मराठी

Prove that : ∣ ∣ ∣ ∣ ∣ ( B + C ) 2 a 2 B C ( C + a ) 2 B 2 C a ( a + B ) 2 C 2 a B ∣ ∣ ∣ ∣ ∣ = ( a − B ) ( B − C ) ( C − a ) ( a + B + C ) ( a 2 + B 2 + C 2 ) - Mathematics

Advertisements
Advertisements

प्रश्न

Prove that :

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

Advertisements

उत्तर

\[\text{ Let LHS }= \Delta = \begin{vmatrix} \left( b + c \right)^2 & a^2 & bc\\ \left( c + a \right)^2 & b^2 & ca\\ \left( a + b \right)^2 & c^2 & ab \end{vmatrix}\] 
\[ = \begin{vmatrix} \left( b + c \right)^2 - \left( c + a \right)^2 & a^2 - b^2 & bc - ca\\ \left( c + a \right)^2 - \left( a + b \right)^2 & b^2 - c^2 & ca - ab\\ \left( a + b \right)^2 & c^2 & ab \end{vmatrix} \left[\text{ Applying }R_1 \to R_1 - R_2\text{ and }R_2 \to R_2 - R_3 \right]\] 
\[ = \begin{vmatrix} \left( b - a \right)\left( b + 2c + a \right) & \left( a + b \right) \left( a - b \right)b & c\left( b - a \right)\\ \left( c - b \right)\left( b + 2a + c \right) & \left( b - c \right) \left( b + c \right) & a\left( c - b \right)\\ \left( a + b \right)^2 & c^2 & ab \end{vmatrix}\] 
\[ = \left( a - b \right)\left( b - c \right)\begin{vmatrix} - \left( b + 2c + a \right) & a + b & - c \\ - \left( b + 2a + c \right) & b + c & - a\\ \left( a + b \right)^2 & c^2 & ab \end{vmatrix} \left[\text{ Applying }x^2 - y^2 = \left( x + y \right)\left( x - y \right)\text{ and taking out }\left( a - b \right)\text{ common from }R_1\text{ and }\left( b - c \right)\text{ from }R_2 \right]\] 
\[ = \left( a - b \right)\left( b - c \right) \begin{vmatrix} - 2\left( b + c + a \right) & a + b & - c \\ - 2\left( b + a + c \right) & b + c & - a\\ \left( a + b \right)^2 - c^2 & c^2 & ab \end{vmatrix} \left[\text{ Applying }C_1 \to C_1 - C_2 \right]\]
\[ = \left( a - b \right)\left( b - c \right) \begin{vmatrix} - 2\left( b + c + a \right) & a + b & - c \\ - 2\left( b + a + c \right) & b + c & - a\\ \left( a + b + c \right) \left( a + b - c \right) & c^2 & ab \end{vmatrix} \left[\text{ Applying }x^2 - y^2 = \left( x + y \right)\left( x - y \right)\text{ in }C_1 \right]\] 
\[ = \left( a - b \right)\left( b - c \right)\left( a + b + c \right) \begin{vmatrix} - 2 & a + b & - c \\ - 2 & b + c & - a\\ \left( a + b - c \right) & c^2 & ab \end{vmatrix} \left[\text{ Taking out }\left( a + b + c \right)\text{ common from }C_1 \right]\] 
\[ = \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\begin{vmatrix} - 2 & a + b & - c \\ 0 & c - a & c - a\\ \left( a + b - c \right) & c^2 & ab \end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\] 
\[ = \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\left( c - a \right)\begin{vmatrix} - 2 & a + b & - c \\ 0 & 1 & 1\\\left( a + b - c \right) & c^2 & ab \end{vmatrix} \left[\text{ Taking out }\left( c - a \right)\text{ common from }R_2 \right]\] 
\[ = \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\left( c - a \right)\begin{vmatrix} - 2 & a + b + c & - c \\ 0 & 0 & 1\\\left( a + b - c \right) & c^2 - ab & ab \end{vmatrix} \left[\text{ Applying }C_2 \to C_2 - C_3 \right]\] 
\[ = \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\left( c - a \right) \left\{ \left( - 1 \right)\begin{vmatrix} - 2 & a + b + c \\\left( a + b - c \right) & c^2 - ab \end{vmatrix} \right\} \left[\text{ Expanding along }R_2 \right]\] 
\[ = - \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\left( c - a \right)\left\{ - 2 c^2 + 2ab - a^2 - b^2 - 2ab + c^2 \right\}\] 
\[ = - \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\left( c - a \right)\left( - a^2 - b^2 - c^2 \right)\] 
\[ = \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\left( c - a \right)\left( a^2 + b^2 + c^2 \right)\]
\[ = RHS\] 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Determinants - Exercise 6.2 [पृष्ठ ५९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 6 Determinants
Exercise 6.2 | Q 20 | पृष्ठ ५९

संबंधित प्रश्‍न

Solve the system of linear equations using the matrix method.

