English

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

Advertisements
Advertisements

Question

Prove that :

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

Solution

\[\text{ Let LHS }= \Delta = \begin{vmatrix} a^2 & a^2 - \left( b - c \right)^2 & bc\\ b^2 & b^2 - \left( c - a \right)^2 & ca\\ c^2 & c^2 - \left( a - b \right)^2 & ab \end{vmatrix}\]
\[ \Rightarrow ∆ = \begin{vmatrix} a^2 & - \left( b - c \right)^2 & bc\\ b^2 & - \left( c - a \right)^2 & ca\\ c^2 & - \left( a - b \right)^2 & ab \end{vmatrix} \left[\text{ Applying }C_2 \to C_2 - C_1 \right]\] 
\[ = \left( - 1 \right)\begin{vmatrix} a^2 & \left( b - c \right)^2 & bc\\ b^2 & \left( c - a \right)^2 & ca\\ c^2 & \left( a - b \right)^2 & ab \end{vmatrix}\] 
\[ = - \begin{vmatrix} a^2 & b^2 + c^2 & bc\\ b^2 & c^2 + a^2 & ca\\ c^2 & a^2 + b^2 & ab \end{vmatrix} \left[\text{ Applying }C_2 \to C_2 - 2 C_3 \right]\] 
\[ = - \begin{vmatrix} a^2 + b^2 + c^2 & b^2 + c^2 & bc\\ b^2 + c^2 + a^2 & c^2 + a^2 & ca\\ c^2 + a^2 + b^2 & a^2 + b^2 & ab \end{vmatrix} \left[\text{ Applying }C_1 \to C_1 + C_2 \right]\] 
\[ = - \left( a^2 + b^2 + c^2 \right)\begin{vmatrix} 1 & b^2 + c^2 & bc\\1 & c^2 + a^2 & ca\\1 & a^2 + b^2 & ab \end{vmatrix}\]
\[ = - \left( a^2 + b^2 + c^2 \right)\begin{vmatrix} 1 & b^2 + c^2 & bc\\0 & \left( c^2 + a^2 \right) - \left( b^2 + c^2 \right) & ca - bc\\0 & \left( a^2 + b^2 \right) - \left( b^2 + c^2 \right) & ab - bc \end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \text{ and }R_3 \to R_3 - R_1 \right]\] 
\[ = \left( \left( a^2 + b^2 + c^2 \right) \right)\begin{vmatrix} 1 & b^2 + c^2 & bc\\0 & a^2 - b^2 & c\left( a - b \right)\\0 & a^2 - c^2 & b \left( a - c \right) \end{vmatrix}\]
\[ = - \left( a^2 + b^2 + c^2 \right)\left( a - b^{} \right)\left( a - c \right)\begin{vmatrix} 1 & b^2 + c^2 & bc\\0 & a + b^{} & c\\0 & a^{} + c^{} & b \end{vmatrix} \left[\text{ Taking }\left( a - b \right)\text{ common from }R_2\text{ and }\left( a - c \right)\text{ common from }R_3 \right]\] 
\[ = \left( a^2 + b^2 + c^2 \right)\left( a - b^{} \right)\left( c - a \right) \times \left\{ 1 \times \begin{vmatrix} a + b^{} & c\\ a^{} + c^{} & b \end{vmatrix} \right\} \left[ \because \left( c - a \right) = - \left( a - c \right) \right] \left[\text{ Expanding along }C_1 \right]\] 
\[ = \left( a^2 + b^2 + c^2 \right)\left( a - b^{} \right) \left( c - a \right) \left( ab + b^2 - ac - c^2 \right) \] 
\[= \left( a^2 + b^2 + c^2 \right)\left( a - b \right)\left( c - a \right)\left\{ a\left( b - c \right) + \left( b + c \right)\left( b - c \right) \right\}\] 
\[ = \left( a - b \right)\left( c - a \right)\left( b - c \right)\left( a + b + c \right)\left( a^2 + b^2 + c^2 \right)\]
 = RHS
Hence proved

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Determinants - Exercise 6.2 [Page 59]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.2 | Q 22 | Page 59

RELATED QUESTIONS

Examine the consistency of the system of equations.

x + 2y = 2

2x + 3y = 3


Show that

\[\begin{vmatrix}\sin 10^\circ & - \cos 10^\circ \\ \sin 80^\circ & \cos 80^\circ\end{vmatrix} = 1\]


Evaluate

\[\begin{vmatrix}2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1\end{vmatrix}\] by two methods.

