is zero
= 5 × 0
= 0
Hence | 23 + 3 5 5 15 + 46 5 10 3 + 115 15 5 | = 0
deteminants Exercise 5 .2 Question 17
Answer: | sin 2 A cot A 1 sin 2 B cot B 1 sin 2 C cot C 1 | = 0 where A,B,C are the angles of Δ ABC Hint: We will try to convert some elements of determinant zero or 1
Given:| sin 2 A cot A 1 sin 2 B cot B 1 sin 2 C cot C 1 | Solution: | sin 2 A cot A 1 sin 2 B cot B 1 sin 2 C cot C 1 | On applying R 1 → R 1 − R 2 and R 2 → R 2 − R 3 = | sin 2 A − sin 2 B cot A − cot B 0 sin 2 B − sin 2 C cot B − cot C 0 sin 2 C cot C 1 | ∵ sin 2 A − sin 2 B = sin ( A + B ) ⋅ sin ( A − B ) ∖ ∵ A + B + C = 180 ∘ A − B = 180 − C sin ( 180 − C ) = sin C sin ( 180 − A ) = sin A Now, cot A − cot B cos A sin A − cos B sin B = sin B cos A − cos B sin A sin A sin B ⇒ sin ( B − A ) sin A sin B = sin { − ( A − B ) } sin A sin B = − sin ( A − B ) sin A sin B Now from (1) = | sin ( A + B ) sin ( A − B ) sin ( B − A ) sin A sin B 0 sin ( B + C ) sin ( B − C ) sin ( C − B ) sin B sin C 0 sin 2 C cot C 1 | = | sin ( 180 − C ) sin ( A − B ) − sin ( A − B ) sin A sin B 0 sin ( 180 − A ) sin ( B − C ) − sin ( B − C ) sin B sin C 0 sin 2 C cot C 1 | = | sin ( A ) sin ( B − C ) − sin ( B − C ) sin B sin C 0 sin 2 C cot C 1 | On taking sin ( A − B ) common from R 1 and sin ( B − C ) from R 2 = sin ( A − B ) ⋅ sin ( B − C ) | sin C − 1 sin A sin B 0 sin A − 1 sin B sin C 0 sin 2 C cot C 1 | On expanding w.r.t C 3 = sin ( A − B ) ⋅ sin ( B − C ) [ 0 | sin A − 1 sin B sin C sin 2 C cot C | − 0 | sin C − 1 sin A sin B sin 2 C cot C | + 1 | sin C − 1 sin A sin B sin A − 1 sin B sin C | ] = sin ( A − B ) ⋅ sin ( B − C ) [ 0 − 0 + 1 ( − sin C sin B sin C + sin A sin A sin B ) ] = sin ( A − B ) ⋅ sin ( B − C ) [ − 1 sin B + 1 sin B ] = sin ( A − B ) ⋅ sin ( B − C ) × 0 = 0 Hence
| sin 2 A cot A 1 sin 2 B cot B 1 sin 2 C cot C 1 | = 0 deteminants Exercise MCQs Question 2 sub question 3
Answer:-| a b + c a 2 b c + a b 2 c a + b c 2 | = − ( a + b + c ) ( a − b ) ( b − c ) ( c − a ) Hint : We will try to make some elements of the determinant into zero
Given:-| a b + c a 2 b c + a b 2 c a + b c 2 | Solution :-| a b + c a 2 b c + a b 2 c a + b c 2 | On applying C 1 → C 1 + C 2 = | a + b + c b + c a 2 b + c + a c + a b 2 c + a + b a + b c 2 | On taking common ( a + b + c ) from C 1 = a + b + c | 1 b + c a 2 1 c + a b 2 1 a + b c 2 | On applying C 1 → C 1 − C 2 and C 2 → C 2 − C 3 = a + b + c | 0 ( b + c ) − ( c + a ) a 2 − b 2 0 ( c + a ) − ( a + b ) b 2 − c 2 1 a + b c 2 | = a + b + c | 0 b + c − c − a ( a + b ) ( a − b ) 0 c + a − a − b ( b + c ) ( b − c ) 1 a + b c 2 | ( ∵ a 2 − b 2 = ( a + b ) ( a − b ) ) = a + b + c | 0 b − a ( a + b ) ( a − b ) 0 c − b ( b + c ) ( b − c ) 1 a + b c 2 | = a + b + c | 0 − ( a − b ) ( a + b ) ( a − b ) 0 − ( b − c ) ( b + c ) ( b − c ) 1 a + b c 2 | On taking common ( a − b ) from R 1 and ( b − c ) from R 2 = ( a + b + c ) ( a − b ) ( b − c ) | 0 − 1 ( a + b ) 0 − 1 ( b + c ) 1 a + b c 2 | On expanding w.r.t R 1 = ( a + b + c ) ( a − b ) ( b − c ) [ 0 | − 1 b + c a + b c 2 | − ( − 1 ) | 0 b + c 1 c 2 | + ( a + b ) | 0 − 1 1 a + b | ] = ( a + b + c ) ( a − b ) ( b − c ) [ 0 + 1 { 0 × c 2 − 1 ( b + c ) } + ( a + b ) { 0 ⋅ ( a + b ) − ( 1 ) ( − 1 ) } ] = ( a + b + c ) ( a − b ) ( b − c ) { 1 ( − b − c ) + ( a + b ) ( 1 ) } = ( a + b + c ) ( a − b ) ( b − c ) ( − b − c + a + b ) = ( a + b + c ) ( a − b ) ( b − c ) ( a − c ) = − ( a + b + c ) ( a − b ) ( b − c ) ( c − a ) Hence
| a b + c a 2 b c + a b 2 c a + b c 2 | = − ( a + b + c ) ( a − b ) ( b − c ) ( c − a ) determinants Exercise 5.2 Question 2 sub question 4
Answer:-| 1 a b c 1 b c a 1 c a b | = ( a − b ) ( b − c ) ( c − a ) Hint: We will try to make some elements of the determinant into zero
Given:| 1 a b c 1 b c a 1 c a b | Solution :| 1 a b c 1 b c a 1 c a b | On applying R 1 → R 1 − R 2 and R 2 → R 2 − R 3 = | 0 a − b b c − c a 0 b − c c a − a b 1 c a b | = | 0 a − b − c ( a − b ) 0 b − c − a ( b − c ) 1 c a b | On taking common ( a − b ) from R 1 and ( b − c ) from R 2 = ( a − b ) ( b − c ) | 0 1 − c 0 1 − a 1 c a b | On expanding w.r.t R 1 = ( a − b ) ( b − c ) [ 0 | 1 − a c a b | − 1 | 0 − a 1 a b | + ( − c ) | 0 1 1 c | ] = ( a − b ) ( b − c ) [ 0 − 1 { 0 × a b − ( − a ) × 1 } − c { 0 × c − 1 × 1 } ] = ( a − b ) ( b − c ) [ − 1 ( + a ) − c ( − 1 ) ] = ( a − b ) ( b − c ) ( − a + c ) = ( a − b ) ( b − c ) ( c − a ) Hence | 1 a b c 1 b c a 1 c a b | = ( a − b ) ( b − c ) ( c − a ) determinants Exercise 5.2 Question 2 sub question 5
Answer:| x + λ x x x x + λ x x x x + λ | = λ 2 ( 3 x + λ ) Hint: We will try to make some elements of the determinant into zero
Given:| x + λ x x x x + λ x x x x + λ | Solution:| x + λ x x x x + λ x x x x + λ | On applying R 1 → R 1 + R 2 + R 3 = | 3 x + λ 3 x + λ 3 x + λ x x + λ x x x x + λ | On taking common ( 3 x + λ ) from R 1 = ( 3 x + λ ) | 1 1 1 x x + λ x x x x + λ | On applying C 1 → C 1 − C 2 and C 2 → C 2 − C 3 = ( 3 x + λ ) | 0 0 1 x − ( x + λ ) ( x + λ ) − x x x − x x − ( x + λ ) x + λ | = ( 3 x + λ ) | 0 0 1 x − x − λ x + λ − x x 0 x − x − λ x + λ | = ( 3 x + λ ) | 0 0 1 − λ λ x 0 − λ x + λ | On expanding w.r.t R 1 = ( 3 x + λ ) [ 0 | λ x − λ x + λ | − 0 | − λ x 0 x + λ | + 1 | − λ λ 0 − λ | ] = ( 3 x + λ ) [ 0 − 0 + 1 [ ( − λ ) ( − λ ) − 0 × λ ] ] = ( 3 x + λ ) { 1 ( λ 2 − 0 ) } = λ 2 ( 3 x + λ ) Hence| x + λ x x x x + λ x x x x + λ | = λ 2 ( 3 x + λ ) determinants Exercise 5.2 Question 2 sub question 6
Answer:| a b c c a b b c a | = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) Hint : We will try to make some elements of the determinant into zero
Given :| a b c c a b b c a | Solution: | a b c c a b b c a | On applying R 1 → R 1 + R 2 + R 3 = | a + b + c a + b + c a + b + c c a b b c a | On taking common ( a + b + c ) from R 1 = ( a + b + c ) | 1 1 1 c a b b c a | On applying C 1 → C 1 − C 2 and C 2 → C 2 − C 3 = ( a + b + c ) | 0 0 1 c − a a − b b b − c c − a a | On expanding w.r.t R 1 = ( a + b + c ) [ 0 | a − b b c − a a | − 0 | b − a b 0 a | + 1 | c − a a − b b − c c − a | = ( a + b + c ) [ 0 − 0 + 1 { ( c − a ) ( c − a ) − ( b − c ) ( a − b ) } ] = ( a + b + c ) { ( c 2 − a c − a c + a 2 ) − ( b a − b 2 + a c − b c ) } = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) Hence | a b c c a b b c a | = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) determinants Exercise 5.2 Question 2 sub question 7
Answer:| x 1 1 1 x 1 1 1 x | = ( x + 2 ) ( x − 1 ) 2 Hint: We will try to make some elements of the determinant into zero
Given:| x 1 1 1 x 1 1 1 x | Solution: | x 1 1 1 x 1 1 1 x | On applying R 1 → R 1 + R 2 + R 3 = | x + 2 x + 2 x + 2 1 x 1 1 1 x | On taking common ( x + 2 ) from R 1 = ( x + 2 ) | 1 1 1 1 x 1 1 1 x | On applying C 1 → C 1 − C 2 and C 2 → C 2 − C 3 = ( x + 2 ) | 0 0 1 1 − x x − 1 1 1 1 − x x | On expanding w.r.t R 1 = ( x + 2 ) [ 0 | x − 1 1 1 − x x | − 0 | 1 − x 1 0 x | + 1 | 1 − x x − 1 0 1 − x | ] = ( x + 2 ) [ 0 − 0 + 1 { ( 1 − x ) ( 1 − x ) − 0 ( x − 1 ) } ] = ( x + 2 ) [ 1 { − ( x − 1 ) 2 − 0 } ] = ( x + 2 ) ( x − 1 ) 2 Hence
| x 1 1 1 x 1 1 1 x | = ( x + 2 ) ( x − 1 ) 2 determinants Exercise 5.2 Question 2 sub question 8
Answer:| 0 x y 2 x z 2 x 2 y 0 y z 2 x 2 z z y 2 0 | = 2 x 3 y 3 z 3 Hint: We will take common from
C 1 , C 2 a n d C 3 Given:| 0 x y 2 x z 2 x 2 y 0 y z 2 x 2 z z y 2 0 | Solution: | 0 x y 2 x z 2 x 2 y 0 y z 2 x 2 z z y 2 0 | On taking common x 2 from C 1 , y 2 from C 2 and z 2 from C 3 = x 2 y 2 z 2 | 0 x x y 0 y z z 0 | On expanding w.r.t R 1 = x 2 y 2 z 2 [ 0 | 0 y z 0 | − x | y y z 0 | + x | y 0 z z | ] = x 2 y 2 z 2 [ 0 − x ( 0 x y − y x z ) + x ( y x z − 0 x z ) ] = x 2 y 2 z 2 [ − x ( 0 − y z ) + x ( y z − 0 ) ] = x 2 y 2 z 2 ( x y z + x y z ) = x 2 y 2 z 2 × 2 x y z = 2 x 3 y 3 z 3 Hence| 0 x y 2 x z 2 x 2 y 0 y z 2 x 2 z z y 2 0 | = 2 x 3 y 3 z 3 determinants Exercise 5.2 Question 2 sub question 9
Answer:
| a + x y z x a + y z x y a + z | = a 2 ( a + x + y + z ) Hint:
We will try to make some elements of the determinant into zero Given :
| a + x y z x a + y z x y a + z | Solution:
| a + x y z x a + y z x y a + z | On applying C 1 → C 1 + C 2 + C 3 = | a + x + y + z y z a + x + y + z a + y z a + x + y + z y a + z | On taking common ( a + x + y + z ) from C 1 = ( a + x + y + z ) | 1 y z 1 a + y z 1 y a + z | On applying R 1 → R 1 − R 2 and R 2 → R 2 − R 3 = ( a + x + y + z ) | 0 y − ( a + y ) z − z 0 a + y − y z − ( a + z ) 1 y a + z | = ( a + x + y + z ) | 0 y − a − y 0 0 a z − a − z 1 y a + z | = ( a + x + y + z ) | 0 − a 0 0 a − a 1 y a + z | On expanding w.r.t R 1 = ( a + x + y + z ) [ 0 | a − a y a + z | − ( − a ) | 0 − a 1 a + z | + 0 | 0 a 1 y | ] = ( a + x + y + z ) [ 0 + a { 0 × ( a + z ) − ( − a ) 1 } + 0 ] = ( a + x + y + z ) { a ( 0 + a ) } = ( a + x + y + z ) a 2 = a 2 ( a + x + y + z ) Hence
| a + x y z x a + y z x y a + z | = a 2 ( a + x + y + z ) deteminants Exercise 5.2 Question 10
