Determinants NCERT Solutions
ncert exercise 4.1
Evaluate the determinants in Exercises 1 and 2.
Question: 1. `|(2,4),(-5, -1)|`
Solution:
Let `|A|=|(2,4),(-5,-1)|`
`:.|A|= -2-(-20)`
`=-2+20=18`
Question: 2 .(i)`|A|= |(cos theta, -sin theta), (sin theta, cos theta)|`
Solution:(i)
Let `|A|=|(Costheta, -sintheta), (sintheta, Costheta)|`
`:.|A|=cos^2theta-(-sin^2theta)` `=cos^2theta+sin^2theta=1`
(ii)`|(x^2-x+1, x-1), (x+1, x+1)|`
Solution:
`:.|A|=(x+1)(x^2-x+1)` `-(x+1)(x-1)`
`=x^3+1-(x^2-1)`
`=x^3+1-x^2+1`
`=x^3-x^2+2.`
Question: 3.
ifA`=[(1, 2), (4, 2)]`then show that |24| = 4 |A|
Solution:
Here`A = [(1,2), (4,2)]`then `2A = [(2,4),(8,4)]`
`:.|2A| = 8 - 32 = -24 `
Also `|A|= 2 -8 =-6`
`:. 4|A| = 4xx -6 = - 24`
Thus `|2A| =4|A|`
Question: 4 .
If A = `[(1, 0, 1),(0, 1, 2), (0, 0, 4)]`then show that `|3A| = 27 |A|`
Solution:
Here `A=[(1, 0, 1),(0, 1, 2), (0, 0, 4)]`
then 3A =`[(3, 0, 3),(0, 3, 6), (0, 0, 12)]`
`:.|3A| = 3|(3, 6), (0, 12)|`
`-0|(0, 6), (0, 12)|+3|(0, 3), (0, 0)|`
`= 3(36-0)-(0-0)+ 3(0-0)`
`=108 - 0 + 0 = 108`
Also `|A|=1|(1, 2), (0,4)|-0|(0, 2),(0, 4)|` `+1|(0, 1),(0, 0)|`
`=1(4-0)-0(0-0)+ 1(0-0)`
= 4 - 0 + 0 = 4
`:.27 |A|= 27xx4=108`
Thus `|3A|= 27|A|`.
Question:5 .Evaluate the determinants:
(i)`|(3, -1, -2),(0, 0, -1), (3, -5, 0)|`
Solution:
(i) Let `|(3, -1, -2),(0, 0, -1), (3, -5, 0)|`
`:.|A| =3|(0, -1),(-5, 0)|` `-(-1)|(0, -1),(3, 0)|` `+(-2)|(0, 0),(3, 5)|`
`=3(0-5)+1(0+3)-2(0-0)`
`=-15+3-0=-12`
(ii)`|(3, -1, -2),(0, 0, -1), (3, -5, 0)|`
Solution:
Let `|(3, -1, -2),(0, 0, -1), (3, -5, 0)|`
`:.|A|=3|(1, -2),(3, 1)|` `- (-4)|(1, -2), (2, 1)|` `+5|(1, 1), (2, 3)|`
`=3(1+6)+4(1+4)+5(3-2)`
`21+20+5+=46`
(iii)`|(0, 1, 2), (-1, 0, -3),(-2, 3, 0)|`
Solution:
`|(0, 1, 2), (-1, 0, -3),(-2, 3, 0)|`
`:. |A|=0 |(0,-3),(3, 0)|-1|(-1, -3), (-2, 0)|` `+2|(-1, 0), (-2, 3)|`
`= 0(0+9)-1(0-6)` `+2(-3-0)`
`=0+6-6=0`
(iv)`|(2, -1, -2), (0, 2, -1), (3, -5, 0)|`
Solution:
Let |A|=`|(2, -1, -2), (0, 2, -1), (3, -5, 0)|`
`:.|A| =2|(2, -1), (-5, 0)|` `-(-1)|(0, -1), (3, 0)|` `+(-2)|(0, 2),(3, -5)|`
`= 2(0-5)+1 (0+3)-2 (0-6)`
`= - 10 + 3+12 =5`
Question: 6 .
If A = `|(1, 1, -2),(2, 1, -3), (5, 4, -9)|` find |A|
Solution:
Here A= `|(1, 1, -2),(2, 1, -3), (5, 4, -9)|`
then |A| =`|(1, 1, -2),(2, 1, -3), (5, 4, -9)|`
`:.|A|= 1|(1, -3),(4, -9)|-1|(2, -3), (5, -9)|` `+(-2)|(2, 1),(5, 4)|`
`=1(-9+12)-1(-18+15)` `-2(8-5)`
`=3+3-6=6-6=0`
Question:7 .
find values of x if
(i)`|(2, 4), (5, 1)|` `=|(2x, 4),(6, x)|`
Solution:
Here `|(2, 4), (5, 1)|` `=|(2x, 4),(6, x)|`
`:. (2-20)=(2x^2-24)`
`=> 2x^2= -18+24`
`=>2x^2 = 6`
`=>x^2=3`
`=>x=+-sqrt3`
(ii)`|(2, 3), (4, 5)| =|(x, 3),(2x, 5)|`
Solution:
Here `|(2, 3), (4, 5)| =|(x, 3),(2x, 5)|`
`:.(10-12)=(5x-6x)`
`=>-2 = -x =>x=2`
Question:8 . If `|(x, 2), (18, x)|=` `|(6, 2), (18, 6)|`
then x is equal to
(A)6
(B)`+-6`
(C)-6
(D)0
Solution:
Here `|(x, 2), (18, x)|`= `|(6, 2), (18, 6)|`
`=> (x^2-36)= (36- 36)`
`=> x^2 -36 = 0 `
`=> x^2 = 36 => x=+-6`
Thus answer is (B)
Reference: