English

The Value of the Determinant ∣ ∣ ∣ ∣ X X + Y X + 2 Y X + 2 Y X X + Y X + Y X + 2 Y X ∣ ∣ ∣ ∣ is (A) 9x2(X + Y) (B) 9y2(X + Y) (C) 3y2(X + Y) (D) 7x2(X + Y)

Advertisements
Advertisements

Question

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 


Options

  • 9x2(x + y)

  • 9y2(x + y)

  • 3y2(x + y)

  • 7x2(x + y)

MCQ
Advertisements

Solution

\[\begin{vmatrix}x & x + y & x + 2y \\ x + 2y & x & x + y \\ x + y & x + 2y & x\end{vmatrix}\]
\[ = \begin{vmatrix}- 2y & y & y \\ x + 2y & x & x + y \\ - y & 2y & - y\end{vmatrix} \left[\text{ Applying }R_1 \to R_1 - R_2\text{ and }R_3 \to R_3 - R_2 \right]\]
\[ = y^2 \begin{vmatrix}- 2 & 1 & 1 \\ x + 2y & x & x + y \\ - 1 & 2 & - 1\end{vmatrix} \left[\text{ Taking }\left( y \right)\text{ common from }R_1\text{ and from }R_3 \right]\]
\[ = y^2 \begin{vmatrix}- 2 & - 3 & 3 \\ x + 2y & 3x + 4y & - y \\ - 1 & 0 & 0\end{vmatrix} \left[\text{ Applying }C_2 \to C_2 + 2 C_1\text{ and }C_3 \to C_3 - C_1 \right]\]
\[ = y^2 \left[ - 1\left( 3y - 9x - 12y \right) \right]\]
\[ = y^2 \left[ 9y + 9x \right]\]
\[ = 9 y^2 \left( y + x \right)\]

 

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Determinants - Exercise 6.7 [Page 96]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 5 Determinants
Exercise 6.7 | Q 29 | Page 96

RELATED QUESTIONS

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.

x + 2y = 2

2x + 3y = 3


Solve the system of linear equations using the matrix method.

5x + 2y = 3

3x + 2y = 5


Evaluate the following determinant:

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


Evaluate the following determinant:

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


Evaluate :

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


Evaluate the following:

\[\begin{vmatrix}0 & x y^2 & x z^2 \\ x^2 y & 0 & y z^2 \\ x^2 z & z y^2 & 0\end{vmatrix}\]


Evaluate the following:

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


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


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


\[If \begin{vmatrix}p & b & c \\ a & q & c \\ a & b & r\end{vmatrix} = 0,\text{ find the value of }\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}, p \neq a, q \neq b, r \neq c .\]

 


​Solve the following determinant equation:

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

 


​Solve the following determinant equation:

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

Using determinants show that the following points are collinear:

(3, −2), (8, 8) and (5, 2)


Using determinants, find the equation of the line joining the points

(3, 1) and (9, 3)


Prove that :

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

 


Prove that :

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

 


Prove that :

\[\begin{vmatrix}\left( a + 1 \right) \left( a + 2 \right) & a + 2 & 1 \\ \left( a + 2 \right) \left( a + 3 \right) & a + 3 & 1 \\ \left( a + 3 \right) \left( a + 4 \right) & a + 4 & 1\end{vmatrix} = - 2\]

 


Prove that :

\[\begin{vmatrix}1 & 1 + p & 1 + p + q \\ 2 & 3 + 2p & 4 + 3p + 2q \\ 3 & 6 + 3p & 10 + 6p + 3q\end{vmatrix} = 1\]

 


Prove that

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

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


Find the value of the determinant
\[\begin{bmatrix}4200 & 4201 \\ 4205 & 4203\end{bmatrix}\]


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

 

Find the value of x from the following : \[\begin{vmatrix}x & 4 \\ 2 & 2x\end{vmatrix} = 0\]


If \[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}\] , write the value of x.


If x ∈ N and \[\begin{vmatrix}x + 3 & - 2 \\ - 3x & 2x\end{vmatrix}\]  = 8, then find the value of x.


The value of the determinant

\[\begin{vmatrix}a^2 & a & 1 \\ \cos nx & \cos \left( n + 1 \right) x & \cos \left( n + 2 \right) x \\ \sin nx & \sin \left( n + 1 \right) x & \sin \left( n + 2 \right) x\end{vmatrix}\text{ is independent of}\]

 


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


\[\begin{vmatrix}\log_3 512 & \log_4 3 \\ \log_3 8 & \log_4 9\end{vmatrix} \times \begin{vmatrix}\log_2 3 & \log_8 3 \\ \log_3 4 & \log_3 4\end{vmatrix}\]


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 


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 + 7y = 4
x + 2y = −1


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


Show that the following systems of linear equations is consistent and also find their solutions:
2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3


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

If \[A = \begin{bmatrix}2 & 3 & 1 \\ 1 & 2 & 2 \\ 3 & 1 & - 1\end{bmatrix}\] , find A–1 and hence solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2y – z = 8.


If \[A = \begin{bmatrix}2 & 4 \\ 4 & 3\end{bmatrix}, X = \binom{n}{1}, B = \binom{ 8}{11}\]  and AX = B, then find n.

The number of solutions of the system of equations:
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5


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

Let A = `[(1,sin α,1),(-sin α,1,sin α),(-1,-sin α,1)]`, where 0 ≤ α ≤ 2π, then:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×