English

∣ ∣ ∣ ∣ B + C a A B C + a B C C a + B ∣ ∣ ∣ ∣ = 4 a B C - Mathematics

Advertisements
Advertisements

Question

\[\begin{vmatrix}b + c & a & a \\ b & c + a & b \\ c & c & a + b\end{vmatrix} = 4abc\]

Advertisements

Solution

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

\[ = \begin{vmatrix}0 & - 2c & - 2b \\ b & c + a & b \\ c & c & a + b\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }R_1 - ( R_2 + R_3 ) \right]\]

\[ = \begin{vmatrix}0 & - 2c & - 2b \\ b & c + a - b & 0 \\ c & 0 & a + b - c\end{vmatrix} \left[\text{ Applying }C_2 \text{ to }C_2 - C_1\text{ and }C_3 \text{ to }C_3 - C_1 \right]\]

\[ = 0\begin{vmatrix}c + a - b & 0 \\ 0 & a + b - c\end{vmatrix} - ( - 2c)\begin{vmatrix}b & 0 \\ c & a + b - c\end{vmatrix} - 2b\begin{vmatrix}b & c + a - b \\ c & 0\end{vmatrix} \left[\text{ Expanding along }R_1 \right]\]

\[ = 2c[b(a + b - c) - 0] - 2b[0 - c(c + a - b)]\]

\[ = 2bc[a + b - c] - 2bc[b - c - a]\]

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

\[ = 4abc\]

Hence proved.

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Determinants - Exercise 6.2 [Page 60]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.2 | Q 32 | Page 60

RELATED QUESTIONS

If `|[x+1,x-1],[x-3,x+2]|=|[4,-1],[1,3]|`, then write the value of x.


Solve the system of linear equations using the matrix method.

x − y + z = 4

2x + y − 3z = 0

x + y + z = 2


The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method.


If A \[\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}\] , then show that |3 A| = 27 |A|.

 

Find the value of x, if

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


Evaluate :

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


Evaluate the following:

\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix}\]


Show that x = 2 is a root of the equation

\[\begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 3 & 2x & x + 2\end{vmatrix} = 0\]  and solve it completely.
 

 


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

 (0, 0), (6, 0) and (4, 3)


If the points (3, −2), (x, 2), (8, 8) are collinear, find x using determinant.


Prove that :

\[\begin{vmatrix}b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c\end{vmatrix} = 3abc - a^3 - b - c^3\]

 


Prove that :

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

 


Prove that :

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

 


3x + y = 19
3x − y = 23


3x + ay = 4
2x + ay = 2, a ≠ 0


2x + 3y = 10
x + 6y = 4


5x + 7y = − 2
4x + 6y = − 3


2y − 3z = 0
x + 3y = − 4
3x + 4y = 3


2x + y − 2z = 4
x − 2y + z = − 2
5x − 5y + z = − 2


A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission 

Month Sale of units Total commission
drawn (in Rs)
  A B C  
Jan 90 100 20 800
Feb 130 50 40 900
March 60 100 30 850


Find out the rates of commission on items A, B and C by using determinant method.


Solve each of the following system of homogeneous linear equations.
2x + 3y + 4z = 0
x + y + z = 0
2x − y + 3z = 0


For what value of x, the following matrix is singular?

\[\begin{bmatrix}5 - x & x + 1 \\ 2 & 4\end{bmatrix}\]

 


Write the value of the determinant 
\[\begin{bmatrix}2 & 3 & 4 \\ 2x & 3x & 4x \\ 5 & 6 & 8\end{bmatrix} .\]

 


Write the value of the determinant 

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

 


Evaluate \[\begin{vmatrix}4785 & 4787 \\ 4789 & 4791\end{vmatrix}\]


If \[A + B + C = \pi\], then the value of \[\begin{vmatrix}\sin \left( A + B + C \right) & \sin \left( A + C \right) & \cos C \\ - \sin B & 0 & \tan A \\ \cos \left( A + B \right) & \tan \left( B + C \right) & 0\end{vmatrix}\]  is equal to 


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 maximum value of  \[∆ = \begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin\theta & 1 \\ 1 + \cos\theta & 1 & 1\end{vmatrix}\] is (θ is real)

 





Solve the following system of equations by matrix method:

\[\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0\]

 


Show that each one of the following systems of linear equation is inconsistent:
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3


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: 

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.


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


On her birthday Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got ₹ 10 more. However, if there were 16 children more, everyone would have got ₹ 10 less. Using the matrix method, find the number of children and the amount distributed by Seema. What values are reflected by Seema’s decision?


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


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 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


If `|(x + 1, x + 2, x + a),(x + 2, x + 3, x + b),(x + 3, x + 4, x + c)|` = 0, then a, b, care in


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×