English

X + Y − Z = 0 X − 2y + Z = 0 3x + 6y − 5z = 0 - Mathematics

Advertisements
Advertisements

Question

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

Solution

Using the equations we get 
\[D = \begin{vmatrix}1 & 1 & - 1 \\ 1 & - 2 & 1 \\ 3 & 6 & - 5\end{vmatrix}\] 
\[ \Rightarrow 1\left( 10 - 6 \right) - 1\left( - 5 - 3 \right) - 1\left( 6 + 6 \right) = 0\] 
\[ D_1 = \begin{vmatrix}0 & 1 & - 1 \\ 0 & - 2 & 1 \\ 0 & 6 & - 5\end{vmatrix}\] 

\[ \Rightarrow 0\left( 10 - 6 \right) - 1\left( 0 - 0 \right) - 1\left( 0 + 0 \right) = 0\] 
\[ D_2 = \begin{vmatrix}1 & 0 & - 1 \\ 1 & 0 & 1 \\ 3 & 0 & - 5\end{vmatrix}\] 
\[ \Rightarrow 1\left( 0 - 0 \right) - 0\left( - 5 - 3 \right) - 1\left( 0 - 0 \right) = 0\] 
\[ D_3 = \begin{vmatrix}1 & 1 & 0 \\ 1 & - 2 & 0 \\ 3 & 6 & 0\end{vmatrix}\] 
\[ \Rightarrow 1\left( 0 - 0 \right) - 1\left( 0 - 0 \right) + 0\left( 6 + 6 \right) = 0\] 
\[\therefore D = D_1 = D_2\]
Hence, the system of linear equations has infinitely many solutions.

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Determinants - Exercise 6.4 [Page 85]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.4 | Q 28 | Page 85

RELATED QUESTIONS

Examine the consistency of the system of equations.

3x − y − 2z = 2

2y − z = −1

3x − 5y = 3


Examine the consistency of the system of equations.

5x − y + 4z = 5

2x + 3y + 5z = 2

5x − 2y + 6z = −1


Show that

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


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


For what value of x the matrix A is singular? 

\[A = \begin{bmatrix}x - 1 & 1 & 1 \\ 1 & x - 1 & 1 \\ 1 & 1 & x - 1\end{bmatrix}\]


Evaluate the following determinant:

\[\begin{vmatrix}1 & - 3 & 2 \\ 4 & - 1 & 2 \\ 3 & 5 & 2\end{vmatrix}\]


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

\[\begin{vmatrix}1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2\end{vmatrix}\]


​Solve the following determinant equation:

\[\begin{vmatrix}1 & x & x^2 \\ 1 & a & a^2 \\ 1 & b & b^2\end{vmatrix} = 0, a \neq b\]

 


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

 (0, 0), (6, 0) and (4, 3)


Using determinants prove that the points (ab), (a', b') and (a − a', b − b') are collinear if ab' = a'b.

 

Using determinants, find the area of the triangle with vertices (−3, 5), (3, −6), (7, 2).


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

(3, 1) and (9, 3)


Prove that :

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

 


Prove that :

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

 


2x − y = 17
3x + 5y = 6


x + y + z + 1 = 0
ax + by + cz + d = 0
a2x + b2y + x2z + d2 = 0


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


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.


Solve each of the following system of homogeneous linear equations.
x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0


If A is a singular matrix, then write the value of |A|.

 

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

 


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 \[\begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8\] , write the value of x.


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


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 


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:
 x + y + z = 6
x + 2z = 7
3x + y + z = 12


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.

A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.


Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prices at the rate of Rs. xy and z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total price money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with a total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of xy and z. What values are described in this equations?


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


Let \[X = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}, A = \begin{bmatrix}1 & - 1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}3 \\ 1 \\ 4\end{bmatrix}\] . If AX = B, then X is equal to

 


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


x + y = 1
x + z = − 6
x − y − 2z = 3


Using determinants, find the equation of the line joining the points (1, 2) and (3, 6).


If `alpha, beta, gamma` are in A.P., then `abs (("x" - 3, "x" - 4, "x" - alpha),("x" - 2, "x" - 3, "x" - beta),("x" - 1, "x" - 2, "x" - gamma)) =` ____________.


If the system of equations x + ky - z = 0, 3x - ky - z = 0 & x - 3y + z = 0 has non-zero solution, then k is equal to ____________.


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`


Choose the correct option:

If a, b, c are in A.P. then the determinant `[(x + 2, x + 3, x + 2a),(x + 3, x + 4, x + 2b),(x + 4, x + 5, x + 2c)]` is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×