हिंदी

Evaluate : ∣ ∣ ∣ ∣ a B C C a B B C a ∣ ∣ ∣ ∣ - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate :

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

उत्तर

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

\[ = a( a^2 - bc) - b(ca - b^2 ) + c( c^2 - ba)\]

\[ = a^3 - abc - bca + b^3 + c^3 - abc\]

\[ = a^3 + b^3 + c^3 - 3abc = (a + b + c)( a^2 + b^2 + c^2 - ab - bc - ca)\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Determinants - Exercise 6.2 [पृष्ठ ५८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 6 Determinants
Exercise 6.2 | Q 6 | पृष्ठ ५८

संबंधित प्रश्न

Find the value of a if `[[a-b,2a+c],[2a-b,3c+d]]=[[-1,5],[0,13]]`


Examine the consistency of the system of equations.

5x − y + 4z = 5

2x + 3y + 5z = 2

5x − 2y + 6z = −1


Solve system of linear equations, using matrix method.

5x + 2y = 4

7x + 3y = 5


Evaluate

\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}^2 .\]


Show that

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


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


Without expanding, prove that

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


​Solve the following determinant equation:

\[\begin{vmatrix}x + 1 & 3 & 5 \\ 2 & x + 2 & 5 \\ 2 & 3 & x + 4\end{vmatrix} = 0\]

 


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:

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


Prove that :

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


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 + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11


2y − 3z = 0
x + 3y = − 4
3x + 4y = 3


3x + y = 5
− 6x − 2y = 9


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.


State whether the matrix 
\[\begin{bmatrix}2 & 3 \\ 6 & 4\end{bmatrix}\] is singular or non-singular.


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


If \[\begin{vmatrix}x + 1 & x - 1 \\ x - 3 & x + 2\end{vmatrix} = \begin{vmatrix}4 & - 1 \\ 1 & 3\end{vmatrix}\], then write the value of x.

If \[\begin{vmatrix}3x & 7 \\ - 2 & 4\end{vmatrix} = \begin{vmatrix}8 & 7 \\ 6 & 4\end{vmatrix}\] , find the value of x.


If  \[∆_1 = \begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{vmatrix}, ∆_2 = \begin{vmatrix}1 & bc & a \\ 1 & ca & b \\ 1 & ab & c\end{vmatrix},\text{ then }\]}




If \[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}\] , then x = 

 


The value of the determinant  

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




Solve the following system of equations by matrix method:
6x − 12y + 25z = 4
4x + 15y − 20z = 3
2x + 18y + 15z = 10


Solve the following system of equations by matrix method:
\[\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10\]
\[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10\]
\[\frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13\]


Show that the following systems of linear equations is consistent and also find their solutions:
5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5


Show that each one of the following systems of linear equation is inconsistent:
4x − 5y − 2z = 2
5x − 4y + 2z = −2
2x + 2y + 8z = −1


\[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\], find AB. Hence, solve the system of equations: x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

The prices of three commodities P, Q and R are Rs x, y and z per unit respectively. A purchases 4 units of R and sells 3 units of P and 5 units of Q. B purchases 3 units of Q and sells 2 units of P and 1 unit of R. Cpurchases 1 unit of P and sells 4 units of Q and 6 units of R. In the process A, B and C earn Rs 6000, Rs 5000 and Rs 13000 respectively. If selling the units is positive earning and buying the units is negative earnings, find the price per unit of three commodities by using matrix method.

 

Two factories decided to award their employees for three values of (a) adaptable tonew techniques, (b) careful and alert in difficult situations and (c) keeping clam in tense situations, at the rate of ₹ x, ₹ y and ₹ z per person respectively. The first factory decided to honour respectively 2, 4 and 3 employees with a total prize money of ₹ 29000. The second factory decided to honour respectively 5, 2 and 3 employees with the prize money of ₹ 30500. If the three prizes per person together cost ₹ 9500, then
i) represent the above situation by matrix equation and form linear equation using matrix multiplication.
ii) Solve these equation by matrix method.
iii) Which values are reflected in the questions?


A shopkeeper has 3 varieties of pens 'A', 'B' and 'C'. Meenu purchased 1 pen of each variety for a total of Rs 21. Jeevan purchased 4 pens of 'A' variety 3 pens of 'B' variety and 2 pens of 'C' variety for Rs 60. While Shikha purchased 6 pens of 'A' variety, 2 pens of 'B' variety and 3 pens of 'C' variety for Rs 70. Using matrix method, find cost of each variety of pen.

 

x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0


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


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


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


Solve the following equations by using inversion method.

x + y + z = −1, x − y + z = 2 and x + y − z = 3


Prove that (A–1)′ = (A′)–1, where A is an invertible matrix.


If the system of equations x + λy + 2 = 0, λx + y – 2 = 0, λx + λy + 3 = 0 is consistent, then


If the system of linear equations x + 2ay + az = 0; x + 3by + bz = 0; x + 4cy + cz = 0 has a non-zero solution, then a, b, c ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×