मराठी

Using Properties of Determinants, Prove the Following: - Mathematics

Advertisements
Advertisements

प्रश्न

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`.

बेरीज
Advertisements

उत्तर

Δ = `|(a, b,c),(a-b, b-c, c-a),(b+c, c+a, a+b)|`

 

= `|(a,b,c),(a-b, b-c, c-a),(a+b+c, c+a+b, a+b+c)|   ...[ "Applying" R_3 -> R_3 + R_1]` 

 

= `(a +b+c)   |(a,b,c),(a-b, b-c, c-a),(1,1,1)|   ...["Taking" (a +b +c) "common"]`

 

= `(a +b+c)  |(a, b ,c),(-b , -c, -a),(1, 1, 1)|    ...["Applying" R_2 -> R_2 - R_1]`

 

= `(a +b+c) |(a-c, b-c, c),(-b +a , -c+a, -a),(0, 0, 1)|     ...[C_1 -> C_1 - C_3  "and"  C_2 ->C_2 - C_3]`

 

= `(a +b+c) [{(a-c) (a-c) -(b-c)(a-b)}]`

 

= `(a +b+c)  ...[{(a - c)^2 - (ab - ac - b^2 + bc)]`

 

= `(a +b+c)  [(a -c ^2- (ab - ac - b^2 + bc)]`

 

= `(a +b+c)  (a^2 + b^2 + c^2 - ab - bc - ca)`

 

= `a^3 + b^3 + c^3 - 3abc`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2018-2019 (March) 65/1/3

संबंधित प्रश्‍न

 

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`


Using the property of determinants and without expanding, prove that:

`|(2,7,65),(3,8,75),(5,9,86)| = 0`


Using the property of determinants and without expanding, prove that:

`|(b+c, q+r, y+z),(c+a, r+p, z +x),(a+b, p+q, x + y )| = 2|(a,p,x),(b,q,y),(c, r,z)|`


By using properties of determinants, show that:

`|(-a^2, ab, ac),(ba, -b^2, bc),(ca,cb, -c^2)| = 4a^2b^2c^2`


Using properties of determinants, prove that:

`|(x, x^2, 1+px^3),(y, y^2, 1+py^3),(z, z^2, 1+pz^2)|` = (1 + pxyz) (x – y) (y – z) (z – x), where p is any scalar.


Using properties of determinants, prove that:

`|(1, 1+p, 1+p+q),(2, 3+2p, 4+3p+2q),(3,6+3p,10+6p+3q)| =  1`                 


Using properties of determinants, prove that 

`|(a^2 + 2a,2a + 1,1),(2a+1,a+2, 1),(3, 3, 1)| = (a - 1)^3`


Using properties of determinants, prove the following :

\[\begin{vmatrix}1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1\end{vmatrix} = \left( 1 - a^3 \right)^2\].

Prove the following using properties of determinants :

\[\begin{vmatrix}a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b\end{vmatrix} = 2\left( a + b + c \right) {}^3\]


Without expanding the determinants, show that `|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|` = 0


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 `|(2, 3, 5),(400, 600, 1000),(48, 47, 18)|` by using properties


Prove that: `|(y^2z^2, yz, y + z),(z^2x^2, zx, z + x),(x^2y^2, xy, x + y)|` = 0


Prove that: `|(y + z, z, y),(z, z + x, x),(y, x, x + y)|` = 4xyz


The maximum value of Δ = `|(1, 1, 1),(1, 1 + sin theta, 1),(1 + cos theta, 1, 1)|` is ______. (θ is real number)


The value of the determinant `|(x , x + y, x + 2y),(x + 2y, x, x + y),(x + y, x + 2y, x)|` is ______.


`|(x + 1, x + 2, x + "a"),(x + 2, x + 3, x + "b"),(x + 3, x + 4, x + "c")|` = 0, where a, b, c are in A.P.


Let Δ = `|("a", "p", x),("b", "q", y),("c", "r", z)|` = 16, then Δ1 = `|("p" + x, "a" + x, "a" + "p"),("q" + y, "b" + y, "b" + "q"),("r" + z, "c" + z, "c" + "r")|` = 32.


The determinant `abs (("a","bc","a"("b + c")),("b","ac","b"("c + a")),("c","ab","c"("a + b"))) =` ____________


If `abs ((2"x",5),(8, "x")) = abs ((6,-2),(7,3)),`  then the value of x is ____________.


In a triangle the length of the two larger sides are 10 and 9, respectively. If the angles are in A.P., then the length of the third side can be ______.


Without expanding evaluate the following determinant.

`|(1, a, a + 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)|`


By using properties of determinant prove that `|(x+y,y+z,z+x),(z,x,y),(1,1,1)|` = 0


By using properties of determinant prove that

`|(x+y,y+z,z+x),(z,x,y),(1,1,1)|=0`


Without expanding determinant find the value of `|(10,57,107),(12,64,124),(15,78,153)|`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×