Determinants NCERT Solutions
NCERT Exercise 4.2
Using the property of determinants and without expanding in Exercises 1 to 5 prove that.
Question :1 .
`|(x, a, x+a), (y, b, y+b),(z, c, z+c)|`=0
Solution:
Let`|(x, a, x+a), (y, b, y+b),(z, c, z+c)|`=0
Operating `C_1-> C_1+C_2` we get
`|(x+a, a, x+a), (y+b, b, y+b),(z+c, c, z+c)|`=0`[∵C_1 and C_2` are identical]
Question: 2 .
`|(a-b, b-c, c-a), (b-c, c-a, a-b), (c-a, a-b, b-c)|`=0
Solution:
Let |A|= `|(a-b, b-c, c-a), (b-c, c-a, a-b), (c-a, a-b, b-c)|`=0
Operating `R_1->R_1+R_2+R_3`
We get = `|(0, 0, 0), (b-c, c-a, a-b),(c-a, a-b, b-c)|`= 0
[∵ Each element of `R_1` is zero]
Question: 3 . `|(2, 7, 65),(3, 8, 75),(5, 9, 86)|`=0
Solution:
Let `|A| =|(2, 7, 65),(3, 8, 75),(5, 9, 86)|`
Operating `C_1-> C_1+9C_2`, we get
=`|(65, 7, 65), (75, 8, 75), (86, 9, 86)|= 0`
[∵ `c_1 and C_3` are identical]
Question: 4 .
`|(1, bc, a(b+c)), (1, ca, b(c+a)), (1, ab, c(a+b))|`= 0
Solution:
Let |A| = `|(1,bc, (ab+ac)),(1, ca,(bc+ab)),(1, ab, (ca+bc))|`
Oprating `C_3-> C_2+C_3` ,we get
=`|(1, bc, ab+bc+ac),(1, ca, ab+bc+ac), (1, ab, ab+bc+ac)|`
Taking (ab+bc+ac)common from `C_3` we get
=(ab+bc+ac)`|(1, bc, 1), (1,ca, 1), (1, ab, 1)|`
= `(ab+bc+ac)xx0` = 0
[∵ `C_1 and c_3` are identical]
Question:5 .
`|(b+c, q+r, y+z), (c+a, r+p, z+x), (a+b, p+q, x+y)|`
= 2 `|(a, p, x), (b, q, y), (c, r, z)|`
Solution:
Let |A| `|(b+c, q+r, y+z), (c+a, r+p, z+x), (a+b, p+q, x+y)|`
Operating `R_1-> R_1 +R_2- R_3` we get
`|(2c, 2r, 2x), (c+a, r+p, z+x), (a+b, p+q, x+y)|`
Taking 2 common from `R_1` we get
`=2 |(c, r, x), (c+a, r+p, z+x), (a+b, p+q, x+y)|`
Operationg `R_2-> R_2-R_1` we get
=2 `|(c, r, z), (a, p, x),(a+b, p+q, x+y)|`
Opreting `R_3-> R_3- R_2` , we get
= `2|(c, r, z), (a, p, x), (b, q, y)|`
`-2|(a, p, x), (c, r, z), (b, q, y)|`
[∵ `R_1harr R_2`]
=`2|(a, p, x), (b, q, y), (c, r, z)|`
[∵ `R_2harr R_3`]
R.H.S.
By using properties of determinants in Exercies 6 to 14 show that:
Question:6 .
`|(0, a, -b), (-a, 0, -c), (b, c, 0)|`
Solution:
L.H.S.=`|(0, a, -b), (-a, 0, -c), (b, c, 0)|`
Operating`R_2 ->bR_2 and R_3-> aR_3`
`=1/(ab)|(0, a, -b),(-ab, 0, -bc), (ab, ac, 0)|`
Operating `R_2-> R_2 + R_3` we get
`=1/(ab)|(0, a, -b), (0, ac, -bc), (ab, ac, o)|`
Expanding using first column, we get
=`1/(ab)[0|(ac, -bc), (ac, 0)|` `-0|(a, -b), (ac, 0)|+ab|(a, -b), (ac, -bc)|`
`=1/(ab)[ab(-abc+abc)]` `=1/(ab)xx0` = 0 = R.H.S.
Question: 7.
`|(-a^2, ab, ac), (ba, -b^2, bc), (ca, cb, -c^2)|` `= 4a^2b^2c^2`.
Solution:
L.H.S.`|(-a^2, ab, ac), (ba, -b^2, bc), (ca, cb, -c^2)|`
Taking a commonn from `C_1, b` from `C_2` and c from `C_3` we get
`=abc|(-a, a, a), (b, -b, b), (c, c, -c)|`
Taking a commonn from `R_1, b` from `R_2` and c from `R_3` we get
`=(abc)(abc)|(-1, 1, 1), (1, -1, 1), (1, 1, -1)|`
Operating `R_1-> R_1+R_3 and R_2 ->R_2 - R_3` we get
`=a^2b^2c^2|(0, 0, 2), (0, -2, 2)(1, 1, -1)|`
Expanding using first column, we get
`a^2b^2c^2[0|(-2, 2), -0` `|(0, 2), (1, -1)|+1` `|(0, 2),(-2, 2)|]`
`a^2b^2c^2[0-0+1(0+4)]` `=4a^2b^2c^2`=R.H.S.
Question:8 .
(i)`|(1, a, a^2), (1, b, b^2), (1, c, c^2)|` `=(a-b)(b-c)(c-a)`
Solution:
(i)L.H.S.`|(1, a, a^2), (1, b, b^2), (1, c, c^2)|`
Operation `R_2 -> R_2-R_2 and R_3 -> R_3 - R_1` we get
`= |(1, a, a^2), (0, b-a, b^2-a^2), (0, c-a, c^2-a^2)|`
Expanding using first column, we get
`= 1|(b-a, b^2, -a^2), (c-a, c^2, -a^2)|`
`=(b-a)(c^2-a^2)-(c-a)(b^2-a^2)`
`=(b-a)(c+a)(c-a)-(c-a)` `(b+a)(b-a)`
`=(b-a)(c+a)(c-a)` `-(c-a)(b+a)(b-a)`
`=(b-a)(cxxa)[c+a-b-a]`
`=(b-a)(c-a)(c-b)`
`=(a-b)(b-c)(c-a)` =RHS
(ii) `|(1,1,1),(a,b,c),(a^3,b^3,c^3)|`
`=(a-b)(b-c)(c-a)(a+b+c)`
Solution:
Here, LHS=`|(1,1,1),(a,b,c),(a^3,b^3,c^3)|`
Operating, `C_2->C_2-C_1` and `C_3->C_3-C_1` we get
`=|(1,0,0),(a, b-c, c-a),(a^3, b^3-c^3, c^3-a^3)|`
Expanding using first row, we get
`=1|(b-a, c-a), (b^3-a^3, c^3-a^3)|`
`=(b-a)(c^3-a^3)` `-(b^3-a^3)(c-a)`
`=(b-a)(c-a)(c^2+a^2+ac)` `-(b-a)(b^2+a^2+ab)(c-a)`
`=(b-a)(c-a)` `[c^2+ac-b^2-a^2-ab]`
`=(b-a)(c-a)` `[c^2+ac-b^2-ab]`
`=(b-a)(c-a)` `[(c^2-b^2)+(ac+ab)]`
`=(b-a)(c-a)` `[(c+b)(c-b)+a(c-b)]`
`=(b-a)(c-a)(c-b)` `[c+b+a]`
`=(a-b)(b-c)(c-a)` `(a+b+c)`=RHS
Reference: