हिंदी

Prove that : ∣ ∣ ∣ ∣ ∣ a 2 B C a C + C 2 a 2 + a B B 2 a C a B B 2 + B C C 2 ∣ ∣ ∣ ∣ ∣ = 4 a 2 B 2 C 2

Advertisements
Advertisements

प्रश्न

Prove that :

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

उत्तर

\[\text{ Let LHS }= \Delta = \begin{vmatrix} a^2 & bc & ac + c^2 \\ a^2 + ab & b^2 & ac\\ab & b^2 + bc & c^2 \end{vmatrix}\] 
\[\Delta = abc \begin{vmatrix} a & c & a + c\\a + b & b & a\\b & b + c & c \end{vmatrix} \left[\text{ Taking out a, b and c common from }C_1 , C_2\text{ and }C_3 \right]\] 
\[ = abc \begin{vmatrix} a & c & 0\\a + b & b & - 2b\\b & b + c &  - 2b \end{vmatrix} \left[\text{ Applying }C_3 \to C_3 - C_2 - C_1 \right]\] 
\[ = \left( abc \right)\left( - 2b \right) \begin{vmatrix} a & c & 0\\a + b & b & 1\\b & b + c & 1 \end{vmatrix}\left[\text{ Taking ( - 2b) common from }C_3 \right]\] 
\[ = \left( abc \right)\left( - 2b \right) \begin{vmatrix} a & c & 0\\a & - c & 0\\b & b + c & 1 \end{vmatrix} \left[\text{ Applying }R_2 \to \hspace{0.167em} R_2 - R_3 \right]\] 
\[ = \left( abc \right)\left( - 2b \right) \times 1\begin{vmatrix} a & c \\a & - c \end{vmatrix} \left[\text{ Expanding along }C_3 \right]\] 
\[ = \left( abc \right)\left( - 2b \right)\left( - 2ac \right)\] 
\[ = 4 a^2 b^2 c^2 \] 
\[ = RHS\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Determinants - Exercise 6.2 [पृष्ठ ५९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 5 Determinants
Exercise 6.2 | Q 24 | पृष्ठ ५९

संबंधित प्रश्न

Examine the consistency of the system of equations.

3x − y − 2z = 2

2y − z = −1

3x − 5y = 3


Evaluate

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

 

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


Find the value of x, if
\[\begin{vmatrix}2 & 4 \\ 5 & 1\end{vmatrix} = \begin{vmatrix}2x & 4 \\ 6 & x\end{vmatrix}\]


Find the value of x, if

\[\begin{vmatrix}3x & 7 \\ 2 & 4\end{vmatrix} = 10\] , find the value of x.


Find the integral value of x, if \[\begin{vmatrix}x^2 & x & 1 \\ 0 & 2 & 1 \\ 3 & 1 & 4\end{vmatrix} = 28 .\]


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

\[\begin{vmatrix}a + b & 2a + b & 3a + b \\ 2a + b & 3a + b & 4a + b \\ 4a + b & 5a + b & 6a + b\end{vmatrix}\]


Prove the following identity:

\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix} = a^2 \left( a + x + y + z \right)\]

 


​Solve the following determinant equation:

\[\begin{vmatrix}x + a & b & c \\ a & x + b & c \\ a & b & x + c\end{vmatrix} = 0\]

 


Using determinants show that the following points are collinear:

(2, 3), (−1, −2) and (5, 8)


Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, −6) and (5, 4).


Prove that :

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

 


3x + y = 19
3x − y = 23


6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8


xy = 5
y + z = 3
x + z = 4


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


Write the value of the determinant 

\[\begin{vmatrix}a & 1 & b + c \\ b & 1 & c + a \\ c & 1 & a + b\end{vmatrix} .\]

 


If \[A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 0 \\ - 1 & 0\end{bmatrix}\] , find |AB|.

 

If \[A = \left[ a_{ij} \right]\]   is a 3 × 3 diagonal matrix such that a11 = 1, a22 = 2 a33 = 3, then find |A|.

 

If A and B are non-singular matrices of the same order, write whether AB is singular or non-singular.


If \[\begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8\] , write 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 \[\begin{vmatrix}x & x + y & x + 2y \\ x + 2y & x & x + y \\ x + y & x + 2y & x\end{vmatrix}\] is 



Solve the following system of equations by matrix method:
 2x + 6y = 2
3x − z = −8
2x − y + z = −3


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.

 

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.

 

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


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


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


The system of equations:
x + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ
has a unique solution, if
(a) λ = 5, µ = 13
(b) λ ≠ 5
(c) λ = 5, µ ≠ 13
(d) µ ≠ 13


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


Solve the following equations by using inversion method.

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


Solve the following system of equations by using inversion method

x + y = 1, y + z = `5/3`, z + x = `4/3`


The value of λ, such that the following system of equations has no solution, is

`2x - y - 2z = - 5`

`x - 2y + z = 2`

`x + y + lambdaz = 3`


The system of linear equations

3x – 2y – kz = 10

2x – 4y – 2z = 6

x + 2y – z = 5m

is inconsistent if ______.


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×