4x – 3y = 3

3x – 5y = 7


Evaluate the following determinant:

\[\begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}\]


Find the value of x, if

\[\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}\]


Evaluate the following determinant:

\[\begin{vmatrix}1 & - 3 & 2 \\ 4 & - 1 & 2 \\ 3 & 5 & 2\end{vmatrix}\]


Evaluate the following determinant:

\[\begin{vmatrix}1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25\end{vmatrix}\]


Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2\end{vmatrix}\]


Evaluate :

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


\[If ∆ = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix}, ∆_1 = \begin{vmatrix}1 & 1 & 1 \\ yz & zx & xy \\ x & y & z\end{vmatrix},\text{ then prove that }∆ + ∆_1 = 0 .\]


Prove that

\[\begin{vmatrix}\frac{a^2 + b^2}{c} & c & c \\ a & \frac{b^2 + c^2}{a} & a \\ b & b & \frac{c^2 + a^2}{b}\end{vmatrix} = 4abc\]


​Solve the following determinant equation:

\[\begin{vmatrix}1 & 1 & x \\ p + 1 & p + 1 & p + x \\ 3 & x + 1 & x + 2\end{vmatrix} = 0\]

​Solve the following determinant equation:
\[\begin{vmatrix}15 - 2x & 11 - 3x & 7 - x \\ 11 & 17 & 14 \\ 10 & 16 & 13\end{vmatrix} = 0\]

If a, b, c are real numbers such that
\[\begin{vmatrix}b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a\end{vmatrix} = 0\] , then show that either
\[a + b + c = 0 \text{ or, } a = b = c\]


If \[a, b\] and c  are all non-zero and 

\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c\end{vmatrix} =\] 0, then prove that 
\[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} +\]1
= 0

 


Find the area of the triangle with vertice at the point:

(3, 8), (−4, 2) and (5, −1)


Using determinants show that the following points are collinear:

(1, −1), (2, 1) and (4, 5)


Using determinants, find the value of k so that the points (k, 2 − 2 k), (−k + 1, 2k) and (−4 − k, 6 − 2k) may be collinear.


If the points (3, −2), (x, 2), (8, 8) are collinear, find x using determinant.


Prove that :

\[\begin{vmatrix}b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c\end{vmatrix} = 3abc - a^3 - b - c^3\]

 


Prove that :

\[\begin{vmatrix}\left( a + 1 \right) \left( a + 2 \right) & a + 2 & 1 \\ \left( a + 2 \right) \left( a + 3 \right) & a + 3 & 1 \\ \left( a + 3 \right) \left( a + 4 \right) & a + 4 & 1\end{vmatrix} = - 2\]

 


Prove that :

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

 


5x − 7y + z = 11
6x − 8y − z = 15
3x + 2y − 6z = 7


2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11


Write the value of the determinant \[\begin{vmatrix}2 & - 3 & 5 \\ 4 & - 6 & 10 \\ 6 & - 9 & 15\end{vmatrix} .\]


If the matrix \[\begin{bmatrix}5x & 2 \\ - 10 & 1\end{bmatrix}\]  is singular, find the value of x.


Find the value of x from the following : \[\begin{vmatrix}x & 4 \\ 2 & 2x\end{vmatrix} = 0\]


If \[A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\] , then for any natural number, find the value of Det(An).


If\[f\left( x \right) = \begin{vmatrix}0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0\end{vmatrix}\]





Let \[f\left( x \right) = \begin{vmatrix}\cos x & x & 1 \\ 2\sin x & x & 2x \\ \sin x & x & x\end{vmatrix}\] \[\lim_{x \to 0} \frac{f\left( x \right)}{x^2}\]  is equal to


Solve the following system of equations by matrix method:
 x − y + z = 2
2x − y = 0
2y − z = 1


Solve the following system of equations by matrix method:
 x + y + z = 6
x + 2z = 7
3x + y + z = 12


2x − y + z = 0
3x + 2y − z = 0
x + 4y + 3z = 0


For the system of equations:
x + 2y + 3z = 1
2x + y + 3z = 2
5x + 5y + 9z = 4


On her birthday Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got ₹ 10 more. However, if there were 16 children more, everyone would have got ₹ 10 less. Using the matrix method, find the number of children and the amount distributed by Seema. What values are reflected by Seema’s decision?


Transform `[(1, 2, 4),(3, -1, 5),(2, 4, 6)]` into an upper triangular matrix by using suitable row transformations


Solve the following by inversion method 2x + y = 5, 3x + 5y = −3


`abs (("a"^2, 2"ab", "b"^2),("b"^2, "a"^2, 2"ab"),(2"ab", "b"^2, "a"^2))` is equal to ____________.


A set of linear equations is represented by the matrix equation Ax = b. The necessary condition for the existence of a solution for this system is


If c < 1 and the system of equations x + y – 1 = 0, 2x – y – c = 0 and – bx+ 3by – c = 0 is consistent, then the possible real values of b are


If a, b, c are non-zeros, then the system of equations (α + a)x + αy + αz = 0, αx + (α + b)y + αz = 0, αx+ αy + (α + c)z = 0 has a non-trivial solution if


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×