Advertisements
Advertisements
प्रश्न
Solve the following equation: `|(x + 2, x + 6, x - 1),(x + 6, x - 1,x + 2),(x - 1, x + 2, x + 6)|` = 0
उत्तर
`|(x + 2, x + 6, x - 1),(x + 6, x - 1,x + 2),(x - 1, x + 2, x + 6)|` = 0
Applying R2 → R2 – R1 and R3 → R3 – R1, we get
`|(x + 2, x + 6, x - 1),(4, -7, 3),(-3, -4, 7)|` = 0
∴ (x + 2)( – 49 + 12) – (x + 6)(28 + 9) + (x – 1) ( – 16 – 21) = 0
∴ (x + 2) ( – 37) – (x + 6) (37) + (x – 1) (– 37) = 0
∴ – 37(x + 2 + x + 6 x + x – 1) = 0
∴ 3x + 7 = 0
∴ x = `(-7)/3`
APPEARS IN
संबंधित प्रश्न
Using properties of determinants, prove that `|[2y,y-z-x,2y],[2z,2z,z-x-y],[x-y-z,2x,2x]|=(x+y+z)^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:
`|(2,7,65),(3,8,75),(5,9,86)| = 0`
By using properties of determinants, show that:
`|(1,1,1),(a,b,c),(a^3, b^3,c^3)|` = (a-b)(b-c)(c-a)(a+b+c)
By using properties of determinants, show that:
`|(y+k,y, y),(y, y+k, y),(y, y, y+k)| = k^2(3y + k)`
By using properties of determinants, show that:
`|(1,x,x^2),(x^2,1,x),(x,x^2,1)| = (1-x^3)^2`
Using properties of determinants, prove the following:
Without expanding the determinants, show that `|(x"a", y"b", z"c"),("a"^2, "b"^2, "c"^2),(1, 1, 1)| = |(x, y, z),("a", "b", "c"),("bc", "ca", "ab")|`
Without expanding evaluate the following determinant:
`|(2, 3, 4),(5, 6, 8),(6x, 9x, 12x)|`
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
Select the correct option from the given alternatives:
Let D = `|(sintheta*cosphi, sintheta*sinphi, costheta),(costheta*cosphi, costheta*sinphi, -sintheta),(-sintheta*sinphi, sintheta*cosphi, 0)|` then
Select the correct option from the given alternatives:
The value of a for which system of equation a3x + (a + 1)3 y + (a + 2)3z = 0 ax + (a +1)y + (a + 2)z = 0 and x + y + z = 0 has non zero Soln. is
Answer the following question:
Without expanding determinant show that
`|(x"a", y"b", z"c"),("a"^2, "b"^2, "c"^2),(1, 1, 1)| = |(x, y, z),("a", "b", "c"),("bc", "ca", "ab")|`
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)|`
In a third order matrix B, bij denotes the element in the ith row and jth column. If
bij = 0 for i = j
= 1 for > j
= – 1 for i < j
Then the matrix is
`f : {1, 2, 3) -> {4, 5}` is not a function, if it is defined by which of the following?
The value of the determinant `|(1, cos(β - α), cos(γ - α)),(cos(α - β), 1, cos(γ - β)),(cos(α - γ), cos(β - γ), 1)|` is equal to ______.
Without expanding evaluate the following determinant.
`|(1,"a","b+c"),(1,"b","c+a"),(1,"c","a+b")|`