मराठी

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

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

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

\[ = \frac{1}{abc}\begin{vmatrix}- abc & a b^2 + abc & a c^2 + abc \\ a^2 b + abc & - abc & c^2 b + abc \\ a^2 c + abc & b^2 c + abc & - abc\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }aR_1 , R_2 \text{ to }bR_2\text{ and }R_3 \text{ to }cR_3\text{ and then dividing by abc }\right]\]

\[ = \frac{abc}{abc}\begin{vmatrix}- bc & ab + ac & ac + ab \\ ab + bc & - ac & cb + ab \\ ac + bc & bc + ac & - ab\end{vmatrix} \left[\text{ Taking out a, b and c common from the three columns }\right]\]

\[\begin{vmatrix}ab + bc + ca & ab + bc + ca & ab + bc + ca \\ ab + bc & - ac & cb + ab \\ ac + bc & bc + ac & - ab\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }R_1 + R_2 + R_3 \right]\]

\[ = (ab + bc + ca)\begin{vmatrix}1 & 1 & 1 \\ ab + bc & - ac & cb + ab \\ ac + bc & bc + ac & - ab\end{vmatrix}\]

\[ = (ab + bc + ca)\begin{vmatrix}0 & 0 & 1 \\ 0 & - (ab + bc + ac) & cb + ab \\ ac + bc + ab & bc + ac + ab & - ab\end{vmatrix} \left[\text{ Applying }C_1 \text{ to }C_1 - C_3 \text{ and }C_2 \text{ to }C_2 - C_3 \right]\]

\[ = (ab + bc + ca)\begin{vmatrix}0 & - (ab + bc + ac) \\ ac + bc + ab & bc + ac + ab\end{vmatrix}\]

\[ = (ab + bc + ca)(ab + bc + ac )^2 \]

\[ = (ab + bc + ca )^3\]

Hence proved.

 
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Determinants - Exercise 6.2 [पृष्ठ ५८]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 5 Determinants
Exercise 6.2 | Q 36 | पृष्ठ ५८

संबंधित प्रश्‍न

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.

5x + 2y = 4

7x + 3y = 5


Solve the system of linear equations using the matrix method.

2x + y + z = 1

x – 2y – z = `3/2`

3y – 5z = 9


Solve the system of linear equations using the matrix method.

x − y + 2z = 7

3x + 4y − 5z = −5

2x − y + 3z = 12


Evaluate the following determinant:

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


For what value of x the matrix A is singular? 
\[ A = \begin{bmatrix}1 + x & 7 \\ 3 - x & 8\end{bmatrix}\]


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

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


Evaluate :

\[\begin{vmatrix}x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda\end{vmatrix}\]


\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & a \\ 1 & 1 & 1 + a\end{vmatrix} = a^3 + 3 a^2\]


​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
\[a + b + c = 0 \text{ or, } a = b = c\]


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

(3, 8), (−4, 2) and (5, −1)


If the points (x, −2), (5, 2), (8, 8) are collinear, find x using determinants.


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


Prove that :

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


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

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


x + 2y = 5
3x + 6y = 15


If \[A = \begin{bmatrix}0 & i \\ i & 1\end{bmatrix}\text{  and }B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\] , find the value of |A| + |B|.


If \[A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}\text{ and B} = \begin{bmatrix}1 & - 4 \\ 3 & - 2\end{bmatrix},\text{ find }|AB|\]


Find the value of the determinant \[\begin{vmatrix}243 & 156 & 300 \\ 81 & 52 & 100 \\ - 3 & 0 & 4\end{vmatrix} .\]


For what value of x is the matrix  \[\begin{bmatrix}6 - x & 4 \\ 3 - x & 1\end{bmatrix}\]  singular?


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


Using the factor theorem it is found that a + bb + c and c + a are three factors of the determinant 

\[\begin{vmatrix}- 2a & a + b & a + c \\ b + a & - 2b & b + c \\ c + a & c + b & - 2c\end{vmatrix}\]
The other factor in the value of the determinant is


If ω is a non-real cube root of unity and n is not a multiple of 3, then  \[∆ = \begin{vmatrix}1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^n \\ \omega^n & \omega^{2n} & 1\end{vmatrix}\] 


Let \[A = \begin{bmatrix}1 & \sin \theta & 1 \\ - \sin \theta & 1 & \sin \theta \\ - 1 & - \sin \theta & 1\end{bmatrix},\text{ where 0 }\leq \theta \leq 2\pi . \text{ Then,}\]




If \[x, y \in \mathbb{R}\], then the determinant 

\[∆ = \begin{vmatrix}\cos x & - \sin x  & 1 \\ \sin x & \cos x & 1 \\ \cos\left( x + y \right) & - \sin\left( x + y \right) & 0\end{vmatrix}\]



The value of \[\begin{vmatrix}1 & 1 & 1 \\ {}^n C_1 & {}^{n + 2} C_1 & {}^{n + 4} C_1 \\ {}^n C_2 & {}^{n + 2} C_2 & {}^{n + 4} C_2\end{vmatrix}\] is


Show that each one of the following systems of linear equation is inconsistent:
2x + 3y = 5
6x + 9y = 10


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


If \[A = \begin{bmatrix}3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}\] , find A−1 and hence solve the following system of equations: 

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

Two schools P and Q want to award their selected students on the values of Discipline, Politeness and Punctuality. The school P wants to award ₹x each, ₹y each and ₹z each the three respectively values to its 3, 2 and 1 students with a total award money of ₹1,000. School Q wants to spend ₹1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for three values as before). If the total amount of awards for one prize on each value is ₹600, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.


Let a, b, c be positive real numbers. The following system of equations in x, y and z 

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1, \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1, - \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \text { has }\]
(a) no solution
(b) unique solution
(c) infinitely many solutions
(d) finitely many solutions

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

 

x + y = 1
x + z = − 6
x − y − 2z = 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`


What is the nature of the given system of equations

`{:(x + 2y = 2),(2x + 3y = 3):}`


The number of real values λ, such that the system of linear equations 2x – 3y + 5z = 9, x + 3y – z = –18 and 3x – y + (λ2 – |λ|z) = 16 has no solution, is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×