English

The Number of Solutions of the System of Equations 2x + Y − Z = 7 X − 3y + 2z = 1 X + 4y − 3z = 5 is (A) 3 (B) 2 (C) 1 (D) 0 - Mathematics

Advertisements
Advertisements

Question

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

Options

  • 3

  • 2

  • 1

  • 0

MCQ
Advertisements

Solution

\[(d) 0\]
The given system of equations can be written in matrix form as follows:
\[\begin{bmatrix}2 & 1 & - 1 \\ 1 & - 3 & 2 \\ 1 & 4 & - 3\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}7 \\ 1 \\ 5\end{bmatrix}\]
\[AX = B \]
Here,
\[A = \begin{bmatrix}2 & 1 & - 1 \\ 1 & - 3 & 2 \\ 1 & 4 & - 3\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}7 \\ 1 \\ 5\end{bmatrix}\]
Now,
\[\left| A \right|=2 \left( 9 - 8 \right) - 1\left( - 3 - 2 \right) - 1\left( 4 + 3 \right)\]
\[ = 2 + 5 - 7\]
\[ = 0\]
\[ {\text{  Let }C}_{ij} {\text{  be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right]. \text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 3 & 2 \\ 4 & - 3\end{vmatrix} = 1, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 2 \\ 1 & - 3\end{vmatrix} = 5, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & - 3 \\ 1 & 4\end{vmatrix} = 7\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & - 1 \\ 4 & - 3\end{vmatrix} = - 1, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & - 1 \\ 1 & - 3\end{vmatrix} = - 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 1 \\ 1 & 4\end{vmatrix} = - 7\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & - 1 \\ - 3 & 2\end{vmatrix} = - 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & - 1 \\ 1 & 2\end{vmatrix} = - 5, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 1 \\ 1 & - 3\end{vmatrix} = - 7\]
\[adj A = \begin{bmatrix}1 & 5 & 7 \\ - 1 & - 5 & - 7 \\ - 1 & - 5 & - 7\end{bmatrix}^T \]
\[ = \begin{bmatrix}1 & - 1 & - 1 \\ 5 & - 5 & - 5 \\ 7 & - 7 & - 7\end{bmatrix}\]
\[ \Rightarrow \left( adj A \right) B = \begin{bmatrix}1 & - 1 & - 1 \\ 5 & - 5 & - 5 \\ 7 & - 7 & - 7\end{bmatrix}\begin{bmatrix}7 \\ 1 \\ 5\end{bmatrix}\]
\[ = \begin{bmatrix}7 - 1 - 5 \\ 35 - 5 - 25 \\ 49 - 7 - 35\end{bmatrix}\]
\[ = \begin{bmatrix}1 \\ 5 \\ 7\end{bmatrix} \neq 0\]
So, the given system of equations has no solution.

shaalaa.com
  Is there an error in this question or solution?
Chapter 8: Solution of Simultaneous Linear Equations - Exercise 8.4 [Page 21]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 8 Solution of Simultaneous Linear Equations
Exercise 8.4 | Q 2 | Page 21

RELATED QUESTIONS

Examine the consistency of the system of equations.

3x − y − 2z = 2

2y − z = −1

3x − 5y = 3


Solve the system of linear equations using the matrix method.

2x – y = –2

3x + 4y = 3


Evaluate the following determinant:

\[\begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}\]


Find the value of x, if
\[\begin{vmatrix}2 & 4 \\ 5 & 1\end{vmatrix} = \begin{vmatrix}2x & 4 \\ 6 & x\end{vmatrix}\]


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

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


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

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


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

\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]


Prove that:

`[(a, b, c),(a - b, b - c, c - a),(b + c, c + a, a + b)] = a^3 + b^3 + c^3 -3abc`


Prove the following identities:
\[\begin{vmatrix}x + \lambda & 2x & 2x \\ 2x & x + \lambda & 2x \\ 2x & 2x & x + \lambda\end{vmatrix} = \left( 5x + \lambda \right) \left( \lambda - x \right)^2\]


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


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


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.
 

 


​Solve the following determinant equation:

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

 


​Solve the following determinant equation:
\[\begin{vmatrix}15 - 2x & 11 - 3x & 7 - x \\ 11 & 17 & 14 \\ 10 & 16 & 13\end{vmatrix} = 0\]

If \[a, b\] and c  are all non-zero and 

\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c\end{vmatrix} =\] 0, then prove that 
\[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} +\]1
= 0

 


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

 (0, 0), (6, 0) and (4, 3)


Using determinants show that the following points are collinear:

(2, 3), (−1, −2) and (5, 8)


Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, −6) and (5, 4).


Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?


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


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

(1, 2) and (3, 6)


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 = 17
3x + 5y = 6


2x − y = − 2
3x + 4y = 3


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


If w is an imaginary cube root of unity, find the value of \[\begin{vmatrix}1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w\end{vmatrix}\]


If \[A = \left[ a_{ij} \right]\]   is a 3 × 3 diagonal matrix such that a11 = 1, a22 = 2 a33 = 3, then find |A|.

 

If |A| = 2, where A is 2 × 2 matrix, find |adj A|.


If \[D_k = \begin{vmatrix}1 & n & n \\ 2k & n^2 + n + 2 & n^2 + n \\ 2k - 1 & n^2 & n^2 + n + 2\end{vmatrix} and \sum^n_{k = 1} D_k = 48\], then n equals

 


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:
4x − 2y = 3
6x − 3y = 5


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.


If \[\begin{bmatrix}1 & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ - 1 \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}\] , find x, y and z.


Write the value of `|(a-b, b- c, c-a),(b-c, c-a, a-b),(c-a, a-b, b-c)|`


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


If the system of linear equations

2x + y – z = 7

x – 3y + 2z = 1

x + 4y + δz = k, where δ, k ∈ R has infinitely many solutions, then δ + k is equal to ______.


Let P = `[(-30, 20, 56),(90, 140, 112),(120, 60, 14)]` and A = `[(2, 7, ω^2),(-1, -ω, 1),(0, -ω, -ω + 1)]` where ω = `(-1 + isqrt(3))/2`, and I3 be the identity matrix of order 3. If the determinant of the matrix (P–1AP – I3)2 is αω2, then the value of α is equal to ______.


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×