Answer:-Δ = | 1 x x 2 1 y y 2 1 z z 2 | + Δ 1 = | 1 1 1 y z z x x y x y z | = 0 Hint: First we solve Δ Δ 1 Given:Δ = | 1 x x 2 1 y y 2 1 z z 2 | , Δ 1 = | 1 1 1 y z z x x y x y z | Solution: Let us solve for Δ = | 1 x x 2 1 y y 2 1 z z 2 | Δ = | 1 x x 2 1 y y 2 1 z z 2 | On applying R 1 → R 1 − R 2 and R 2 → R 2 − R 3 Δ = | 0 x − y x 2 − y 2 0 y − z y 2 − z 2 1 z z 2 | Δ = | 0 x − y ( x + y ) ( x − y ) 0 y − z ( y + z ) ( y − z ) 1 z z 2 | a 2 − b 2 = ( a + b ) ( a − b ) On taking common ( x − y ) from R 1 and ( y − z ) from R 2 Δ = ( x − y ) ( y − z ) | 0 1 ( x + y ) 0 1 ( y + z ) 1 z z 2 | Δ = ( x − y ) ( y − z ) | 0 0 1 1 1 z ( x + y ) ( y + z ) z 2 | Δ = ( x − y ) ( y − z ) | 0 0 1 1 1 z ( x + y ) ( y + z ) z 2 | On expanding w.r.t R 1 Δ = ( x − y ) ( y − z ) [ 0 | 1 z y + z z 2 | − 0 | 1 z x + y z 2 | + 1 | 1 1 x + y y + z | ] Δ = ( x − y ) ( y − z ) [ 0 − 0 + 1 { 1 ( y + z ) − 1 ( x + y ) } ] Δ = ( x − y ) ( y − z ) ( y + z − x − y ) Δ = ( x − y ) ( y − z ) ( z − x ) … … … ( 1 ) Δ 1 = | 1 1 1 y z z x x y x y z | Δ 1 = | 1 1 1 y z z x x y x y z | On applying C 1 → C 1 − C 2 and C 2 → C 2 − C 3 Δ 1 = | 0 0 1 y z − z x z x − x y x y x − y y − z z | Δ 1 = | 0 0 1 − z ( x − y ) − x ( y − z ) x y x − y y − z z | On taking common ( x − y ) from C 1 and ( y − z ) from C 2 Δ 1 = ( x − y ) ( y − z ) | 0 0 1 − z − x x y 1 1 z | R 1 Δ 1 = ( x − y ) ( y − z ) [ 0 | − x x y 1 z | − 0 | − z x y 1 z | + 1 ∣ − z − x 1 1 ] Δ 1 = ( x − y ) ( y − z ) [ 0 − 0 + 1 { ( − z ) ( 1 ) − ( − x ) ( 1 ) } ] Δ 1 = ( x − y ) ( y − z ) ( − z + x ) Δ 1 = − ( x − y ) ( y − z ) ( z − x ) Δ + Δ 1 = ( x − y ) ( y − z ) ( z − x ) + { − ( x − y ) ( y − z ) ( z − x ) } Δ + Δ 1 = 0 = R ⋅ H ⋅ S Δ + Δ 1 = 0
determinants Exercise 5.2 Question 11
Answer:| a b c a − b b − c c − a b + c c + a a + b | = a 3 + b 3 + c 3 − 3 a b c Hint: We will try to make some elements of the determinant zero
Given:| a b c a − b b − c c − a b + c c + a a + b | Solution: L.H.S
| a b c a − b b − c c − a b + c c + a a + b | On applying C 1 → C 1 + C 2 + C 3 = | a + b + c b c a − b + b − c + c − a b − c c − a b + c + c + a + a + b c + a a + b | = | a + b + c b c 0 b − c c − a 2 a + 2 b + 2 c c + a a + b | = | a + b + c b c 0 b − c c − a 2 ( a + b + c ) c + a a + b | On taking common ( a + b + c ) from C 1 = ( a + b + c ) | 1 b c 0 b − c c − a 2 c + a a + b | On expanding wr.t R 1 = ( a + b + c ) [ 1 | b − c c − a c + a a + b | − b | 0 c − a 2 a + b | + c ∣ 0 b − c 2 c + a ] = ( a + b + c ) [ 1 { ( b − c ) ( a + b ) − ( c − a ) ( c + a ) } − b { 0 × ( a + b ) − 2 ( c − a ) } + c { 0 × ( c + a ) − 2 ( b − c ) } ] = ( a + b + c ) [ 1 { ( a b − c a + b 2 − b c ) − ( c 2 − a 2 ) } − b { 0 − 2 c + 2 a } + c { 0 − 2 b + 2 c } ] = ( a + b + c ) [ a b − c a + b 2 − b c − c 2 + a 2 + 2 b c − 2 a b − 2 b c + 2 c 2 ] = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) = a 3 + b 3 + c 3 − 3 a b c ∵ x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) = R H S Hence it is proved that
| a b c a − b b − c c − a b + c c + a a + b | = a 3 + b 3 + c 3 − 3 a b c determinants Exercise 5.2 Question 12
Answer:| b + c a − b a c + a b − c b a + b c − a c | = 3 a b c − a 3 − b 3 − c 3 Hint: We will try to make some elements of the determinant zero
Given:| b + c a − b a c + a b − c b a + b c − a c | Solution: L.H.S
| b + c a − b a c + a b − c b a + b c − a c | On applying R 1 → R 1 + R 2 + R 3 = | b + c + c + a + a + b ( a − b + b − c + c − a ) a + b + c c + a b − c b a + b c − a c | = | 2 a + 2 b + 2 c 0 a + b + c c + a b − c b a + b c − a c | = | 2 ( a + b + c ) 0 a + b + c c + a b − c b a + b c − a c | On taking common ( a + b + c ) from R 1 = ( a + b + c ) | 2 0 1 c + a b − c b a + b c − a c | On expanding w.r.t R 1 = ( a + b + c ) [ 2 | b − c b c − a c | − 0 | c + a b a + b c | + 1 | c + a b − c a + b c − a | = ( a + b + c ) [ 1 { ( b − c ) ( c ) − ( b ) ( c − a ) } − 0 + 1 { ( c + a ) ( c − a ) − ( b − c ) ( a + b ) } ] = ( a + b + c ) [ 2 ( b c − c 2 − b c + a b ) + 1 { c 2 − a 2 − ( a b − a c + b 2 − b c ) } ] = ( a + b + c ) { 2 ( − c 2 + a b ) + 1 ( c 2 − a 2 − a b + a c − b 2 + b c ) } = ( a + b + c ) ( − 2 c 2 + 2 a b + c 2 − a 2 − a b + a c − b 2 + b c ) = ( a + b + c ) ( c 2 − a 2 − b 2 − a b + a c + b c ) = − ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) = − ( a 3 + b 3 + c 3 − 3 a b c ) = − a 3 − b 3 − c 3 + 3 a b c ∵ x 3 + y 3 + z 3 − 3 x y z = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) = 3 a b c − a 3 − b 3 − c 3 =RHS Hence it is proved that
| b + c a − b a c + a b − c b a + b c − a c | = 3 a b c − a 3 − b 3 − c 3 deteminants Exercise 5.2 Question 13
Answer: | a + b b + c c + a b + c c + a a + b c + a a + b b + c | = 2 | a b c b c a c a b | Hint: We will make
R 1 of L.H.S according to R.H.S
Given:| a + b b + c c + a b + c c + a a + b c + a a + b b + c | = 2 | a b c b c a c a b | Solution: L.H.S
| a + b b + c c + a b + c c + a a + b c + a a + b b + c | = 2 | a b c b c a c a b | On applying R 1 → R 1 − R 2 + R 3 = | ( a + b ) − ( b + c ) + ( c + a ) ( b + c ) − ( c + a ) + ( a + b ) ( c + a ) − ( a + b ) + ( b + c ) b + c c + a a + b c + a a + b b + c | = | a + b − b − c + c + a b + c − c − a + a + b c + a − a − b + b + c b + c c + a a + b c + a a + b b + c | = | 2 a 2 b 2 c b + c c + a a + b c + a a + b b + c | On taking 2 common from R 1 = 2 | a b c b + c c + a a + b c + a a + b b + c | On applying R 3 → R 3 − R 1 = 2 | a b c b + c c + a a + b c + a − a a + b − b b + c − c | = 2 | a b c b + c c + a a + b c a b | On applying R 2 → R 2 − R 3 = 2 | a b c b + c − c c + a − a a + b − b c a b | = 2 | a b c b c a c a b | = R ⋅ H ⋅ S Hence it is proved that
| a + b b + c c + a b + c c + a a + b c + a a + b b + c | = 2 | a b c b c a c a b | deteminants Exercise 5.2 Question 14
Answer:| a + b + 2 c a b c b + c + 2 a b c a c + a + 2 b | = 2 ( a + b + c ) 3 Hint We will try to make some elements of the determinant zero
Given:| a + b + 2 c a b c b + c + 2 a b c a c + a + 2 b | = 2 ( a + b + c ) 3 Solution: L.H.S | a + b + 2 c a b c b + c + 2 a b c a c + a + 2 b | On applying C 1 → C 1 + C 2 + C 3 = | 2 a + 2 b + 2 c a b 2 a + 2 b + 2 c b + c + 2 a b 2 a + 2 b + 2 c a c + a + 2 b | = | 2 ( a + b + c ) a b 2 ( a + b + c ) b + c + 2 a b 2 ( a + b + c ) a c + a + 2 b | On taking common 2 ( a + b + c ) from C 1 = 2 ( a + b + c ) | 1 a b 1 b + c + 2 a b 1 a c + a + 2 b | On applying R 1 → R 1 − R 2 , R 2 → R 2 − R 3 = 2 ( a + b + c ) | 0 a − ( b + c + 2 a ) b − b 0 b + c + 2 a − a b − ( c + a + 2 b ) 1 a c + a + 2 b | = 2 ( a + b + c ) | 0 a − b − c − 2 a 0 0 b + c + a b − c − a − 2 b 1 a c + a + 2 b | = 2 ( a + b + c ) | 0 − a − b − c 0 0 a + b + c − a − b − c 1 a c + a + 2 b | = 2 ( a + b + c ) | 0 − ( a + b + c ) 0 0 a + b + c − ( a + b + c ) 1 a c + a + 2 b | On taking common ( a + b + c ) from R 2 = 2 ( a + b + c ) ( a + b + c ) | 0 − ( a + b + c ) 0 0 1 − 1 1 a c + a + 2 b | = 2 ( a + b + c ) 2 | 0 − ( a + b + c ) 0 0 1 − 1 1 a c + a + 2 b | On expanding w.r.t R 1 = 2 ( a + b + c ) 2 [ 0 | 1 − 1 a c + a + 2 b | − { − ( a + b + c ) } | 0 − 1 1 c + a + 2 b | + 0 ∣ 0 1 1 a ] = 2 ( a + b + c ) 2 [ 0 + ( a + b + c ) { 0 − ( − 1 ) 1 } + 0 ] = 2 ( a + b + c ) 2 [ ( a + b + c ) ( 1 ) ] = 2 ( a + b + c ) 2 ( a + b + c ) = 2 ( a + b + c ) 3 = R ⋅ H ⋅ S Hence it is proved that
| a + b + 2 c a b c b + c + 2 a b c a c + a + 2 b | = 2 ( a + b + c ) 3 determinants Exercise 5.2 Question 15
Answer:| a − b − c 2 a 2 a 2 b b − c − a 2 b 2 c 2 c c − a − b | = ( a + b + c ) 3 Hint: We will try to make some elements of the determinant zero
Given:| a − b − c 2 a 2 a 2 b b − c − a 2 b 2 c 2 c c − a − b | = ( a + b + c ) 3 Solution: L.H.S | a − b − c 2 a 2 a 2 b b − c − a 2 b 2 c 2 c c − a − b | On applying R 1 → R 1 + R 2 + R 3 = | a − b − c + 2 b + 2 c 2 a + b − c − a + 2 c 2 a + 2 b + c − a − b 2 b b − c − a 2 b 2 c 2 c c − a − b | = | a + b + c a + b + c a + b + c 2 b b − c − a 2 b 2 c 2 c c − a − b | On taking common ( a + b + c ) from C 1 = ( a + b + c ) | 1 1 1 2 b b − c − a 2 b 2 c 2 c c − a − b | On applying C 1 → C 1 − C 2 , C 2 → C 2 − C 3 = ( a + b + c ) | 0 0 1 2 b − ( b − c − a ) b − c − a − 2 b 2 b 2 c − 2 c 2 c − ( c − a − b ) c − a − b | = ( a + b + c ) | 0 0 1 2 b − b + c + a − b − c − a 2 b 2 c − 2 c 2 c − c + a + b c − a − b | = ( a + b + c ) | 0 0 1 a + b + c − ( a + b + c ) 2 b 0 a + b + c c − a − b | On taking common ( a + b + c ) from C 2 = ( a + b + c ) ( a + b + c ) | 0 0 1 a + b + c − 1 2 b 0 1 c − a − b | On expanding w.r.t R 1 = ( a + b + c ) 2 [ 0 | − 1 2 b 1 c − a − b | − 0 | a + b + c 2 b 0 c − a − b | + 1 | a + b + c − 1 0 1 | ] = ( a + b + c ) 2 [ 0 − 0 + 1 { ( a + b + c ) × 1 − ( − 1 ) × 0 } ] = ( a + b + c ) 2 [ 1 ( a + b + c ) ] = ( a + b + c ) 2 ( a + b + c ) = ( a + b + c ) 3 = R ⋅ H ⋅ S Hence it is proved that
| a − b − c 2 a 2 a 2 b b − c − a 2 b 2 c 2 c c − a − b | = ( a + b + c ) 3 determinants Exercise 5.2 Question 16
Answer: Answer:| 1 b + c b 2 + c 2 1 c + a c 2 + a 2 1 a + b a 2 + b 2 | = ( a − b ) ( b − c ) ( c − a ) Hint : We will try to make some elements of the determinant zero
Given:| 1 b + c b 2 + c 2 1 c + a c 2 + a 2 1 a + b a 2 + b 2 | = ( a − b ) ( b − c ) ( c − a ) Solution: L.H.S | 1 b + c b 2 + c 2 1 c + a c 2 + a 2 1 a + b a 2 + b 2 | If the rows and columns of a determinant are interchanged, the value of the determinant remains the same.
= | 1 1 1 b + c c + a a + b b 2 + c 2 c 2 + a 2 a 2 + b 2 | On applying C 1 → C 1 − C 2 and C 2 → C 2 − C 3 = | 0 0 1 b + c − ( c − a ) c + a − ( a + b ) a + b b 2 + c 2 − ( c 2 + a 2 ) c 2 + a 2 − ( a 2 + b 2 ) a 2 + b 2 | = | 0 0 1 b + c − c − a c + a − a − b a + b b 2 + c 2 − c 2 − a 2 c 2 + a 2 − a 2 − b 2 a 2 + b 2 | = | 0 0 1 b − a c − b a + b b 2 − a 2 c 2 − b 2 a 2 + b 2 | = | 0 0 1 b − a c − b a + b ( b − a ) ( b + a ) ( c − b ) ( c + b ) a 2 + b 2 | ∵ x 2 − y 2 = ( x − y ) ( x + y ) On taking common ( b − a ) from C 1 and ( c − b ) from C 2 = ( b − a ) ( c − b ) | 0 0 1 1 1 a + b ( b + a ) ( c + b ) a 2 + b 2 | On expanding w.r.t R 1 = ( b − a ) ( c − b ) [ 0 | 1 a + b c + b a 2 + b 2 | − 0 | 1 a + b b + a a 2 + b 2 | + 1 | 1 1 b + a c + b | ] = ( − ) ( a − b ) ( − ) ( b − c ) [ 0 − 0 + 1 { ( c + b ) 1 − 1 ( b + a ) } ] = ( a − b ) ( b − c ) [ 1 { ( c + b − b − a ) } ] = ( a − b ) ( b − c ) ( c − a ) = R H . S Hence it is proved that
| 1 b + c b 2 + c 2 1 c + a c 2 + a 2 1 a + b a 2 + b 2 | = ( a − b ) ( b − c ) ( c − a ) determinants Exercise 5.2 Question 17
Answer:| a a + b a + 2 b a + 2 b a a + b a + b a + 2 b a | = 9 ( a + b ) b 2 Hint We will try to make some elements of the determinant zero
Given :| a a + b a + 2 b a + 2 b a a + b a + b a + 2 b a | = 9 ( a + b ) b 2 Solution: L.H.S | a a + b a + 2 b a + 2 b a a + b a + b a + 2 b a | On applying R 1 → R 1 + R 2 + R 3 = | a + a + 2 b + a + b a + b + a + a + 2 b a + 2 b + a + b + a a + 2 b a a + b a + b a + 2 b a | = | 3 a + 3 b 3 a + 3 b 3 a + 3 b a + 2 b a a + b a + b a + 2 b a | Taking common ( 3 a + 3 b ) from R 1 = ( 3 a + 3 b ) | 1 1 1 a + 2 b a a + b a + b a + 2 b a | On applying R 2 → R 2 − R 3 = 3 ( a + b ) | 1 1 1 a + 2 b − ( a + b ) a − ( a + 2 b ) a + b − a a + b a + 2 b a | = 3 ( a + b ) | 1 1 1 a + 2 b − a − b a − a − 2 b b a + b a + 2 b a | = 3 ( a + b ) | 1 1 1 b − 2 b b a + b a + 2 b a | Taking common (b) from R 2 = 3 ( a + b ) b | 1 1 1 1 − 2 1 a + b a + 2 b a | On applying C 1 → C 1 − C 2 and C 2 → C 2 − C 3 = 3 ( a + b ) b | 0 0 1 1 − ( − 2 ) − 2 − 1 1 a + b − ( a + 2 b ) a + 2 b − a a | = 3 ( a + b ) b | 0 0 1 1 + 2 − 2 − 1 1 a + b − a − 2 b a + 2 b − a a | = 3 ( a + b ) b | 0 0 1 3 − 3 1 − b 2 b a | On expanding w.r.t R 1 = 3 ( a + b ) b [ 0 | − 3 1 2 b a | − 0 | 3 1 − b a | + 1 | 3 − 3 − b 2 b | ] = 3 ( a + b ) b [ 0 − 0 + 1 { 3 × 2 b − ( − 3 ) ( − b ) } ] = 3 ( a + b ) b [ 1 ( 6 b − 3 b ) ] = 3 ( a + b ) b × 3 b = 9 ( a + b ) b 2 Hence it is proved that
| a a + b a + 2 b a + 2 b a a + b a + b a + 2 b a | = 9 ( a + b ) b 2 deteminants Exercise 5.2 Question 18
Answer:| 1 a b c 1 b c a 1 c a b | = | 1 a a 2 1 b b 2 1 c c 2 | Hint We will make column (3) of L.H.S abc
Given:| 1 a b c 1 b c a 1 c a b | = | 1 a a 2 1 b b 2 1 c c 2 | Solution: L.H.S | 1 a b c 1 b c a 1 c a b | On multiplying by 'a' in R 1 = by b ' in R 2 and by 'c' in R m = 1 a b c | a a 2 a b c b b 2 a b c c c 2 a b c | Taking common ( a b c ) from C 3 = a b c a b c | a a 2 1 b b 2 1 c c 2 1 | = | a a 2 1 b b 2 1 c c 2 1 | If any two rows or columns of a determinant are interchanged, then sign of the determinant is changed.
C 2 ↔ C 3 = ( − ) | a 1 a 2 b 1 b 2 c 1 c 2 | C 1 ↔ C 2 = ( − ) ( − ) | 1 a a 2 1 b b 2 1 c c 2 | = | 1 a a 2 1 b b 2 1 c c 2 | = R ⋅ H ⋅ S Hence it is proved that
| 1 a b c 1 b c a 1 c a b | = | 1 a a 2 1 b b 2 1 c c 2 | deteminants Exercise 5.2 Question 19.
Answer:| z x y z 2 x 2 y 2 z 4 x 4 y 4 | = | x y z x 2 y 2 z 2 x 4 y 4 z 4 | = | x 2 y 2 z 2 x 4 y 4 z 4 x y z | = x y z ( x − y ) ( y − z ) ( z − x ) ( x + y + z ) Hint: We will prove all the determinant equal by using interchange column or row property
Given:| z x y z 2 x 2 y 2 z 4 x 4 y 4 | = | x y z x 2 y 2 z 2 x 4 y 4 z 4 | = | x 2 y 2 z 2 x 4 y 4 z 4 x y z | = x y z ( x − y ) ( y − z ) ( z − x ) ( x + y + z ) Solution: Let us first solve for
| z x y z 2 x 2 y 2 z 4 x 4 y 4 | If any two rows or columns of a determinant are interchanged, then sign of the determinant is changed.
C 1 ↔ C 2 = ( − ) | x z y x 2 z 2 y 2 x 4 z 4 y 4 | C 1 ↔ C 2 = ( − ) ( − ) | x y z x 2 y 2 z 2 x 4 y 4 z 4 | = | x y z x 2 y 2 z 2 x 4 y 4 z 4 | R 1 ↔ R 2 = ( − ) | x 2 y 2 z 2 x y z x 4 y 4 z 4 | R 2 ↔ R 3 = ( − ) ( − ) | x 2 y 2 z 2 x 4 y 4 z 4 x y z | = | x 2 y 2 z 2 x 4 y 4 z 4 x y z | .....2
F r o m e q u a t i o n ( 1 ) a n d ( 2 ) w e g e t | z x y z 2 x 2 y 2 z 4 x 4 y 4 | = | x y z x 2 y 2 z 2 x 4 y 4 z 4 | = | x 2 y 2 z 2 x 4 y 4 z 4 x y z | … (3)
N o w c o n s i d e r − | x 2 y 2 z 2 x 4 y 4 z 4 x y z | T a k i n g c o m m o n x f r o m C 1 , y f r o m C 2 a n d z f r o m C 3 = x y z | x y z x 3 y 3 z 3 1 1 1 | O n a p p l y i n g C 1 → C 1 − C 2 a n d C 2 → C 2 − C 3 = x y z | x − y y − z z x 3 − y 3 y 3 − z 3 z 3 0 0 1 | = x y z | x − y y − z z ( x − y ) ( x 2 + x y + y 2 ) ( y − z ) ( y 2 + y z + z 2 ) z 3 0 0 1 | ∵ x 3 − y 3 = ( x − y ) ( x 2 + x y + y 2 ) On taking common ( x − y ) common from C 1 and ( y − z ) from C 2 = x y z ( x − y ) ( y − z ) | 1 1 z ( x 2 + x y + y 2 ) ( y 2 + y z + z 2 ) z 3 0 0 1 | On expanding w.r.t R 3 = x y z ( x − y ) ( y − z ) [ 0 | 1 z y 2 + y z + z 2 z 3 | − 0 | 1 z x 2 + x y + y 2 z 3 | + 1 | 1 1 x 2 + x y + y 2 y 2 + y z + z 2 | ] = x y z ( x − y ) ( y − z ) [ 0 − 0 + 1 { ( y 2 + y z + z 2 ) − ( x 2 + x y + y 2 ) } ] = x y z ( x − y ) ( y − z ) ( y 2 + y z + z 2 − x 2 − x y − y 2 ) = x y z ( x − y ) ( y − z ) ( z 2 − x 2 + y z − x y ) = x y z ( x − y ) ( y − z ) { ( z + x ) ( z − x ) + y ( z − x ) } = x y z ( x − y ) ( y − z ) ( z − x ) ( z + x + y ) = x y z ( x − y ) ( y − z ) ( z − x ) ( x + y + z ) From equation (1), (2),(3) and (4)
Hence it is proved that
| z x y z 2 x 2 y 2 z 4 x 4 y 4 | = | x y z x 2 y 2 z 2 x 4 y 4 z 4 | = | x 2 y 2 z 2 x 4 y 4 z 4 x y z | = x y z ( x − y ) ( y − z ) ( z − x ) ( x + y + z ) determinants Exercise 5.2 Question 20
Answer:| ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b | = ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) ( a 2 + b 2 + c 2 ) Hint We will try to make
( a − b ) and
( b − c ) any two elements of determinant
Given:| ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b | = ( a − b ) ( b − c ) ( c − a ) ( a 2 + b 2 + c 2 ) Solution: L.H.S | ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b | On applying R 1 → R 1 − R 2 and R 2 → R 2 − R 3 = | ( b + c ) 2 − ( c + a ) 2 a 2 − b 2 b c − c a ( c + a ) 2 − ( a + b ) 2 b 2 − c 2 c a − a b ( a + b ) 2 c 2 a b | x 2 − y 2 = ( x + y ) ( x − y ) = | ( b + c + c + a ) ( b + c − c − a ) ( a + b ) ( a − b ) − c ( a − b ) ( c + a + a + b ) ( c + a − a − b ) ( b + c ) ( b − c ) − a ( b − c ) ( a + b ) 2 c 2 a b | = | ( a + b + 2 c ) ( b − a ) ( a + b ) ( a − b ) − c ( a − b ) ( 2 a + b + c ) ( c − b ) ( b + c ) ( b − c ) − a ( b − c ) ( a + b ) 2 c 2 a b | = | − ( a + b + 2 c ) ( a − b ) ( a + b ) ( a − b ) − c ( a − b ) − ( 2 a + b + c ) ( b − c ) ( b + c ) ( b − c ) − a ( b − c ) ( a + b ) 2 c 2 a b | Taking common ( a − b ) from R 1 , ( b − c ) from R 2 = ( a − b ) ( b − c ) | − ( a + b + 2 c ) ( a + b ) − c − ( 2 a + b + c ) ( b + c ) − a ( a + b ) 2 c 2 a b | On applying R 1 → R 1 − R 2 = ( a − b ) ( b − c ) | − a − b − 2 c + 2 a + b + c a + b − b − c − c + a − ( 2 a + b + c ) ( b + c ) − a ( a + b ) 2 c 2 a b | = ( a − b ) ( b − c ) | a − c a − c a − c − ( 2 a + b + c ) ( b + c ) − a ( a + b ) 2 c 2 a b | Taking common ( a − c ) from R 1 = ( a − b ) ( b − c ) ( a − c ) | 1 1 1 − ( 2 a + b + c ) ( b + c ) − a ( a + b ) 2 c 2 a b | On applying C 1 → C 1 − C 2 and C 2 → C 2 − C 3 = ( a − b ) ( b − c ) ( a − c ) | 0 0 1 − 2 a − b − c − b − c b + c + a − a ( a + b ) 2 − c 2 c 2 − a b a b | = ( a − b ) ( b − c ) ( a − c ) | 0 0 1 − 2 a − 2 b − 2 c a + b + c − a ( a + b + c ) ( a + b − c ) c 2 − a b a b | = ( a − b ) ( b − c ) ( a − c ) | 0 0 1 − 2 ( a + b + c ) a + b + c − a ( a + b + c ) ( a + b − c ) c 2 − a b a b | Taking common(a+b+c)from
