Advertisements
Advertisements
प्रश्न
Solve: `|(x,2,-1),(2,5,x),(-1,2,x)|` = 0
उत्तर
Expanding along R1 we get
`x|(5,x),(2,x)| - 2|(2,x),(-1,x)| - 1|(2,5),(-1,2)|` = 0
⇒ x(5x - 2x) - 2(2x + x) - 1(4 + 5) = 0
⇒ x(3x) - 2(3x) - 1(9) = 0
⇒ 3x2 - 6x - 9 = 0
⇒ x2 - 2x - 3 = 0 ....(Divided by 3)
⇒ (x - 3)(x + 1) = 0
⇒ x = 3 or -1
APPEARS IN
संबंधित प्रश्न
Evaluate the following determinants: `|(2"i", 3),(4, -"i")|`
Evaluate the following determinant:
`|("a", "h", "g"),("h", "b", "f"),("g", "f","c")|`
The value of `|(2x + y,x,y),(2y+z,y,z),(2z+x,z,x)|` is
The value of `|(5,5,5),(4x,4y,4z),(-3x,-3y,-3z)|` is
If any three rows or columns of a determinant are identical then the value of the determinant is:
Evaluate the following determinant:
`|(a,h,g),(h,b,f),(g,f,c)|`
Evaluate the following determinants:
`|(4,7),(-7,0)|`
Find the value of x if `|(x, 2x, 3),(-1, 1, -4),(3, -3, 5)|` = 29
Evaluate the following determinant:
`[(4, 7) ,(-7, 0)]`
Evaluate the following determinant:
`|(a, h, g),(h, b, f),(g, f, c)|`