हिंदी

3x + Y + Z = 2 2x − 4y + 3z = − 1 4x + Y − 3z = − 11 - Mathematics

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

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

\[D = \begin{vmatrix}3 & 1 & 1 \\ 2 & - 4 & 3 \\ 4 & 1 & - 3\end{vmatrix}\] 
\[ = 3\left( 12 - 3 \right) - 2\left( - 3 - 1 \right) + 4\left( 3 + 4 \right)\] 
\[ = 27 + 8 + 28\] 
\[ = 63\] 
\[ D_1 = \begin{vmatrix}2 & 1 & 1 \\ - 1 & - 4 & 3 \\ - 11 & 1 & - 3\end{vmatrix}\] 
\[ = 2\left( 12 - 3 \right) + 1\left( - 3 - 1 \right) - 11\left( 3 + 4 \right)\] 
\[ = 18 - 4 - 77\] 
\[ = - 63\] 
\[ D_2 = \begin{vmatrix}3 & 2 & 1 \\ 2 & - 1 & 3 \\ 4 & - 11 & - 3\end{vmatrix}\] 
\[ = 3\left( 3 + 33 \right) - 2\left( - 6 + 11 \right) + 4\left( 6 + 1 \right)\] 
\[ = 108 - 10 + 28\] 
\[ = 126\] 
\[ D_3 = \begin{vmatrix}3 & 1 & 2 \\ 2 & - 4 & - 1 \\ 4 & 1 & - 11\end{vmatrix}\] 
\[ = 3\left( 44 + 1 \right) - 2\left( - 11 - 2 \right) + 4\left( - 1 + 8 \right)\] 
\[ = 135 + 26 + 28\] 
\[ = 189\] 
\[Now, \] 
\[x = \frac{D_1}{D} = \frac{- 63}{63} = - 1\] 
\[y = \frac{D_2}{D} = \frac{126}{63} = 2\] 
\[z = \frac{D_3}{D} = \frac{189}{63} = 3\] 
\[ \therefore x = - 1, y = 2\text{ and }z = 3\] 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Determinants - Exercise 6.4 [पृष्ठ ८४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 6 Determinants
Exercise 6.4 | Q 11 | पृष्ठ ८४

संबंधित प्रश्न

Examine the consistency of the system of equations.

x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4


Evaluate :

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


Prove the following identity:

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

 


Using determinants show that the following points are collinear:

(2, 3), (−1, −2) and (5, 8)


Find the value of \[\lambda\]  so that the points (1, −5), (−4, 5) and \[\lambda\]  are collinear.


Using determinants, find the value of k so that the points (k, 2 − 2 k), (−k + 1, 2k) and (−4 − k, 6 − 2k) may be collinear.


Find values of k, if area of triangle is 4 square units whose vertices are 

(−2, 0), (0, 4), (0, k)


Prove that

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

2x − y = − 2
3x + 4y = 3


x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1


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


3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20.


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


Write the value of 

\[\begin{vmatrix}\sin 20^\circ & - \cos 20^\circ\\ \sin 70^\circ& \cos 70^\circ\end{vmatrix}\]

Find the value of the determinant \[\begin{vmatrix}2^2 & 2^3 & 2^4 \\ 2^3 & 2^4 & 2^5 \\ 2^4 & 2^5 & 2^6\end{vmatrix}\].


If  \[∆_1 = \begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{vmatrix}, ∆_2 = \begin{vmatrix}1 & bc & a \\ 1 & ca & b \\ 1 & ab & c\end{vmatrix},\text{ then }\]}




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


The value of \[\begin{vmatrix}5^2 & 5^3 & 5^4 \\ 5^3 & 5^4 & 5^5 \\ 5^4 & 5^5 & 5^6\end{vmatrix}\]

 


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 value of the determinant  

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




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)

 





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:
 5x + 2y = 3
 3x + 2y = 5


Solve the following system of equations by matrix method:
3x + 7y = 4
x + 2y = −1


Solve the following system of equations by matrix method:
 5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25


Show that the following systems of linear equations is consistent and also find their solutions:
5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5


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


Use product \[\begin{bmatrix}1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4\end{bmatrix}\begin{bmatrix}- 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2\end{bmatrix}\]  to solve the system of equations x + 3z = 9, −x + 2y − 2z = 4, 2x − 3y + 4z = −3.


The sum of three numbers is 2. If twice the second number is added to the sum of first and third, the sum is 1. By adding second and third number to five times the first number, we get 6. Find the three numbers by using matrices.


The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.

 

Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award ₹x each, ₹y each and ₹z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹2,200. School Q wants to spend ₹3,100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each values is ₹1,200, using matrices, find the award money for each value.
Apart from these three values, suggest one more value which should be considered for award.


The system of linear equations:
x + y + z = 2
2x + y − z = 3
3x + 2y + kz = 4 has a unique solution if


If A = `[[1,1,1],[0,1,3],[1,-2,1]]` , find A-1Hence, solve the system of equations: 

x +y + z = 6

y + 3z = 11

and x -2y +z = 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 A = `[(1, -1, 2),(3, 0, -2),(1, 0, 3)]`, verify that A(adj A) = (adj A)A


If `|(2x, 5),(8, x)| = |(6, -2),(7, 3)|`, then value of x is ______.


A set of linear equations is represented by the matrix equation Ax = b. The necessary condition for the existence of a solution for this system is


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


Let `θ∈(0, π/2)`. If the system of linear equations,

(1 + cos2θ)x + sin2θy + 4sin3θz = 0

cos2θx + (1 + sin2θ)y + 4sin3θz = 0

cos2θx + sin2θy + (1 + 4sin3θ)z = 0

has a non-trivial solution, then the value of θ is

 ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×