C 1 = ( a + b + c ) ( a − b ) ( b − c ) ( a − c ) | 0 0 1 − 2 a + b + c − a ( a + b − c ) c 2 − a b a b | = − ( a + b + c ) ( a − b ) ( b − c ) ( c − a ) | 0 0 1 ( a + b − c ) c 2 − a b a b | On expanding w.r.t R 1 = − ( a + b + c ) ( a − b ) ( b − c ) ( c − a ) [ 0 | a + b + c − a c 2 − a b a b | − 0 | − 2 − a a + b − c a b | + 1 | − 2 a + b + c a + b − c c 2 − a b | ] = − ( a + b + c ) ( a − b ) ( b − c ) ( c − a ) [ 0 − 0 + 1 { − 2 ( c 2 − a b ) − ( a + b − c ) ( a + b + c ) } ] = − ( a + b + c ) ( a − b ) ( b − c ) ( c − a ) [ − 2 c 2 + 2 a b − { ( a + b ) 2 − c 2 } ] = − ( a + b + c ) ( a − b ) ( b − c ) ( c − a ) [ − 2 c 2 + 2 a b − { a 2 + b 2 + 2 a b − c 2 } ] ∵ ( a + b ) 2 = a 2 + b 2 + 2 a b = − ( a + b + c ) ( a − b ) ( b − c ) ( c − a ) ( − 2 c 2 + 2 a b − a 2 − b 2 − 2 a b − c 2 ) = − ( a + b + c ) ( a − b ) ( b − c ) ( c − a ) ( − a 2 − b 2 − c 2 ) = ( − ) ( − ) ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) ( a 2 + b 2 + c 2 ) = ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) ( a 2 + b 2 + c 2 ) Hence it is proved that
| ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b | = ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) ( a 2 + b 2 + c 2 ) deteminants Exercise 5.2 Question 21
Answer:| ( a + 1 ) ( a + 2 ) a + 2 1 ( a + 2 ) ( a + 3 ) a + 3 1 ( a + 3 ) ( a + 4 ) a + 4 1 | = − 2
Hint: First we will make two elements of C 3 zero. then we will expand it
Given: | ( a + 1 ) ( a + 2 ) a + 2 1 ( a + 2 ) ( a + 3 ) a + 3 1 ( a + 3 ) ( a + 4 ) a + 4 1 | = − 2
Solution:
L.H.S | ( a + 1 ) ( a + 2 ) a + 2 1 ( a + 2 ) ( a + 3 ) a + 3 1 ( a + 3 ) ( a + 4 ) a + 4 1 |
On applying R 1 → R 1 − R 2 and R 2 → R 2 − R 3 = | ( a + 1 ) ( a + 2 ) − ( a + 2 ) ( a + 3 ) ( a + 2 ) − ( a + 3 ) 0 ( a + 2 ) ( a + 3 ) − ( a + 3 ) ( a + 4 ) ( a + 3 ) − ( a + 4 ) 0 ( a + 3 ) ( a + 4 ) a + 4 1 | = | ( a + 2 ) ( a + 1 − a − 3 ) a + 2 − a − 3 0 ( a + 3 ) ( a + 2 − a − 4 ) a + 3 − a − 4 0 ( a + 3 ) ( a + 4 ) a + 4 1 | = | ( a + 2 ) ( − 2 ) − 1 0 ( a + 3 ) ( − 2 ) − 1 0 ( a + 3 ) ( a + 4 ) a + 4 1 |
On expanding w.r.t C 3 = 0 | ( a + 3 ) ( − 2 ) − 1 ( a + 3 ) ( a + 4 ) a + 4 | − 0 | ( a + 2 ) ( − 2 ) − 1 ( a + 3 ) ( a + 4 ) a + 4 | + 1 ∣ ( a + 2 ) ( − 2 ) − 1 ( a + 3 ) ( − 2 ) − 1 ∣ = 0 − 0 + 1 { ( a + 2 ) ( − 2 ) ( − 1 ) − ( − 1 ) ( − 2 ) ( a + 3 ) } = 1 { 2 ( a + 2 ) − 2 ( a + 3 ) } = 2 a + 4 − 2 a − 6 = − 2 = R ⋅ H ⋅ S
Hence it is proved that
| ( a + 1 ) ( a + 2 ) a + 2 1 ( a + 2 ) ( a + 3 ) a + 3 1 ( a + 3 ) ( a + 4 ) a + 4 1 | = − 2
deteminants Exercise 5.2 Question 22
Answer: Answer:| a 2 a 2 − ( b − c ) 2 b c b 2 b 2 − ( c − a ) 2 c a c 2 c 2 − ( a − b ) 2 a b | = ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) ( a 2 + b 2 + c 2 ) Hint We will try to convert some elements of the determinant ( a − b ) , ( b − c ) and ( c − a ) to get the final answer Given:| a 2 a 2 − ( b − c ) 2 b c b 2 b 2 − ( c − a ) 2 c a c 2 c 2 − ( a − b ) 2 a b | = ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) ( a 2 + b 2 + c 2 ) Solution: L.H.S | a 2 a 2 − ( b − c ) 2 b c b 2 b 2 − ( c − a ) 2 c a c 2 c 2 − ( a − b ) 2 a b | On applying C 1 → C 2 − 2 C 1 − 2 C 3 = | a 2 a 2 − ( b − c ) 2 − 2 a 2 − 2 b c b c b 2 b 2 − ( c − a ) 2 − 2 b 2 − 2 c a c a c 2 c 2 − ( a − b ) 2 − 2 c 2 − 2 a b a b | = | a 2 − a 2 − ( b − c ) 2 − 2 b c b c b 2 − b 2 − ( c − a ) 2 − 2 c a c a c 2 − c 2 − ( a − b ) 2 − 2 a b a b | = | a 2 − { a 2 + ( b − c ) 2 + 2 b c } b c b 2 − { b 2 + ( c − a ) 2 + 2 c a } c a c 2 − { c 2 + ( a − b ) 2 + 2 a b } a b | = | a 2 − ( a 2 + b 2 + c 2 − 2 b c + 2 b c ) b c b 2 − ( b 2 + c 2 + a 2 − 2 c a + 2 c a ) c a c 2 − ( c 2 + a 2 + b 2 − 2 a b + 2 a b ) a b | = | a 2 − ( a 2 + b 2 + c 2 ) b c b 2 − ( a 2 + b 2 + c 2 ) c a c 2 − ( a 2 + b 2 + c 2 ) a b | Taking common − ( a 2 + b 2 + c 2 ) from C 2 = − ( a 2 + b 2 + c 2 ) | a 2 1 b c b 2 1 c a c 2 1 a b | On applying R 2 → R 2 − R 1 and R 3 → R 3 − R 1 = − ( a 2 + b 2 + c 2 ) | a 2 1 b c b 2 − a 2 0 c ( a − b ) c 2 − a 2 0 b ( a − c ) | = − ( a 2 + b 2 + c 2 ) | a 2 1 b c − ( a + b ) ( a − b ) 0 c ( a − b ) ( c + a ) ( c − a ) 0 − b ( c − a ) | On taking common ( a − b ) from R 2 and ( c − a ) from R 3 = − ( a 2 + b 2 + c 2 ) ( a − b ) ( c − a ) | a 2 1 b c − ( a + b ) 0 c ( c + a ) 0 − b | On expanding w.r.t R 1 = − ( a 2 + b 2 + c 2 ) ( a − b ) ( c − a ) [ a 2 | 0 c 0 − b | − 1 | − ( a + b ) c c + a − b | + b c ∣ − ( a + b ) 0 c + a 0 ] ] = − ( a 2 + b 2 + c 2 ) ( a − b ) ( c − a ) [ a 2 ( 0 − 0 ) − 1 { ( − b ) ( − ( a + b ) − c ( c + a ) ) + b c ( 0 − 0 ) } ] = − ( a 2 + b 2 + c 2 ) ( a − b ) ( c − a ) [ 0 − 1 { b ( a + b ) − c 2 − a c } ] = − ( a 2 + b 2 + c 2 ) ( a − b ) ( c − a ) { − 1 ( a b + b 2 − c 2 − a c ) } = − ( a 2 + b 2 + c 2 ) ( a − b ) ( c − a ) ( − a b − b 2 + c 2 + a c ) = − ( a 2 + b 2 + c 2 ) ( a − b ) ( c − a ) ( a c − a b − b 2 + c 2 ) = − ( a 2 + b 2 + c 2 ) ( a − b ) ( c − a ) { − a ( b − c ) − ( b 2 − c 2 ) } = − ( a 2 + b 2 + c 2 ) ( a − b ) ( c − a ) { − a ( b − c ) − ( b + c ) ( b − c ) } = − ( a 2 + b 2 + c 2 ) ( a − b ) ( c − a ) ( b − c ) ( − a − b − c ) = ( − ) ( − ) ( a 2 + b 2 + c 2 ) ( a − b ) ( c − a ) ( b − c ) ( a + b + c ) = ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) ( a 2 + b 2 + c 2 ) = R H . S Hence it is proved that
| a 2 a 2 − ( b − c ) 2 b c b 2 b 2 − ( c − a ) 2 c a c 2 c 2 − ( a − b ) 2 a b | = ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) ( a 2 + b 2 + c 2 ) determinants Exercise 5.2 Question 23
Answer:| 1 a 2 + b c a 3 1 b 2 + c a b 3 1 c 2 + a b c 3 | = − ( a − b ) ( b − c ) ( c − a ) ( a 2 + b 2 + c 2 ) Hint We will make some elements of the determinant zero
Given:| 1 a 2 + b c a 3 1 b 2 + c a b 3 1 c 2 + a b c 3 | = − ( a − b ) ( b − c ) ( c − a ) ( a 2 + b 2 + c 2 ) Solution: L.H.S | 1 a 2 + b c a 3 1 b 2 + c a b 3 1 c 2 + a b c 3 | On applying R 1 → R 1 − R 2 and R 2 → R 2 − R 3 = | 0 a 2 + b c − ( b 2 + c a ) a 3 − b 3 0 b 2 + c a − ( c 2 + a b ) b 3 − c 3 1 c 2 + a b c 3 | = | 0 a 2 + b c − b 2 − c a ( a − b ) ( a 2 + a b + b 2 ) 0 b 2 + c a − c 2 − a b ( b − c ) ( b 2 + b c + c 2 ) 1 c 2 + a b c 3 | … . . ( 1 ) Consider a 2 + b c − b 2 − c a = a 2 − b 2 + b c − c a = ( a − b ) ( a + b ) − ( a − b ) = ( a − b ) ( a + b − c ) Similarly b 2 + c a − c 2 − a b = ( b − c ) ( b + c − a ) ∵ a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) Now from ( 1 ) = | 0 ( a − b ) ( a + b − c ) ( a − b ) ( a 2 + a b + b 2 ) 0 ( b − c ) ( b + c − a ) ( b − c ) ( b 2 + b c + c 2 ) 1 c 2 + a b c 3 | On taking common ( a − b ) from R 1 and ( b − c ) from R 2 = ( a − b ) ( b − c ) | 0 ( a + b − c ) ( a 2 + a b + b 2 ) 0 ( b + c − a ) ( b 2 + b c + c 2 ) 1 c 2 + a b c 3 | On applying R 1 → R 1 − R 2 = ( a − b ) ( b − c ) | 0 ( a + b − c ) − ( b + c − a ) ( a 2 + a b + b 2 ) − ( b 2 + b c + c 2 ) 0 ( b + c − a ) ( b 2 + b c + c 2 ) 1 c 2 + a b c 3 | = ( a − b ) ( b − c ) | 0 a + b − c − b − c + a a 2 + a b + b 2 − b 2 − b c − c 2 0 ( b + c − a ) ( b 2 + b c + c 2 ) 1 c 2 + a b c 3 | = ( a − b ) ( b − c ) | 0 2 a − 2 c a 2 + a b − b c − c 2 0 ( b + c − a ) ( b 2 + b c + c 2 ) 1 c 2 + a b c 3 | … . . Consider a 2 + a b − b c − c 2 = a 2 − c 2 + a b − b c = ( a + c ) ( a − c ) + b ( a − c ) = ( a − c ) ( a + c + b ) From (2)
= ( a − b ) ( b − c ) | 0 2 ( a − c ) ( a − c ) ( a + c + b ) 0 b + c − a b 2 + b c + c 2 1 c 2 + a b c 3 | On taking common ( a − c ) from R 1 = ( a − b ) ( b − c ) ( a − c ) | 0 2 ( a + c + b ) 0 b + c − a b 2 + b c + c 2 1 c 2 + a b c 3 | On expanding w.r.t C 1 = − ( a − b ) ( b − c ) ( c − a ) [ 0 | b + c − a b 2 + b c + c 2 c 2 + a b c 3 | − 0 | 2 a + b + c c 2 + a b c 3 | + 1 ∣ 2 a + b + c b + c + a b 2 + b c + c 2 ] = − ( a − b ) ( b − c ) ( c − a ) [ 0 − 0 + { 2 ( b 2 + b c + c 2 ) − ( b + c − a ) ( b + c − a ) } ] = − ( a − b ) ( b − c ) ( c − a ) [ 2 b 2 + 2 b c + 2 c 2 − { ( b + c ) 2 − a 2 } ] = − ( a − b ) ( b − c ) ( c − a ) { 2 b 2 + 2 b c + 2 c 2 − ( b 2 + 2 b c + c 2 − a 2 ) } = − ( a − b ) ( b − c ) ( c − a ) ( 2 b 2 + 2 b c + 2 c 2 − b 2 − 2 b c − c 2 + a 2 ) = − ( a − b ) ( b − c ) ( c − a ) ( a 2 + b 2 + c 2 ) Hence it is proved that
| 1 a 2 + b c a 3 1 b 2 + c a b 3 1 c 2 + a b c 3 | = − ( a − b ) ( b − c ) ( c − a ) ( a 2 + b 2 + c 2 ) determinants Exercise 5.2 Question 24
Answer:| a 2 b c a c + c 2 a 2 + a b b 2 a c a b b 2 + b c c 2 | = 4 a 2 b 2 c 2 Hint We will convert some elements of the determinant into zero
Given: | a 2 b c a c + c 2 a 2 + a b b 2 a c a b b 2 + b c c 2 | = 4 a 2 b 2 c 2 Solution: L.H.S | a 2 b c a c + c 2 a 2 + a b b 2 a c a b b 2 + b c c 2 | = | a 2 b c c ( a + c ) a ( a + b ) b 2 a c a b b ( b + c ) c 2 | Taking out 'a' common from C 1 , and b from C 2 and c from C 3 = a b c | a c a + c a + b b a b b + c c | On applying C 1 → C 1 + C 2 + C 3 = a b c | a + c + a + c c a + c a + b + b + a b a b + b + c + c b + c c | = a b c | 2 a + 2 c c a + c 2 a + 2 b b a 2 b + 2 c b + c c | = a b c | 2 ( a + c ) c a + c 2 ( a + b ) b a 2 ( b + c ) b + c c | On taking 2 common from C 1 = 2 a b c | ( a + c ) c a + c ( a + b ) b a ( b + c ) b + c c | On applying C 1 → C 2 − C 1 , C 3 → C 3 − C 1 = 2 a b c | ( a + c ) c − a − c ( a + c ) − ( a + c ) ∣ ( a + b ) b − a − b a − a − b ( b + c ) ( b + c ) − ( b + c ) c − b − c | = 2 a b c | ( a + c ) − a 0 ( a + b ) − a − b ( b + c ) 0 − b | On taking ( − a ) common from C 2 and ( − b ) from C 3 = 2 a b c ( − a ) ( − b ) | ( a + c ) 1 0 ( a + b ) 1 1 ( b + c ) 0 1 | On expanding w.r.t R 1 = 2 a 2 b 2 c [ ( a + c ) | 1 1 0 1 | − 1 | a + b 1 b + c 1 | + 0 | a + b 1 b + c 0 | ] = 2 a 2 b 2 c [ ( a + c ) ( 1 − 0 ) − 1 ( a + b − b − c ) + 0 ] = 2 a 2 b 2 c [ a + c − 1 ( a − c ) ] = 2 a 2 b 2 c ( a + c − a + c ) = 2 a 2 b 2 c × 2 c = 4 a 2 b 2 c 2 = R ⋅ H ⋅ S Hence it is proved that
| a 2 b c a c + c 2 a 2 + a b b 2 a c a b b 2 + b c c 2 | = 4 a 2 b 2 c 2 determinants Exercise 5.2 Question 25
Answer: Answer:| x + 4 x x x x + 4 x x x x + 4 | = 16 ( 3 x + 4 ) Hint First we will try to convert some elements into zero
Given:| x + 4 x x x x + 4 x x x x + 4 | = 16 ( 3 x + 4 ) Solution: L.H.S | x + 4 x x x x + 4 x x x x + 4 | On applying R 1 → R 1 + R 2 + R 3 = | 3 x + 4 3 x + 4 3 x + 4 x x + 4 x x x x + 4 | On taking 3 x + 4 common from R 1 = ( 3 x + 4 ) | 1 1 1 x x + 4 x x x x + 4 | On applying C 1 → C 1 − C 2 , C 2 → C 2 − C 3 = ( 3 x + 4 ) | 0 0 1 x − x − 4 x + 4 − x x x − x x − x − 4 x + 4 | = ( 3 x + 4 ) | 0 0 1 − 4 4 x 0 − 4 x + 4 | On taking ( − 4 ) common from C 1 and ( − 4 ) from C 2 = ( 3 x + 4 ) ( − 4 ) ( − 4 ) | 0 0 1 1 − 1 x 0 1 x + 4 | = 16 ( 3 x + 4 ) | 0 0 1 1 − 1 x 0 1 x + 4 | On expanding w.r.t R 1 = 16 ( 3 x + 4 ) [ 0 | − 1 x 1 x + 4 | − 0 | 1 x 0 x + 4 | + 1 | 1 − 1 0 1 | ] = 16 ( 3 x + 4 ) [ 0 − 0 + 1 ( 1 − 0 ) ] = 16 ( 3 x + 4 ) × 1 = 16 ( 3 x + 4 ) = R ⋅ H ⋅ S Hence it is proved that
| x + 4 x x x x + 4 x x x x + 4 | = 16 ( 3 x + 4 ) determinants Exercise 5.2 Question 26
Answer: Answer:| 1 1 + p 1 + p + q 2 3 + 2 p 4 + 3 p + 2 q 3 6 + 3 p 10 + 6 p + 3 q | = 1 Hint we will convert element of
C 1 into zero
Given:| 1 1 + p 1 + p + q 2 3 + 2 p 4 + 3 p + 2 q 3 6 + 3 p 10 + 6 p + 3 q | = 1 Solution: L.H.S | 1 1 + p 1 + p + q 2 3 + 2 p 4 + 3 p + 2 q 3 6 + 3 p 10 + 6 p + 3 q | On applying R 2 → R 2 − 2 R 1 and R 3 → R 3 − 3 R 1 = | 1 1 + p 1 + p + q 0 1 2 + p 0 3 7 + 3 p | On expanding w.r.t C 1 = 1 | 1 2 + p 3 7 + 3 p | − 0 | 1 + p 1 + p + q 3 7 + 3 p | + 0 | 1 + p 1 + p + q 1 2 + p | = 1 { 1 ( 7 + 3 p ) − 3 ( 2 + p ) } − 0 + 0 = 7 + 3 p − 6 − 3 p = 1 = R ⋅ H ⋅ S Hence it is proved that
| 1 1 + p 1 + p + q 2 3 + 2 p 4 + 3 p + 2 q 3 6 + 3 p 10 + 6 p + 3 q | = 1 determinants Exercise 5.2 Question 27
Answer:| a b − c c − b a − c b c − a a − b b − a c | = ( a + b − c ) ( b + c − a ) ( c + a − b ) Hint First we will make elements of
C 1 ( a + b − c ) or 0 Given:| a b − c c − b a − c b c − a a − b b − a c | = ( a + b − c ) ( b + c − a ) ( c + a − b ) Solution: L.H.S | a b − c c − b a − c b c − a a − b b − a c | On applying C 1 → C 1 + C 2 and C 2 → C 2 + C 3 = | a + b − c 0 c − b a + b − c b + c − a c − a 0 b + c − a c | On taking ( a + b − c ) common from C 1 and ( b + c − a ) common from C 2 = ( a + b − c ) ( b + c − a ) | 1 0 c − b 1 1 c − a 0 1 c | On expanding w.r.t R 1 = ( a + b − c ) ( b + c − a ) [ 1 | 1 c − a 1 c | − 0 | 1 c − a 0 c | + ( c − b ) | 1 1 0 1 | ] = ( a + b − c ) ( b + c − a ) [ 1 { c − ( c − a ) } − 0 + ( c − b ) ( 1 − 0 ) ] = ( a + b − c ) ( b + c − a ) ( c − c + a + c − b ) = ( a + b − c ) ( b + c − a ) ( c + a − b ) = R ⋅ H ⋅ S Hence it is proved that
| a b − c c − b a − c b c − a a − b b − a c | = ( a + b − c ) ( b + c − a ) ( c + a − b ) deteminants Exercise 5.2 Question 28
Answer: Answer:| a 2 2 a b b 2 b 2 a 2 2 a b 2 a b b 2 a 2 | = ( a 3 + b 3 ) 2 Hint First we will try to make some elements of
R 1 or C 1 ( a + b ) 2 = a 2 + 2 a b + b 2 Because from R.H.S
( a 3 + b 3 ) 2 = [ ( a + b ) ( a 2 − a b + b 2 ) ] 2 = ( a + b ) 2 ( a 2 − a b + b 2 ) 2 Given: | a 2 2 a b b 2 b 2 a 2 2 a b 2 a b b 2 a 2 | = ( a 3 + b 3 ) 2 Solution: L.H.S | a 2 2 a b b 2 b 2 a 2 2 a b 2 a b b 2 a 2 | On applying R 1 → R 1 + R 2 + R 3 = | a 2 + b 2 + 2 a b 2 a b + a 2 + b 2 b 2 + 2 a b + a 2 b 2 a 2 2 a b 2 a b b 2 a 2 | ∵ ( a + b ) 2 = a 2 + 2 a b + b 2 = | ( a + b ) 2 ( a + b ) 2 ( a + b ) 2 b 2 a 2 2 a b 2 a b b 2 a 2 | On taking ( a + b ) 2 common from R 1 = ( a + b ) 2 | 1 1 1 b 2 a 2 2 a b 2 a b b 2 a 2 | On expanding w.r.t R 1 = ( a + b ) 2 [ 0 | a 2 − 2 a b 2 a b b 2 − a 2 a 2 | − 0 | b 2 − a 2 2 a b 2 a b − b 2 a 2 | + 1 | b 2 − a 2 a 2 − 2 a b 2 a b − b 2 b 2 − a 2 | ] = ( a + b ) 2 [ 0 − 0 + 1 { ( b 2 − a 2 ) ( b 2 − a 2 ) } − ( a 2 − 2 a b ) ( 2 a b − b 2 ) ] = ( a + b ) 2 ( b 4 + a 4 − 2 a 2 b 2 − 2 a 3 b + a 2 b 2 + 4 a 2 b 2 − 2 a b 3 ) = ( a + b ) 2 [ ( a 2 ) 2 + ( b 2 ) 2 + 2 a 2 b 2 − 2 a 3 b − 2 a b 3 + a 2 b 2 ] = ( a + b ) 2 [ ( a 2 ) 2 + ( a b ) 2 + ( b 2 ) 2 − 2 a 3 b − 2 a b 3 + 2 a 2 b 2 ] = ( a + b ) 2 ( a 2 − a b + b 2 ) 2 = ( a 3 + b 3 ) 2 = R ⋅ H . S Hence it is proved that
| a 2 2 a b b 2 b 2 a 2 2 a b 2 a b b 2 a 2 | = ( a 3 + b 3 ) 2 determinants Exercise 5.2 Question 29
Answer: Answer:| a 2 + 1 a b a c a b b 2 + 1 b c c a c b c 2 + 1 | = 1 + a 2 + b 2 + c 2 Hint We will try to convert some elements of determinant into zero
Given:| a 2 + 1 a b a c a b b 2 + 1 b c c a c b c 2 + 1 | = 1 + a 2 + b 2 + c 2 Solution: L.H.S | a 2 + 1 a b a c a b b 2 + 1 b c c a c b c 2 + 1 | = | 1 + a 2 a b a c a b 1 + b 2 b c c a c b 1 + c 2 | Multiply C 1 by a , C 2 by b and C 3 by c = 1 a b c | a ( 1 + a 2 ) a b 2 a c 2 a 2 b b ( 1 + b 2 ) b c 2 c a 2 c b 2 c ( 1 + c 2 ) | On taking a common from R 1 , b from R 2 and c from R 3 = a b c a b c | 1 + a 2 b 2 c 2 a 2 1 + b 2 c 2 a 2 b 2 1 + c 2 | = | 1 + a 2 b 2 c 2 a 2 1 + b 2 c 2 a 2 b 2 1 + c 2 | On applying C 1 → C 1 + C 2 + C 3 = | 1 + a 2 + b 2 + c 2 b 2 c 2 1 + a 2 + b 2 + c 2 1 + b 2 c 2 1 + a 2 + b 2 + c 2 b 2 1 + c 2 | On taking ( 1 + a 2 + b 2 + c 2 ) common from C 1 = ( 1 + a 2 + b 2 + c 2 ) | 1 b 2 c 2 1 1 + b 2 c 2 1 b 2 1 + c 2 | On applying R 1 → R 1 − R 2 , R 2 → R 2 − R 3 = ( 1 + a 2 + b 2 + c 2 ) | 0 b 2 − 1 − b 2 c 2 0 1 + b 2 − b 2 c 2 − 1 − c 2 1 b 2 1 + c 2 | = ( 1 + a 2 + b 2 + c 2 ) | 0 − 1 0 0 1 − 1 1 b 2 1 + c 2 | On expanding w.r.t R 1 = ( 1 + a 2 + b 2 + c 2 ) [ 0 | 1 − 1 b 2 1 + c 2 | − ( − 1 ) | 0 − 1 1 1 + c 2 | + 0 | 0 1 1 b 2 | ] = ( 1 + a 2 + b 2 + c 2 ) [ 0 + 1 { 0 − ( − 1 ) 1 } + 0 ] = ( 1 + a 2 + b 2 + c 2 ) ( 1 × 1 ) = ( 1 + a 2 + b 2 + c 2 ) = R ⋅ H ⋅ S Hence it is proved that
| a 2 + 1 a b a c a b b 2 + 1 b c c a c b c 2 + 1 | = 1 + a 2 + b 2 + c 2 determinants Exercise 5.2 Question 30
Answer: | 1 a a 2 a 2 1 a a a 2 1 | = ( a 3 − 1 ) 2 Hint We will try to convert some elements of determinant into zero
Given:| 1 a a 2 a 2 1 a a a 2 1 | = ( a 3 − 1 ) 2 Solution: L.H.S | 1 a a 2 a 2 1 a a a 2 1 | On applying R 1 → R 1 + R 2 + R 3 = | 1 + a 2 + a a + 1 + a 2 a 2 + a + 1 a 2 1 a a a 2 1 | On taking ( a 2 + a + 1 ) common from R 1 = ( a 2 + a + 1 ) | 1 1 1 a 2 1 a a a 2 1 | On applying C 1 → C 1 − C 2 , C 2 → C 2 − C 3 = ( a 2 + a + 1 ) | 0 0 1 a 2 − 1 1 − a a a − a 2 a 2 − 1 1 | = ( a 2 + a + 1 ) | 0 0 1 ( a + 1 ) ( a − 1 ) − ( a − 1 ) a − a ( a − 1 ) ( a + 1 ) ( a − 1 ) 1 | ∵ a 2 − b 2 = ( a + b ) ( a − b ) On taking common ( a − 1 ) from C 1 and ( a − 1 ) from C 2 = ( a 2 + a + 1 ) ( a − 1 ) ( a − 1 ) | 0 0 1 ( a + 1 ) − 1 a − a ( a + 1 ) 1 | On expanding w.r.t R 1 = ( a 2 + a + 1 ) ( a − 1 ) 2 [ 0 | − 1 a a + 1 1 | − 0 | a + 1 a − a 1 | + 1 | a + 1 − 1 − a a + 1 | ] = ( a − 1 ) 2 ( a 2 + a + 1 ) [ 0 − 0 + 1 { ( a + 1 ) 2 − ( − a ) ( − 1 ) } ] = ( a − 1 ) 2 ( a 2 + a + 1 ) ( a 2 + 2 a + 1 − a ) = ( a − 1 ) 2 ( a 2 + a + 1 ) ( a 2 + a + 1 ) = { ( a − 1 ) ( a 2 + a + 1 ) } 2 = ( a 3 − 1 ) 2 = R ⋅ H ⋅ S Hence it is proved that
