मराठी

Prove that : ∣ ∣ ∣ ∣ ∣ 1 a 2 + B C a 3 1 B 2 + C a B 3 1 C 2 + a B C 3 ∣ ∣ ∣ ∣ ∣ = − ( a − B ) ( B − C ) ( C − a ) ( a 2 + B 2 + C 2 )

Advertisements
Advertisements

प्रश्न

Prove that :

\[\begin{vmatrix}1 & a^2 + bc & a^3 \\ 1 & b^2 + ca & b^3 \\ 1 & c^2 + ab & c^3\end{vmatrix} = - \left( a - b \right) \left( b - c \right) \left( c - a \right) \left( a^2 + b^2 + c^2 \right)\]

 

Advertisements

उत्तर

\[\text{ Let LHS }= \Delta = \begin{vmatrix} 1 & a^2 + bc & a^3 \\1 & b^2 + ca & b^3 \\1 & c^2 + ab & c^3 \end{vmatrix}\] 
\[ \Rightarrow \Delta = \begin{vmatrix} 0 & \left( a^2 + bc \right) - \left( b^2 + ca \right) & a^3 - b^3 \\0 & \left( b^2 + ca \right) - \left( c^2 + ab \right) & b^3 - c^3 \\1 & c^2 + ab & c \end{vmatrix} \left[\text{ Applying }R_1 \to R_1 - R_2\text{ and }R_2 \to R_2 - R_3 \right]\]
\[= \begin{vmatrix} 0 & a^2 - b^2 - ca + bc & a^3 - b^3 \\0 & b^2 - c^2 - ab + ca & b^3 - c^3 \\1 & c^2 + ab & c^3 \end{vmatrix}\] 
\[ = \begin{vmatrix} 0 & \left( a - b \right)  \left( a + b - c \right) &\left( a - b \right)\left( a^2 + ab + b^2 \right)\\0 & \left( b - c \right)\left( b + c - a \right) & \left( b - c \right)\left( b^2 + bc + a^2 \right)\\1 & c^2 + ab & c^3 \end{vmatrix}\] 
\[= \left( a - b \right)\left( b - c \right)\begin{vmatrix} 0 & a + b - c & a^2 + ab + b^2 \\0 & \left( b + c - a \right) & \left( b^2 + bc + c^2 \right)\\1 & c^2 + ab & c^3 \end{vmatrix} \left[\text{ Taking out }\left( a - b \right)\text{ common from }R_1\text{ and }\left( b - c \right)\text{ from }R_2 \right]\] 
\[ = \left( a - b \right)\left( b - c \right)\begin{vmatrix} 0 & a + b - c & a^2 + ab + b^2 \\0 & \left( b + c - a \right) - \left( a + b - c \right) & \left( b^2 + bc + c^2 \right) - \left( a^2 + ab + b^2 \right)\\1 & c^2 + ab & c^3 \end{vmatrix} \left[\text{ Applying }R \hspace{0.167em}_2 \to R_2 \hspace{0.167em} - R_1 \right]\]
\[= \left( a - b \right)\left( b - c \right)\begin{vmatrix} 0 & a + b - c & a^2 + ab + b^2 \\0 & 2 \left( c - a \right) & b\left( c - a \right) + \left( c^2 - a^2 \right)\\1 & c^2 + ab & c^3 \end{vmatrix}\] 
\[ = \left( a - b \right)\left( b - c \right)\left( c - a \right) \begin{vmatrix}0 & a + b - c & a^2 + ab + b^2 \\0 & 2 & a + b + c\\1 & c^2 + ab & c^3 \end{vmatrix}\] 
\[ = \left( a - b \right)\left( b - c \right)\left( c - a \right) \times \left\{ 1 \times \begin{vmatrix} a + b - c & a^2 + ab + b^2 \\ 2 & a + b + c \end{vmatrix} \right\} \left[\text{ Expanding along }C_1 \right]\]
\[= \left( a - b \right)\left( b - c \right)\left( c - a \right) \times \left\{ \left( a + b \right)^2 - c^2 - \left( 2 a^2 + 2ab + 2 b^2 \right) \right\}\] 
\[ = \left( a - b \right)\left( b - c \right)\left( c - a \right)\left\{ \left( a + b \right)^2 - c^2 - \left( a + b \right)^2 - \left( a^2 + b^2 \right) \right\}\] 
\[ = - \left( a - b \right)\left( b - c \right)\left( c - a \right)\left( a^2 + b^2 + c^2 \right)\] 
\[ = RHS\]
Hence proved.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Determinants - Exercise 6.2 [पृष्ठ ५९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 5 Determinants
Exercise 6.2 | Q 23 | पृष्ठ ५९

