हिंदी

Prove the Following Using Properties of Determinants : ∣ ∣ ∣ ∣ a + B + 2 C a B C B + C + 2 a B C a C + a + 2 B ∣ ∣ ∣ ∣ = 2 ( a + B + C ) 3 - Mathematics

Advertisements
Advertisements

प्रश्न

Prove the following using properties of determinants :

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

Advertisements

उत्तर

Let \[\bigtriangleup = \begin{vmatrix}a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b\end{vmatrix}\]

Applying  \[C_1 \to C_1 + C_2 + C_3\] , we get: 

\[\bigtriangleup = \begin{vmatrix}2a + 2b + 2c & a & b \\ 2a + 2b + 2c & b + c + 2a & b \\ 2a + 2b + 2c & a & c + a + 2b\end{vmatrix}\]

\[\Rightarrow \bigtriangleup = 2\left( a + b + c \right)\begin{vmatrix}1 & a & b \\ 1 & b + c + 2a & b \\ 1 & a & c + a + 2b\end{vmatrix}\]

Now, applying

\[R_2 \to R_2 - R_1 \text { and } R_3 \to R_3 - R_1\] , we get:

\[\Rightarrow \bigtriangleup = 2\left( a + b + c \right)\begin{vmatrix}1 & a & b \\ 0 & b + c + a & 0 \\ 0 & 0 & c + a + b\end{vmatrix}\]

\[\Rightarrow \bigtriangleup = 2 \left( a + b + c \right)^3\]

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

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2013-2014 (March) Delhi Set 2

संबंधित प्रश्न

By using properties of determinants, show that:

`|(1,a,a^2),(1,b,b^2),(1,c,c^2)| = (a - b)(b-c)(c-a)`


By using properties of determinants, show that:

`|(a-b-c, 2a,2a),(2b, b-c-a,2b),(2c,2c, c-a-b)| = (a + b + c)^2`


By using properties of determinants, show that:

`|(x+y+2z, x, y),(z, y+z+2z,y),(z,x,z+x+2y)| = 2(x+y+z)^3`


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:

`|(3a, -a+b, -a+c),(-b+a, 3b, -b+c),(-c+a, -c+b, 3c)|`= 3(a + b + c) (ab + bc + ca)


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 determinants, prove that

`|[b+c , a ,a  ] ,[ b , a+c, b ] ,[c , c, a+b ]|` = 4abc 


Without expanding evaluate the following determinant:

`|(1, "a", "b" + "c"),(1, "b", "c" + "a"),(1, "c", "a" + "b")|`


Using properties of determinants, show that `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc.


Prove that `|(x + y, y + z, z + x),(z + x, x + y, y + z),(y + z, z + x, x + y)| = 2|(x, y, z),(z, x, y),(y, z, x)|`


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


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:

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


Evaluate: `|("a" - "b" - "c", 2"a", 2"a"),(2"b", "b" - "c" - "a", 2"b"),(2"c", 2"c", "c" - "a" - "b")|`


If A + B + C = 0, then prove that `|(1, cos"c", cos"B"),(cos"C", 1, cos"A"),(cos"B", cos"A", 1)|` = 0


Find the value of θ satisfying `[(1, 1, sin3theta),(-4, 3, cos2theta),(7, -7, -2)]` = 0


If `[(4 - x, 4 + x, 4 + x),(4 + x, 4 - x, 4 + x),(4 + x, 4 + x, 4 - x)]` = 0, then find values of x.


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 `abs ((alpha, beta, gamma),(alpha^2, beta^2, gamma^2),(beta + gamma, gamma + alpha, alpha + beta)) =` ____________.


The A.M., H.M. and G.M. between two numbers are `144/15`, 15 and 12, but not necessarily in this order then, H.M., G.M. and A.M. respectively are


If `|(α, 3, 4),(1, 2, 1),(1, 4, 1)|` = 0, then the value of α is ______.


Without expanding determinant 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 evaluate the following determinant.

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


Without expanding determinant 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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×