Advertisements
Advertisements
प्रश्न
If `|("x"^"k", "x"^("k" + 2), "x"^("k" + 3)),("y"^"k", "y"^("k" + 2), "y"^("k" + 3)),("z"^"k", "z"^("k" + 2), "z"^("k" + 3))|` = (x - y) (y - z) (z - x)`(1/"x"+ 1/"y" + 1/"z") ` then
विकल्प
k = –3
k = –1
k = 1
k = 3
उत्तर
k = – 1
Explanation:
L.H.S →
`|(x^k, x^(k + 2), x^(k + 3)),(y^k, y^(k + 2), y^(k + 3)),(z^k, z^(k + 2), z^(k + 3))| = x^k y^k z^k |(1, x^2, x^3),(1, y^2, y^3),(1, z^2, z^3)|`
= (xyz)k(x – y) (y – z) (z – x) (xy + yz + zx)
R.H.S. →
= `(x - y) (y - z) (z - x) (1/x+ 1/y + 1/z)`
= `(x - y) (y - z) (z - x) ((xy + yz + zx)/(xyz))`
L.H.S. = R.H.S.
∴ `(xyz)^k (x - y) (y - z) (z - x) (xy + yz + zx) = (x - y) (y - z) (z - x) ((xy + yz + zx)/(xyz))`
∴ `(xyz)^k cancel((x - y)) cancel((y - z)) cancel((z - x)) (xy + yz + zx) = cancel((x - y)) cancel((y - z)) cancel((z - x)) ((xy + yz + zx)/(xyz))`
∴ `(xyz)^k cancel((xy + yz + zx)) = cancel((xy + yz + zx))/(xyz)`
∴ (xyz)k = `1/"xyz"`
∴ (xyz)k = xyz- 1
∴ k = – 1
APPEARS IN
संबंधित प्रश्न
Using properties of determinants prove the following: `|[1,x,x^2],[x^2,1,x],[x,x^2,1]|=(1-x^3)^2`
If ` f(x)=|[a,-1,0],[ax,a,-1],[ax^2,ax,a]| ` , using properties of determinants find the value of f(2x) − f(x).
Using properties of determinants, prove that
`|[x+y,x,x],[5x+4y,4x,2x],[10x+8y,8x,3x]|=x^3`
Using properties of determinants, prove that
`|[b+c,c+a,a+b],[q+r,r+p,p+q],[y+z,z+x,x+y]|=2|[a,b,c],[p,q,r],[x,y,z]|`
Using the property of determinants and without expanding, prove that:
`|(a-b,b-c,c-a),(b-c,c-a,a-b),(a-a,a-b,b-c)| = 0`
By using properties of determinants, show that:
`|(x,x^2,yz),(y,y^2,zx),(z,z^2,xy)| = (x-y)(y-z)(z-x)(xy+yz+zx)`
By using properties of determinants, show that:
`|(1+a^2-b^2, 2ab, -2b),(2ab, 1-a^+b^2, 2a),(2b, -2a, 1-a^2-b^2)| = (1+a^2+b^2)`
Using properties of determinants, prove that:
`|(alpha, alpha^2,beta+gamma),(beta, beta^2, gamma+alpha),(gamma, gamma^2, alpha+beta)|` = (β – γ) (γ – α) (α – β) (α + β + γ)
Using properties of determinants, prove that `|(x,x+y,x+2y),(x+2y, x,x+y),(x+y, x+2y, x)| = 9y^2(x + y)`
Using propertiesof determinants prove that:
`|(x , x(x^2), x+1), (y, y(y^2 + 1), y+1),( z, z(z^2 + 1) , z+1) | = (x-y) (y - z)(z - x)(x + y+ z)`
Using properties of determinant prove that
`|(b+c , a , a), (b , c+a, b), (c, c, a+b)|` = 4abc
Solve for x : `|("a"+"x","a"-"x","a"-"x"),("a"-"x","a"+"x","a"-"x"),("a"-"x","a"-"x","a"+"x")| = 0`, using properties of determinants.
If `|(4 + x, 4 - x, 4 - x),(4 - x, 4 + x, 4 - x),(4 - x, 4 - x, 4 + x)|` = 0, then find the values of x.
Without expanding determinants, prove that `|(1, yz, y + z),(1, zx, z + x),(1, xy, x + y)| = |(1, x, x^2),(1, y, y^2),(1, z, z^2)|`.
Using properties of determinant show that
`|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc
Using properties of determinant show that
`|(1, log_x y, log_x z),(log_y x, 1, log_y z),(log_z x, log_z y, 1)|` = 0
Select the correct option from the given alternatives:
The determinant D = `|("a", "b", "a" + "b"),("b", "c", "b" + "c"),("a" + "b", "b" + "c", 0)|` = 0 if
Select the correct option from the given alternatives:
`|("b" + "c", "c" + "a", "a" + "b"),("q" + "r", "r" + "p", "p" + "q"),(y + z, z + x, x + y)|` =
Select the correct option from the given alternatives:
The system 3x – y + 4z = 3, x + 2y – 3z = –2 and 6x + 5y + λz = –3 has at least one Solution when
Answer the following question:
Evaluate `|(101, 102, 103),(106, 107, 108),(1, 2, 3)|` by using properties
Answer the following question:
By using properties of determinant prove that `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|` = 0
Answer the following question:
If `|("a", 1, 1),(1, "b", 1),(1, 1, "c")|` = 0 then show that `1/(1 - "a") + 1/(1 - "b") + 1/(1 - "c")` = 1
Evaluate: `|("a" + x, y, z),(x, "a" + y, z),(x, y, "a" + z)|`
Evaluate: `|(0, xy^2, xz^2),(x^2y, 0, yz^2),(x^2z, zy^2, 0)|`
Find the value of θ satisfying `[(1, 1, sin3theta),(-4, 3, cos2theta),(7, -7, -2)]` = 0
The maximum value of Δ = `|(1, 1, 1),(1, 1 + sin theta, 1),(1 + cos theta, 1, 1)|` is ______. (θ is real number)
If cos2θ = 0, then `|(0, costheta, sin theta),(cos theta, sin theta,0),(sin theta, 0, cos theta)|^2` = ______.
If the determinant `|(x + "a", "p" + "u", "l" + "f"),("y" + "b", "q" + "v", "m" + "g"),("z" + "c", "r" + "w", "n" + "h")|` splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.
The value of the determinant `abs ((alpha, beta, gamma),(alpha^2, beta^2, gamma^2),(beta + gamma, gamma + alpha, alpha + beta)) =` ____________.
Without expanding determinants find the value of `|(10,57,107), (12, 64, 124), (15, 78, 153)|`
Without expanding determinants find the value of `|(10,57,107),(12,64,124),(15,78,153)|`
Without expanding determinants, find the value of `|(10, 57, 107), (12, 64, 124), (15, 78, 153)|`
By using properties of determinants, prove that
`|(x+y, y+z, z+x),(z, x, y),(1, 1, 1)|` = 0
Without expanding evaluate the following determinant.
`|(1, a, b+c), (1, b, c+a), (1, c, a+b)|`
Without expanding determinants, find the value of `|(10, 57, 107),(12, 64, 124),(15, 78, 153)|`
Without expanding evaluate the following determinant.
`|(1, a, b + c),(1, b, c + a),(1, c, a + b)|`