| 1 a a 2 a 2 1 a a a 2 1 | = ( a 3 − 1 ) 2 determinants Exercise 5.2 Question 31
Answer: 2 ( a + b ) ( b + c ) ( c + a ) Hint Use determinant formula
Given:| a + b + c − c − b − c a + b + c − a − b − a a + b + c | = 2 ( a + b ) ( b + c ) ( c + a ) Solution: L.H.S | a + b + c − c − b − c a + b + c − a − b − a a + b + c | Use C 1 → C 1 + C 2 and C 2 → C 2 + C 3 | Δ | = | a + b − b − c − b a + b b + c − a − a − b b + c a + b + c | Common ( a + b ) ( b + c ) from C 1 & C 2 | Δ | = ( a + b ) ( b + c ) | 1 − 1 − b 1 1 − a − 1 1 a + b + c | Use R 1 → R 1 + R 2 and R 2 → R 2 + R 3 | Δ | = ( a + b ) ( b + c ) | 2 0 − b + a 0 2 b + c − 1 1 a + b + c | Expanding w.r.t C 1 | Δ | = ( a + b ) ( b + c ) [ 2 ( 2 a + 2 b + 2 c − b − c ) + ( − 1 ) ( 2 b + 2 a ) ] | Δ | = ( a + b ) ( b + c ) [ 4 a + 4 b + 4 c − 2 b − 2 c − 2 b − 2 a ] | Δ | = 2 ( a + b ) ( b + c ) ( c + a ) determinants Exercise 5.2 Question 32
Answer: 4 a b c Hint: Use determinant formula
Given: | b + c a a b a + c b c c a + b | = 4 a b c Solution: L.H.S | b + c a a b a + c b c c a + b | Use R 1 → R 1 + R 2 + R 3 | Δ | = | 2 ( b + c ) 2 ( c + a ) 2 ( a + b ) b a + c b c c a + b | Now R 2 → R 2 − R 1 and R 3 → R 3 − R 1 and 2 common | Δ | = 2 | b + c c + a a + b − c 0 − a − b − a 0 | Use R 1 → R 1 + R 2 + R 3 | Δ | = 2 | 0 c b − c 0 − a − b − a 0 | Expanding w.r.t C 1 | Δ | = 2 [ − c ( 0 − a b ) + b ( a c − 0 ) ] | Δ | = 2 ( a b c + a b c ) | Δ | = 2 ( 2 a b c ) = 4 a b c = R ⋅ H . S determinants Exercise 5.2 Question 33
Answer: 4 a 2 b 2 c 2 Hint: Use determinant formula
Given: | b 2 + c 2 a b a c b a c 2 + a 2 b c c a c b a 2 + b 2 | = 4 a 2 b 2 c 2 Solution: L.H.S | b 2 + c 2 a b a c b a c 2 + a 2 b c c a c b a 2 + b 2 | Use R 1 → a R 1 , R 2 → b R 2 , R 3 → c R 3 = 1 a b c | a ( b 2 + c 2 ) a 2 b a 2 c b 2 a b ( c 2 + a 2 ) b 2 c c 2 a c 2 b c ( a 2 + b 2 ) | Take common a from R 1 , b from R 2 and c from R 3 = a b c a b c | ( b 2 + c 2 ) a 2 a 2 b 2 ( c 2 + a 2 ) b 2 c 2 c 2 ( a 2 + b 2 ) | Use R 1 → R 1 − R 2 − R 3 = | 0 − 2 c 2 − 2 b 2 b 2 ( c 2 + a 2 ) b 2 c 2 c 2 ( a 2 + b 2 ) | Expanding w.r.t C 1 = − b 2 | − 2 c 2 − 2 b 2 c 2 a 2 + b 2 | + c 2 | − 2 c 2 − 2 b 2 c 2 + a 2 b 2 | = − b 2 ( − 2 a 2 c 2 − 2 b 2 c 2 + 2 b 2 c 2 ) + c 2 ( − 2 c 2 b 2 + 2 b 2 c 2 + 2 a 2 b 2 ) = 2 a 2 b 2 c 2 + 2 a 2 b 2 c 2 = 4 a 2 b 2 c 2 = R . H . S deteminants Exercise 5.2 Question 34
Answer:2 a 3 b 3 c 3 Hint Use determinant formula
Given: | 0 a b 2 a c 2 a 2 b 0 b c 2 a 2 c b 2 c 0 | = 2 a 3 b 3 c 3 Solution: L.H.S | 0 a b 2 a c 2 a 2 b 0 b c 2 a 2 c b 2 c 0 | Use R 1 → R 1 + R 2 + R 3 | Δ | = | a 2 ( b + c ) b 2 ( a + c ) c 2 ( a + b ) a 2 b 0 b c 2 a 2 c b 2 c 0 | Taking out a 2 from C 1 , b 2 from C 2 and c 2 from C 3 = a 2 b 2 c 2 | b + c a + c a + b b 0 b c c 0 | Use R 1 → R 1 − ( R 2 + R 3 ) = a 2 b 2 c 2 | b + c − ( b + c ) a + c − c a + b − b b 0 b c c 0 | = a 2 b 2 c 2 | 0 a a b 0 b c c 0 | Expanding w.r.t R 1 = a 2 b 2 c 2 ( − a ( 0 − b c ) + a ( b c − 0 ) ) = a 2 b 2 c 2 ⋅ 2 a b c = 2 a 3 b 3 c 3 = R ⋅ H . S determinants Exercise 5.2 Question 35
Answer: 4 a b c Hint Use determinant formula Given : | a 2 + b 2 c c c a b 2 + c 2 a a b b c 2 + a 2 b | = 4 a b c Solution: L.H.S | a 2 + b 2 c c c a b 2 + c 2 a a b b c 2 + a 2 b | R 1 → c R 1 , R 2 → c R 2 , R 3 → b R 3 = 1 a b c | a 2 + b 2 c 2 c 2 a 2 b 2 + c 2 a 2 b 2 b 2 c 2 + a 2 | Use R 1 → R 1 − R 2 − R 3 = 1 a b c | 0 − 2 b 2 − 2 a 2 a 2 b 2 + c 2 a 2 b 2 b 2 c 2 + a 2 | Now C 2 → C 2 − C 1 and C 3 → C 3 − C 1 = 1 a b c ∣ 0 − 2 b 2 − 2 a 2 a 2 b 2 + c 2 − a 2 a 2 b 2 b 2 c 2 + a 2 − b 2 Expanding w.r.t C 1 = 1 a b c [ c − a 2 ( − 2 b 2 ( c 2 + a 2 − b 2 ) − a ) + b 2 ( 0 − ( 2 a 2 ) ( b 2 + c 2 − a 2 ) ) ] = 1 a b c [ 2 a 2 b 2 ( c 2 + a 2 − b 2 ) + b 2 ( ( 2 a 2 ) ( b 2 + c 2 − a 2 ) ) ] = 2 a 2 b 2 a b c ( c 2 + a 2 − b 2 + b 2 + c 2 − a 2 ) = 2 a b c ( 2 c 2 ) = 4 a b c = R ⋅ H ⋅ S determinants Exercise 5.2 Question 36
Answer:( a b + b c + c a ) 3 Hint Use determinant formula
Given:| − b c b 2 + b c c 2 + a b a 2 + a c − a c c 2 + a c a 2 + a b b 2 + a b − a b | = ( a b + b c + c a ) 3 Solution: L.H.S | − b c b 2 + b c c 2 + a b a 2 + a c − a c c 2 + a c a 2 + a b b 2 + a b − a b | R 1 → a R 1 , R 2 → b R 2 , R 3 → c R 3 = 1 a b c | − a b c a b 2 + a b c a c 2 + a b c a 2 + a b c − a b c c 2 + a b c a 2 + a b c b 2 c + a b c − a b c | Common a from C 1 , b from C 2 and c from C 3 = a b c a b c | b c a b + a c a c + a b a b + b c − a c b c + a b a c + b c b c + a c − a b | Use R 1 → R 1 + R 2 + R 3 = | a b + b c + a c a b + b c + a c a b + b c + a c a b + b c − a c b c + a b a c + b c b c + a c − a b | ( a b + b c + a c ) common from R 1 = ( a b + b c + a c ) | 1 1 1 a b + b c − a c b c + a b a c + b c b c + a c − a b | Now C 2 → C 2 − C 1 and C 3 → C 3 − C 1 = ( a b + b c + a c ) ∣ 1 0 0 a b + b c − ( a b + b c + a c ) a b + b c + a c a c + b c 0 − ( a b + b c + a c ) ( a b + b c + a c ) common from C 2 & C 3 = ( a b + b c + a c ) 3 | 1 0 0 a b + b c − 1 1 a c + b c 0 − 1 | Expanding w.r.t R 1 = ( a b + b c + a c ) 3 { 1 ( 1 − 0 ) + 0 } = ( a b + b c + a c ) 3 = R ⋅ H ⋅ S determinants Exercise 5.2 Question 37
Answer: ( 5 x + λ ) ( λ − x ) 2 Hint: Use determinant formula
Given: | x + λ 2 x 2 x 2 x x + λ 2 x 2 x 2 x x + λ | = ( 5 x + λ ) ( λ − x ) 2 Solution: L.H.S | x + λ 2 x 2 x 2 x x + λ 2 x 2 x 2 x x + λ | Use C 1 → C 1 + C 2 + C 3 = | 5 x + λ 2 x 2 x 5 x + λ x + λ 2 x 5 x + λ 2 x x + λ | ( 5 x + λ ) common from C 1 = ( 5 x + λ ) | 1 2 x 2 x 1 x + λ 2 x 1 2 x x + λ | Now R 2 → R 2 − R 1 and R 3 → R 3 − R 1 = ( 5 x + λ ) | 1 2 x 2 x 0 x − λ 0 0 0 x − λ | = ( 5 x + λ ) ( − 1 ) ( − 1 ) | 1 2 x 2 x 0 − x + λ 0 0 0 − x + λ | Expanding w.r.t R 1 = ( 5 x + λ ) [ 1 ( λ − x ) 2 − 0 + 0 ] = ( 5 x + λ ) ( λ − x ) 2 = R ⋅ H . S deteminants Exercise 5.2 Question 38
Answer:( 5 x + 4 ) ( 4 − x ) 2 Hint Use determinant formula
Given:| x + 4 2 x 2 x 2 x x + 4 2 x 2 x 2 x x + 4 | = ( 5 x + 4 ) ( 4 − x ) 2 Solution: L.H.S | x + 4 2 x 2 x 2 x x + 4 2 x 2 x 2 x x + 4 | Use C 1 → C 1 + C 2 + C 3 = | 5 x + 4 2 x 2 x 5 x + 4 x + 4 2 x 5 x + 4 2 x x + 4 | ( 5 x + 4 ) common from C 1 = ( 5 x + 4 ) | 1 2 x 2 x 1 x + 4 2 x 1 2 x x + 4 | Now R 2 → R 2 − R 1 and R 3 → R 3 − R 1 = ( 5 x + 4 ) | 0 x − 4 0 0 4 + x − 2 x 2 x − x − 4 1 2 x x + 4 | ( x − 4 ) from R 1 common ( x − 4 ) ( 5 x + 4 ) | 0 x − 4 0 0 4 + x − 2 x 2 x − x − 4 1 2 x x + 4 | Expanding w.r.t
C 1 = ( x − 4 ) ( 5 x + 4 ) [ ( x − 4 ) ( x − 4 ) − 0 ] = ( 5 x + 4 ) ( x − 4 ) 2 = R . H . S deteminants Exercise 5.2 Question 39
Answer:4 x y z Hint Use determinant formula
Given: | y + z z y z z + x x y x x + y | = 4 x y z Solution: L.H.S | y + z z y z z + x x y x x + y | Apply C 1 → C 1 + C 2 + C 3 = | 2 ( y + z ) z y 2 ( x + z ) z + x x 2 ( x + y ) x x + y | 2 common from C 1 = 2 | ( y + z ) z y ( x + z ) z + x x ( x + y ) x x + y | Now C 1 → C 1 − C 2 = 2 | y z y 0 z + x x y x x + y | Now C 3 → C 3 − C 1 = 2 | y z 0 0 z + x x y x x | Now R 3 → R 3 − R 1 = 2 | y z 0 0 z + x x 0 x − z x | Expanding w.r.t
C 1 = 2 y ( z + x ) x − x ( x − z ) − 0 + 0 = 2 y ( z x + x 2 − x 2 + x z ) = 4 x y z = R ⋅ H . S determinants Exercise 5.2 Question 40
Answer: a b c ( a 2 + b 2 + c 2 ) 3 Hint Use determinant formula
Given: | − a ( b 2 + c 2 − a 2 ) 2 b 3 2 c 3 2 a 3 − b ( c 2 + a 2 − b 2 ) 2 c 3 2 a 3 2 b 3 − c ( a 2 + b 2 − c 2 ) | = a b c ( a 2 + b 2 + c 2 ) 3 Solution: L.H.S | − a ( b 2 + c 2 − a 2 ) 2 b 3 2 c 3 2 a 3 − b ( c 2 + a 2 − b 2 ) 2 c 3 2 a 3 2 b 3 − c ( a 2 + b 2 − c 2 ) | Common a from C 1 , b from C 2 and c from C 3 = a b c | − ( b 2 + c 2 − a 2 ) 2 b 2 2 c 2 2 a 2 − ( c 2 + a 2 − b 2 ) 2 c 2 2 a 2 2 b 2 − ( a 2 + b 2 − c 2 ) | Apply C 1 → C 1 + C 2 + C 3 = a b c | − b 2 − c 2 + a 2 + 2 b 2 + 2 c 2 2 b 2 2 c 2 2 a 2 + 2 c 2 − c 2 − a 2 + b 2 − c 2 + a 2 − b 2 2 c 2 2 a 2 + 2 b 2 − a 2 − b 2 + c 2 2 b 2 − a 2 + b 2 − c 2 | = a b c | a 2 + b 2 + c 2 2 b 2 2 c 2 a 2 + b 2 + c 2 − c 2 + a 2 − b 2 2 c 2 a 2 + b 2 + c 2 2 b 2 − a 2 + b 2 − c 2 | = a b c ( a 2 + b 2 + c 2 ) | 1 2 b 2 2 c 2 1 − c 2 + a 2 − b 2 2 c 2 1 2 b 2 − a 2 + b 2 − c 2 | Now R 2 → R 2 − R 1 , R 3 → R 3 − R 1 = a b c ( a 2 + b 2 + c 2 ) | 1 2 b 2 2 c 2 0 − ( a 2 + b 2 + c 2 ) 0 0 0 − ( a 2 + b 2 + c 2 ) | Expanding w.r.t C 1 = a b c ( a 2 + b 2 + c 2 ) ( a 2 + b 2 + c 2 ) 2 = a b c ( a 2 + b 2 + c 2 ) 3 = R ⋅ H ⋅ S Question:41.1 determinants Exercise 5. 2 Question 41
Answer: a 3 + 3 a 2 Hint Use determinant formula
Given:| 1 + a 1 1 1 1 + a 1 1 1 1 + a | = a 3 + 3 a 2 Solution: L.H.S | 1 + a 1 1 1 1 + a 1 1 1 1 + a | Apply C 1 → C 1 + C 2 + C 3 = | a + 3 1 1 a + 3 1 + a 1 a + 3 1 1 + a | ( a + 3 ) common from C 1 [ C 1 → C 1 a + 3 ] = ( a + 3 ) | 1 1 1 1 1 + a 1 1 1 1 + a | Now C 3 → C 3 − C 1 = ( a + 3 ) | 1 1 0 1 1 + a 0 1 1 a | Expanding w.r.t C 3 = ( a + 3 ) { a ∣ 1 1 1 1 + a } = ( a + 3 ) { a ( 1 + a − 1 ) } = ( a + 3 ) a 2 = a 3 + 3 a 2 = R H . S determinants Exercise 5.2 Question 41 sub question 2
Answer: Answer:
( a − 1 ) 3 Hint Use determinant formula
Given:
| a 2 + 2 a 2 a + 1 1 2 a + 1 a + 2 1 3 3 1 | = ( a − 1 ) 3 Solution:
L.H.S | a 2 + 2 a 2 a + 1 1 2 a + 1 a + 2 1 3 3 1 | Apply C 1 → C 1 − C 2 & C 2 → C 2 − C 3 = | a 2 − 1 2 a 1 a − 1 a + 1 1 0 2 1 | ( a − 1 ) common from C 1 = ( a − 1 ) | a + 1 2 a 1 1 a + 1 1 0 2 1 | Apply R 1 → R 1 − R 2 & R 2 → R 2 − R 3 = ( a − 1 ) | a ( a − 1 ) 0 1 a − 1 0 0 2 1 | Expanding w.r.t C 3 = ( a − 1 ) ( 1 ( a 2 − a − a + 1 ) ) = ( a − 1 ) ( a 2 − 2 a + 1 ) = ( a − 1 ) ( a − 1 ) 2 = ( a − 1 ) 3 = R . H ⋅ S deteminants Exercise 5.2 Question 42
Answer: ( x + y + z ) 3 Hint Use determinant formula
Given:| 2 y y − z − x 2 y 2 z 2 z z − x − y x − y − z 2 x 2 x | = ( x + y + z ) 3 L.H.S | 2 y y − z − x 2 y 2 z 2 z z − x − y x − y − z 2 x 2 x | Solution: Apply R 1 → R 1 + R 2 + R 3 = | x + y + z x + y + z x + y + z 2 z 2 z z − x − y x − y − z 2 x 2 x | ( x + y + z ) common from C 1 = ( x + y + z ) | 1 1 1 2 z 2 z z − x − y x − y − z 2 x 2 x | Apply C 2 → C 2 − C 1 & C 3 → C 3 − C 1 = ( x + y + z ) | 1 0 1 2 z 0 − ( x + y + z ) x − y − z x + y + z x + y + z | ( x + y + z ) common from C 2 & C 3 = ( x + y + z ) 3 | 1 0 1 2 z 0 − 1 x − y − z 1 1 | Expanding w.r.t C 3 = ( x + y + z ) 3 [ 1 | 0 − 1 1 1 | − 0 ( 1 + 0 ) 1 ] = ( x + y + z ) 3 | 0 − 1 1 1 | = ( x + y + z ) 3 ( 0 ( 1 ) − ( − 1 ) 1 ) = ( x + y + z ) 3 = R ⋅ H . S determinants Exercise 5.2 Question 43
Answer:( x + y + z ) ( x − z ) 2 Hint Use determinant formula
Given:| y + z x y z + x z x x + y y z | = ( x + y + z ) ( x − z ) 2 Solution: L.H.S | y + z x y z + x z x x + y y z | Apply R 1 → R 1 + R 2 + R 3 = | 2 ( x + y + z ) x + y + z x + y + z z + x z x x + y y z | ( x + y + z ) common from R 1 = ( x + y + z ) | 2 1 1 z + x z x x + y y z | Apply R 1 → R 1 − R 2 = ( x + y + z ) | 1 1 1 x z x x y z | Apply C 2 → C 2 − C 1 & C 3 → C 3 − C 1 = ( x + y + z ) | 1 0 0 x z − x 0 x y − x z − x | Expanding w.r.t C 3 = ( x + y + z ) ( ( z − x ) 2 − 0 ) = ( x + y + z ) ( x − z ) 2 = R ⋅ H ⋅ S determinants Exercise 5.2 Question 44
Answer:a 2 ( a + x + y + z ) Hint Use determinant formula Given:| a + x y z x a + y z x y a + z | = a 2 ( a + x + y + z ) Solution: L.H.S | a + x y z x a + y z x y a + z | Apply C 1 → C 1 + C 2 + C 3 = | a + x + y + z y z a + x + y + z a + y z a + x + y + z y a + z | ( a + x + y + z ) common from C 1 = ( a + x + y + z ) | 1 y z 1 a + y z 1 y a + z | Apply R 1 → R 1 − R 2 & R 2 → R 2 − R 3 = ( a + x + y + z ) | 0 − a 0 0 a − a 1 y a + z | Expanding w.r.t C 3 = ( a + x + y + z ) ( 1 ( a 2 ) ) = ( a + x + y + z ) a 2 = R ⋅ H . S deteminants Exercise 5.2 Question 45
Answer: 2 ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) Hint Use determinant formula
Given:| a 3 2 a b 3 2 b c 3 2 c | = 2 ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) Solution: L.H.S | a 3 2 a b 3 2 b c 3 2 c | Apply R 2 → R 2 − R 1 & R 3 → R 3 − R 1 = | a 3 2 a b 3 − a 3 0 b − a c 3 − a 3 0 c − a | Use a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) = | a 3 2 a ( b − a ) ( b 2 + a b + a 2 ) 0 b − a ( c − a ) ( c 2 + a c + a 2 ) 0 c − a | = ( − 1 ) ( − 1 ) | a 3 2 a ( a − b ) ( b 2 + a b + a 2 ) 0 a − b ( c − a ) ( c 2 + a c + a 2 ) 0 c − a | ( a − b ) ( c − a ) common from R 2 & R 3 = ( a − b ) ( c − a ) | a 3 2 a ( b 2 + a b + a 2 ) 0 1 ( c 2 + a c + a 2 ) 0 1 | Expanding w.r.t C 2 = ( a − b ) ( c − a ) [ 2 ( b 2 + a b + a 2 − c 2 − a c − a 2 ) ] = 2 ( a − b ) ( c − a ) [ ( b + c ) ( b − c ) + a ( b − c ) ] = 2 ( a − b ) ( b − c ) ( c − a ) ( a + b + c ) = R . H . S determinants Exercise 5.2 Question 46
Answer: Proved
Hint Use determinant formula
Given: | a b c x y z p q r | = | x y z p q r a b c | = | y b q x a p z c r | Solution: | a b c x y z p q r | = | x y z p q r a b c | R 1 ↔ R 3 , R 2 ↔ R 1 , R 3 ↔ R 1 = ( − ) ( − ) | x y z a b c p q r | For interchanging R2↔R3
negative sign will be added
Now convert rows in columns and column in row
| x y z p q r a b c | = | x p a y q b z r c | Now R 1 ↔ R 2 , C 2 ↔ C 3 ( − ) ( − ) | y q b x p a z r c | = | y b q x a p z c r | deteminants Exercise 5.2 Question 47
Answer: 0 Hint a , b , c are in AP Given: | x + 1 x + 2 x + a x + 2 x + 3 x + b x + 3 x + 4 x + c | = 0 where a , b , c are in AP Solution: L.H.S | x + 1 x + 2 x + a x + 2 x + 3 x + b x + 3 x + 4 x + c | Now R 3 → − 2 R 2 + R 1 + R 3 = | x + 1 x + 2 x + a x + 2 x + 3 x + b 0 0 a + c − 2 b | As
a b 2 c are in AP , 2 b = a + c Substituting the value in the denominator,
= | x + 1 x + 2 x + a x + 2 x + 3 x + b 0 0 a + c − a − c | = | x + 1 x + 2 x + a x + 2 x + 3 x + b 0 0 0 | Expanding along
R 3 = 0 deteminants Exercise 5.2 Question 48
Answer: 0 Hint α , β , γ are in AP Given:| x − 3 x − 4 x − α x − 2 x − 3 x − β x − 1 x − 2 x − γ | = 0 where α , β , γ are in AP Solution: $$α , β , γ are in AP a 2 − a 1 = a 3 − a 1 β − γ = γ − β β + β = γ + α 2 β = γ + α γ + α − 2 β = 0 ...(1)$$ L.H.S | x − 3 x − 4 x − α x − 2 x − 3 x − β x − 1 x − 2 x − γ | Now R 1 → R 1 + R 3 − 2 R 2 = | 0 0 2 β − ( α + γ ) x − 2 x − 3 x − β x − 1 x − 2 x − γ | $$= | 0 0 2 β − 2 β x − 2 x − 3 x − β x − 1 x − 2 x − γ | … … … from(1) = | 0 0 0 x − 2 x − 3 x − β x − 1 x − 2 x − γ | Expanding along
R 1 ,
= 0 ..hence proved
determinants Exercise 5.2 Question 49
Answer: a + b + c = 0 & a = b = c Hint Use determinant formula
Given: $$ If a , b , c are real no. such that | b + c c + a a + b c + a a + b b + c a + b b + c c + a | = 0 show that either a + b = c or a = b = c Solution: $$ L.H.S | b + c c + a a + b c + a a + b b + c a + b b + c c + a | Apply R 1 → R 1 + R 2 + R 3 = | 2 ( a + b + c ) 2 ( a + b + c ) 2 ( a + b + c ) c + a a + b b + c a + b b + c c + a | $$2 ( a + b + c ) common from R 1 = 2 ( a + b + c ) | 1 1 1 c + a a + b b + c a + b b + c c + a | $$ Apply C 1 → C 1 − C 2 & C 2 → C 2 − C 3 = 2 ( a + b + c ) | 0 0 1 c − b a − c b + c a − c b − a c + a | $$ Expand from R 1 = 2 ( a + b + c ) { 1 ∣ c − b a − c a − c b − a } ( a + b + c ) = 0 or b c − b 2 − a c + a b − ( a − c ) 2 = 0 b c − b 2 − a c + a b = a 2 + c 2 − 2 a c a 2 + b 2 + c 2 − a 2 − a b − a c = 0 2 ( a 2 + b 2 + c 2 ) − 2 a c − 2 a b − 2 b c = 0 ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 = 0 a − b = 0 , b − c = 0 , c − a = 0 a = b , b = c , c = a Hence,
a + b + c = 0 & a = b = c deteminants Exercise 5.2 Question 50
Answer: 2 Hint Use determinant formula
Given: $$| p b c q q c a b r | = 0 Find value of p p − a + q q − b + r r − c & p ≠ a , q ≠ b , r ≠ c Solution: $$ L.H.S | p b c q q c a b r | Apply R 1 → R 1 − R 3 & R 2 → R 2 − R 3 | p − a 0 c − p 0 q − b c − p a 0 r | = 0 $$ Expand from R 1 = p − a | q − b c − p b r | + ( c − r ) | 0 q − b a b | = 0 = ( p − a ) [ r ( q − b ) − b ( c − r ) + ( c − r ) ( − a ( q − b ) ) ] = 0 = r ( q − b ) ( p − a ) ( p − a ) ( q − b ) ( r − c ) − b ( p − a ) ( c − r ) ( p − a ) ( q − b ) ( r − c ) − a ( c − r ) ( q − b ) ( p − a ) ( q − b ) ( r − c ) = 0 $$= r r − c + b q − b + a p − a = 0 = r r − c + b + q − q q − b + a + p − p p − a = 0 = r r − c + b − q q − b + q q − b + a − p p − a − p p − a = 0 = r r − c − 1 + q q − b − 1 − p p − a = 0 = p p − a + q q − b + r r − c = 2 determinants Exercise 5.2 Question 51
