Advertisements
Advertisements
प्रश्न
Without expanding determinants, show that
`|(1, 3, 6),(6, 1, 4),(3, 7, 12)| + |(2, 3, 3),(2, 1, 2),(1, 7, 6)| = 10|(1, 2, 1),(3, 1, 7),(3, 2, 6)|`
उत्तर
L.H.S. = `|(1, 3, 6),(6, 1, 4),(3, 7, 12)| + 4 |(2, 3, 3),(2, 1, 2),(1, 7, 6)|`
In 1st determinant, taking 2 common from C3, we get
L.H.S. = `2|(1,3,3),(6,1,2),(3,7,6)| + 4|(2,3,3),(2,1,2),(1,7,6)|`
= `|(2, 3, 3),(12, 1, 2),(6, 7, 6)| + |(8, 3, 3),(8, 1, 2),(4, 7, 6)|`
= `|(2 + 8, 3, 3),(12 + 8, 1, 2),(6 + 4, 7, 6)|`
= `|(10, 3, 3),(20, 1, 2),(10, 7, 6)|`
Interchanging rows and columns, we get
L.H.S. = `|(10, 20, 10),(3, 1, 7),(3, 2, 6)|`
Taking 10 common from R1, we get
L.H.S. = `10|(1, 2, 1),(3, 1, 7),(3, 2, 6)|`
= R.H.S.
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:
`|(1, bc, a(b+c)),(1, ca, b(c+a)),(1, ab, c(a+b))| = 0`
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 properties of determinants, prove the following :
Using properties of determinants, prove that:
`|(a,b,b+c),(c,a,c+a),(b,c,a+b)|` = (a+b+c)(a-c)2
Using properties of determinants, prove the following:
`|(a, b,c),(a-b, b-c, c-a),(b+c, c+a, a+b)| = a^3 + b^3 + c^3 - 3abc`.
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)|`.
Without expanding the determinants, show that `|("b" + "c", "bc", "b"^2"c"^2),("c" + "a", "ca", "c"^2"a"^2),("a" + "b", "ab", "a"^2"b"^2)|` = 0
Without expanding evaluate the following determinant:
`|(1, "a", "b" + "c"),(1, "b", "c" + "a"),(1, "c", "a" + "b")|`
Using properties of determinant show that
`|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc
Select the correct option from the given alternatives:
If x = –9 is a root of `|(x, 3, 7),(2, x, 2),(7, 6, x)|` = 0 has other two roots are
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: `|(3x, -x + y, -x + z),(x - y, 3y, z - y),(x - z, y - z, 3z)|`
Evaluate: `|("a" - "b" - "c", 2"a", 2"a"),(2"b", "b" - "c" - "a", 2"b"),(2"c", 2"c", "c" - "a" - "b")|`
Evaluate the following determinant without expanding:
`|(5, 5, 5),(a, b, c),(b + c, c + a, a + b)|`
Without expanding evaluate the following determinant.
`|(1,"a","b+c"),(1,"b","c+a"),(1,"c","a+b")|`
Without expanding evaluate the following determinant.
`|(1, a, b+c),(1, b, c+a),(1, c, a+b)|`
Without expanding evaluate the following determinant.
`|(1, a, b + c),(1, b, c + a),(1, c, a + b)|`