 

Evaluate
\[∆ = \begin{vmatrix}0 & \sin \alpha & - \cos \alpha \\ - \sin \alpha & 0 & \sin \beta \\ \cos \alpha & - \sin \beta & 0\end{vmatrix}\]


Find the value of x, if

\[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & 5 \\ 8 & 3\end{vmatrix}\]


Evaluate the following determinant:

\[\begin{vmatrix}1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38\end{vmatrix}\]


Evaluate :

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


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


Without expanding, prove that

\[\begin{vmatrix}a & b & c \\ x & y & z \\ p & q & r\end{vmatrix} = \begin{vmatrix}x & y & z \\ p & q & r \\ a & b & c\end{vmatrix} = \begin{vmatrix}y & b & q \\ x & a & p \\ z & c & r\end{vmatrix}\]


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

(2, 7), (1, 1) and (10, 8)


Find values of k, if area of triangle is 4 square units whose vertices are 
(k, 0), (4, 0), (0, 2)


Prove that :

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

 


Prove that :

\[\begin{vmatrix}a^2 & bc & ac + c^2 \\ a^2 + ab & b^2 & ac \\ ab & b^2 + bc & c^2\end{vmatrix} = 4 a^2 b^2 c^2\]

Prove that :

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

 


Prove that

\[\begin{vmatrix}a^2 + 1 & ab & ac \\ ab & b^2 + 1 & bc \\ ca & cb & c^2 + 1\end{vmatrix} = 1 + a^2 + b^2 + c^2\]

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

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


If w is an imaginary cube root of unity, find the value of \[\begin{vmatrix}1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w\end{vmatrix}\]


Write the value of 

\[\begin{vmatrix}\sin 20^\circ & - \cos 20^\circ\\ \sin 70^\circ& \cos 70^\circ\end{vmatrix}\]

Write the value of  \[\begin{vmatrix}a + ib & c + id \\ - c + id & a - ib\end{vmatrix} .\]


If \[\begin{vmatrix}2x + 5 & 3 \\ 5x + 2 & 9\end{vmatrix} = 0\]


If x ∈ N and \[\begin{vmatrix}x + 3 & - 2 \\ - 3x & 2x\end{vmatrix}\]  = 8, then find the value of x.


Let \[\begin{vmatrix}x^2 + 3x & x - 1 & x + 3 \\ x + 1 & - 2x & x - 4 \\ x - 3 & x + 4 & 3x\end{vmatrix} = a x^4 + b x^3 + c x^2 + dx + e\] 
be an identity in x, where abcde are independent of x. Then the value of e is


The number of distinct real roots of \[\begin{vmatrix}cosec x & \sec x & \sec x \\ \sec x & cosec x & \sec x \\ \sec x & \sec x & cosec x\end{vmatrix} = 0\]  lies in the interval
\[- \frac{\pi}{4} \leq x \leq \frac{\pi}{4}\]


If \[\begin{vmatrix}a & p & x \\ b & q & y \\ c & r & z\end{vmatrix} = 16\] , then the value of \[\begin{vmatrix}p + x & a + x & a + p \\ q + y & b + y & b + q \\ r + z & c + z & c + r\end{vmatrix}\] is


Solve the following system of equations by matrix method:
3x + y = 7
5x + 3y = 12


Solve the following system of equations by matrix method:
 5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25


Show that the following systems of linear equations is consistent and also find their solutions:
2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3


If \[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\] , find A−1. Using A−1, solve the system of linear equations  x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7.

Let \[X = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}, A = \begin{bmatrix}1 & - 1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}3 \\ 1 \\ 4\end{bmatrix}\] . If AX = B, then X is equal to

 


The existence of the unique solution of the system of equations:
x + y + z = λ
5x − y + µz = 10
2x + 3y − z = 6
depends on


x + y = 1
x + z = − 6
x − y − 2z = 3


If A = `[(1, -1, 2),(3, 0, -2),(1, 0, 3)]`, verify that A(adj A) = (adj A)A


The cost of 4 dozen pencils, 3 dozen pens and 2 dozen erasers is ₹ 60. The cost of 2 dozen pencils, 4 dozen pens and 6 dozen erasers is ₹ 90. Whereas the cost of 6 dozen pencils, 2 dozen pens and 3 dozen erasers is ₹ 70. Find the cost of each item per dozen by using matrices


Show that if the determinant ∆ = `|(3, -2, sin3theta),(-7, 8, cos2theta),(-11, 14, 2)|` = 0, then sinθ = 0 or `1/2`.


Using determinants, find the equation of the line joining the points (1, 2) and (3, 6).


Solve the following system of equations x − y + z = 4, x − 2y + 2z = 9 and 2x + y + 3z = 1.


In system of equations, if inverse of matrix of coefficients A is multiplied by right side constant B vector then resultant will be?


Let the system of linear equations x + y + az = 2; 3x + y + z = 4; x + 2z = 1 have a unique solution (x*, y*, z*). If (α, x*), (y*, α) and (x*, –y*) are collinear points, then the sum of absolute values of all possible values of α is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×