Answer: $$x = 1 , 2 , − 3 Hint Use determinant formula
Given: $$ S.T x = 2 is root of equation | x − 6 − 1 2 3 x x − 3 − 3 2 x x + 3 | = 0 Solution: $$| x − 6 − 1 2 3 x x − 3 − 3 2 x x + 3 | = 0 $$ Expand from C 1 = x ( − 3 x 2 − 6 x − 2 x 2 + 6 x ) + ( 2 x + 4 + 3 x − 9 ) + 30 x − 30 + 5 x = 0 = − 5 x 2 + 30 x − 30 + 5 x = 0 = − 5 ( x 3 + 7 x + 6 ) = 0 = x 3 + 7 x + 6 = 0 = ( x − 1 ) ( x 2 + x − 6 ) = 0 = ( x − 1 ) ( x 2 + 3 x − 2 x − 6 ) = 0 = ( x − 1 ) ( x + 3 ) ( x − 2 ) = 0 x = 1 , 2 , − 3 determinants Exercise 5.2 Question 52
Answer: 0 Hint Use determinant formula
Given: $$| x + a b c a x + b c a b x + c | = 0 Solution: $$ Apply C 1 → C 1 + C 2 + C 3 | x + a + b + c b c x + a + b + c x + b c x + a + b + c b x + c | ( x + a + b + c ) common from C 1 = ( x + a + b + c ) | 1 b c 1 x + b c 1 b x + c | $$ Apply R 2 → R 2 − R 1 & R 3 → R 3 − R 1 = ( x + a + b + c ) | 1 b c 0 x 0 0 0 x | Expand from C 1 = ( x + a + b + c ) ( ( x 2 − 0 ) − 0 + 0 ) = 0 x + a + b + c = 0 , x 2 = 0 x = − ( a + b + c ) , x = 0 deteminants Exercise 5.2 Question 52 sub question 2
Answer :
x = − a 3 Hint: Use determinant formula
Given: $$| x + a x x x x + a x x x x + a | = 0 Solution: $$ Apply C 1 → C 1 + C 2 + C 3 | 3 x + a x x 3 x + a x + a x 3 x + a x x + a | ( 3 x + a ) common from C 1 = ( 3 x + a ) | 1 x x 1 x + a x 1 x x + c | $$ Apply R 2 → R 2 − R 1 & R 3 → R 3 − R 1 = ( 3 x + a ) | 1 x x 0 a 0 0 0 a | Expand from C 1 = ( 3 x + a ) a 2 = 0 3 x + a = 0 , a 2 = 0 3 x = − a , a = 0 x = − a 3 , a = 0 deteminants Exercise 5.2 Question 52 sub question 3
Answer: x = 2 3 , 11 3 , 11 3 Hint: Use determinant formula
Given: $$| 3 x − 8 3 3 3 3 x − 8 3 3 3 3 x − 8 | = 0 Solution: $$ L.H.S | 3 x − 8 3 3 3 3 x − 8 3 3 3 3 x − 8 | $$ Apply C 1 → C 1 + C 2 + C 3 | 3 x − 2 3 3 3 x − 2 3 x − 8 3 3 x − 2 3 3 x − 8 | $$( 3 x − 2 ) common from C 1 = ( 3 x − 2 ) | 1 3 3 1 3 x − 8 3 1 3 3 x − 8 | Apply R 2 → R 2 − R 1 & R 3 → R 3 − R 1 = ( 3 x − 2 ) | 1 3 3 0 − 3 x − 11 0 0 0 − 3 x − 11 | Expand from C 1 = ( 3 x − 2 ) ( 3 x − 11 ) 2 = 0 3 x − 2 = 0 , 3 x − 11 = 0 3 x = 2 , 3 x = 11 x = 2 3 , x = 11 3 , 11 3 x = 2 3 , 11 3 , 11 3 deteminants Exercise 5.2 Question 52 sub question 4
Answer: x = a , b Hint Use determinant formula
Given:| 1 x x 2 1 a a 2 1 b b 2 | = 0 Solution: $$L ⋅ H S | 1 x x 2 1 a a 2 1 b b 2 | $$ Apply R 2 → R 2 − R 1 & R 3 → R 3 − R 1 | 1 x x 2 0 a − x a 2 − x 2 0 b − x b 2 − x 2 | $$( a − x ) ( b − x ) common from R 2 & R 3 = ( a − x ) ( b − x ) | 1 x x 2 0 1 a + x 0 1 b + x | Expand from C 1 ( a − x ) ( b − x ) [ b + x − a + x ] = 0 ( a − x ) ( b − x ) [ − a + b ] = 0 a = x b = x a = b − neglect deteminants Exercise 5.2 Question 52 sub question 5
Answer: Answer: x = 1 , 1 , − 9 Hint: Use determinant formula
Given: | x + 1 3 5 2 x + 2 5 2 3 x + 4 | = 0 Solution: Apply C 1 → C 1 + C 2 + C 3 | x + 9 3 5 x + 9 x + 2 5 x + 9 3 x + 4 | = 0 ( x + 9 ) common from C 1 = ( x + 9 ) | 1 3 5 1 x + 2 5 1 3 x + 4 | Apply R 2 → R 2 − R 1 & R 3 → R 3 − R 1 = ( x + 9 ) | 1 3 5 0 x − 1 0 0 0 x − 1 | Expand from C 1 ( x + 9 ) ( x − 1 ) 2 = 0 x + 9 = 0 , ( x − 1 ) 2 = 0 x = − 9 , x = 1 , 1 x = 1 , 1 , − 9 deteminants Exercise 5.2 Question 52 Sub Question 6
Answer: x = b , c , − ( b + c ) Hint: Use determinant formula
Given:| 1 x x 3 1 b b 3 1 c c 3 | = 0 Solution: LHS | 1 x x 3 1 b b 3 1 c c 3 | Apply R 2 → R 2 − R 1 & R 3 → R 3 − R 1 1 x x 3 0 b − x b 3 − x 3 0 c − x c 3 − x 3 ∣ Use b 3 − x 3 = ( b − x ) ( b 2 + b x + x 2 ) | 1 x x 3 0 b − x ( b − x ) ( b 2 + b x + x 2 ) 0 c − x ( c − x ) ( c 2 + c x + x 2 ) | ( b − x ) ( c − x ) common from R 2 & R 3 = ( b − x ) ( c − x ) | 1 x x 3 0 1 ( b 2 + b x + x 2 ) 0 1 ( c 2 + c x + x 2 ) | Expand from C 1 ( b − x ) ( c − x ) ( c 2 + c x + x 2 − b 2 − b x − x 2 ) = 0 b − x = 0 , c − x = 0 , c 2 − b 2 − b x + c x = 0 x = b , c = x , ( c + b ) ( c − b ) − x ( c − b ) = 0 deteminants Exercise 5.2 Question 52 Sub Question 7
Answer :
x = 4 Hint: Use determinant formula
Given: | 15 − 2 x 11 − 3 x 7 − x 11 17 14 10 16 13 | = 0 Solution: Apply R 3 → R 3 − R 2 15 − 2 x 11 − 3 x 7 − x 11 17 14 1 1 1 ∣= 0 Now C 1 → C 1 − C 2 & C 2 → C 2 − C 3 | 4 + x 4 − 2 x 7 − x − 6 3 14 0 0 1 | = 0 Expand from R 3 0 | 4 + x 4 − 2 x − 6 3 | − 0 | 4 + x 7 − x − 6 14 | + 1 | 4 + x 4 − 2 x − 6 3 | = 0 1 ( 3 ( 4 + x ) − ( − 6 ) ( 4 − 2 x ) ) = 0 12 + 3 x + 24 − 12 x = 0 36 = 9 x x = 4 deteminants Exercise 5.2 Question 52 Sub Question 9
Answer: θ = n π + ( − 1 ) n π 6 , n ∈ z or n π Hint: Use determinant formula
Given:| 3 − 2 sin 3 θ − 7 8 cos 2 θ − 11 14 2 | = 0 Solution: Apply C 1 → C 1 + C 2 1 − 2 sin 3 θ 1 8 cos 2 θ 3 14 2 ∣= 0 Apply R 2 → R 2 − R 3 & R 3 → R 3 − R 1 | 1 − 2 sin 3 θ 0 10 cos 2 θ − sin 3 θ 0 20 2 − 3 sin 3 θ | = 0 Expand from C 1 1 [ 10 ( 2 − 3 sin θ ) − 20 ( cos 2 θ − sin 3 θ ) ] = 0 2 − 3 sin 3 θ − 2 ( cos 2 θ − sin 3 θ ) = 0 2 − 3 sin 3 θ − 2 cos 2 θ + 2 sin 3 θ = 0 2 − 3 sin 3 θ − 2 cos 2 θ = 0 2 = 3 sin θ − 4 sin 3 θ + 2 ( 1 − 2 sin 2 θ ) 2 = 3 sin θ − 4 sin 3 θ + 2 − 4 sin 2 θ 0 = sin θ − ( 3 − 4 sin 2 θ − 4 sin θ ) − 4 sin 2 θ − 4 sin θ + 3 = 0 4 sin 2 θ + 4 sin θ − 3 = 0 sin θ = x 4 x 2 + 4 x − 3 = 0 4 x 2 + 6 x − 2 x − 3 = 0 2 x ( 2 x + 3 ) − 1 ( 2 x + 3 ) = 0 2 x + 3 = 0 , 2 x − 1 = 0 , 2 x = 0 x = − 3 2 , x = 1 2 , x = 0 , sin θ = − 3 2 , sin θ = 1 2 , sin θ = 0 As sin θ = 0 , θ = n π & as sin θ = 1 2 , θ = n π + ( − 1 ) n π 6 , n ∈ z deteminants Exercise 5.2 Question 52 .1
Answer: x = − 12 , 0 Hint: Use determinant formula
Given: | 4 − x 4 + x 4 + x 4 + x 4 − x 4 + x 4 + x 4 + x 4 − x | = 0 Solution: Apply C 1 → C 1 + C 2 + C 3 | 12 + x 4 + x 4 + x 12 + x 4 − x 4 + x 12 + x 4 + x 4 − x | = 0 ( 12 + x ) common from C 1 ( 12 + x ) | 1 4 + x 4 + x 1 4 − x 4 + x 1 4 + x 4 − x | = 0 Apply R 2 → R 2 − R 1 & R 3 → R 3 − R 1 ( 12 + x ) | 1 4 + x 4 + x 0 − 2 x 0 0 0 − 2 x | = 0 Expand from C 1 ( 12 + x ) ( 4 x 2 ) = 0 12 + x = 0 , 4 x 2 = 0 x = − 12 , x = 0 determinants Exercise 5.2 Question 53
Answer: 1 + 1 a + 1 b + 1 c = 0 Hint Use determinant formula
Given:| 1 + a 1 1 1 1 + b 1 1 1 1 + c | = 0 p.t 1 a + 1 b + 1 c + 1 = 0 Solution: Divide a to C 1 , b to C 2 and c to C 3 ( a × b × c ) | 1 a + 1 1 b 1 c 1 a 1 b + 1 1 c 1 a 1 b 1 c + 1 | = 0 Apply C 1 → C 1 + C 2 + C 3 (abc) | 1 a + 1 b + 1 c + 1 1 b 1 c 1 a + 1 b + 1 c + 1 1 b + 1 1 c 1 a + 1 b + 1 c + 1 1 b 1 c + 1 | = 0 ( 1 a + 1 b + 1 c + 1 ) common from C 1 ( a b c ) ( 1 a + 1 b + 1 c + 1 ) | 1 1 b 1 c 1 1 b + 1 1 c 1 1 b 1 c + 1 | = 0 Apply R 2 → R 2 − R 1 & R 3 → R 3 − R 1 a b c ( 1 a + 1 b + 1 c + 1 ) | 1 1 b 1 c 0 1 b + 1 − 1 b 1 c − 1 c 0 1 b − 1 b 1 c + 1 − 1 c | a b c | 1 a + 1 b + 1 c + 1 | | 1 1 b 1 c 0 1 0 0 0 1 | Expand from R 3 a b c | 1 a + 1 b + 1 c + 1 | = 0 a , b , c ≠ 0 , 1 a + 1 b + 1 c + 1 = 0 a b c ≠ 0 determinants Exercise 5.2 Question 5
Answer: 2 Hint: Use determinant formula
Given:| a b − y c − z a − x b c − z a − x b − y c | = 0 then using property of determinant. Find a x + b y + c z where x , y , z = 0 Solution: Take x , y , z common from C 1 , C 2 , C 3 ( x y z ) | a x b y − 1 c z − 1 a − 1 x b y c z − 1 a x − 1 b y − 1 c z | = 0 Apply C 1 → C 1 + C 2 + C 3 ∣ a x + b y + c z − 2 b y − 1 c z − 1 a x + b y + c z − 2 b y c z − 1 a x + b y + c z − 2 b y − 1 c z ∣= 0 ( a x + b y + c z − 2 ) | 1 b y − 1 c z − 1 1 b y c z − 1 1 b y − 1 c z | = 0 Apply R 2 → R 2 − R 1 & R 3 → R 3 − R 1 ( a x + b y + c z − 2 ) | 1 b y − 1 c z − 1 0 1 0 0 0 1 | = 0 ( a x + b y + c z − 2 ) ⋅ 1 = 0 or a x + b y + c z = 2 determinants Exercise 5.2 Question 55
Answer : 9 ( 3 x y z + x y + y z + z x ) Given: | 1 1 1 + 3 x 1 + 3 y 1 1 1 1 + 3 z 1 | = 9 ( 3 x y z + x y + y z + z x ) Solution: L.H.S | 1 1 1 + 3 x 1 + 3 y 1 1 1 1 + 3 z 1 | Apply R 1 → R 1 − R 3 & R 2 → R 2 − R 3 | 0 − 3 z 3 x 3 y − 3 z 0 1 1 + 3 z 1 | 3 common from R 1 & R 2 3 × 3 | 0 − z x y − z 0 1 1 + 3 z 1 | Expand from R 1 9 [ z ( y ) + x ( y + 3 z y + z ) 9 ( 3 x y z + x y + y z + z x )
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Chapter-wise RD Sharma Class 12 Solutions