संबंधित प्रश्‍न

If `|[2x,5],[8,x]|=|[6,-2],[7,3]|`, write the value of x.


Examine the consistency of the system of equations.

2x − y = 5

x + y = 4


Examine the consistency of the system of equations.

x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4


Solve the system of linear equations using the matrix method.

2x + 3y + 3z = 5

x − 2y + z = −4

3x − y − 2z = 3


If A = `[(2,-3,5),(3,2,-4),(1,1,-2)]` find A−1. Using A−1 solve the system of equations:

2x – 3y + 5z = 11

3x + 2y – 4z = –5

x + y – 2z = –3


Evaluate

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

 

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


Evaluate the following determinant:

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


Evaluate the following determinant:

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


Evaluate :

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


​Solve the following determinant equation:

\[\begin{vmatrix}1 & x & x^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{vmatrix} = 0, b \neq c\]

 


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


Prove that :

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

 


Prove that :

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

Prove that

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

Prove that

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

3x − y + 2z = 3
2x + y + 3z = 5
x − 2y − z = 1


An automobile company uses three types of steel S1S2 and S3 for producing three types of cars C1C2and C3. Steel requirements (in tons) for each type of cars are given below : 

  Cars
C1
C2 C3
Steel S1 2 3 4
S2 1 1 2
S3 3 2 1

Using Cramer's rule, find the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively.


Find the real values of λ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions
\[2 \lambda x - 2y + 3z = 0\] 
\[ x + \lambda y + 2z = 0\] 
\[ 2x + \lambda z = 0\]

 


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


Find the value of the determinant \[\begin{vmatrix}2^2 & 2^3 & 2^4 \\ 2^3 & 2^4 & 2^5 \\ 2^4 & 2^5 & 2^6\end{vmatrix}\].


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 xyare different from zero and \[\begin{vmatrix}1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z\end{vmatrix} = 0\] , then the value of x−1 + y−1 + z−1 is





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:
5x + 7y + 2 = 0
4x + 6y + 3 = 0


Solve the following system of equations by matrix method:

3x + 4y + 7z = 14

2x − y + 3z = 4

x + 2y − 3z = 0


Solve the following system of equations by matrix method:
 8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5


Use product \[\begin{bmatrix}1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4\end{bmatrix}\begin{bmatrix}- 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2\end{bmatrix}\]  to solve the system of equations x + 3z = 9, −x + 2y − 2z = 4, 2x − 3y + 4z = −3.


Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prices at the rate of Rs. xy and z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total price money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with a total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of xy and z. What values are described in this equations?


If \[\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ - 1 \\ 0\end{bmatrix}\], find x, y and z.

The system of equation x + y + z = 2, 3x − y + 2z = 6 and 3x + y + z = −18 has


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


If `|(2x, 5),(8, x)| = |(6, 5),(8, 3)|`, then find x


If `|(2x, 5),(8, x)| = |(6, -2),(7, 3)|`, then value of x is ______.


The value (s) of m does the system of equations 3x + my = m and 2x – 5y = 20 has a solution satisfying the conditions x > 0, y > 0.


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


The greatest value of c ε R for which the system of linear equations, x – cy – cz = 0, cx – y + cz = 0, cx + cy – z = 0 has a non-trivial solution, is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×