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
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.
APPEARS IN
RELATED QUESTIONS
Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3
Find the value of x, if
\[\begin{vmatrix}3 & x \\ x & 1\end{vmatrix} = \begin{vmatrix}3 & 2 \\ 4 & 1\end{vmatrix}\]
Find the integral value of x, if \[\begin{vmatrix}x^2 & x & 1 \\ 0 & 2 & 1 \\ 3 & 1 & 4\end{vmatrix} = 28 .\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda\end{vmatrix}\]
Solve the following determinant equation:
Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, −6) and (5, 4).
If the points (3, −2), (x, 2), (8, 8) are collinear, find x using determinant.
Find values of k, if area of triangle is 4 square units whose vertices are
(k, 0), (4, 0), (0, 2)
x − 2y = 4
−3x + 5y = −7
Prove that :
Prove that :
x − y + 3z = 6
x + 3y − 3z = − 4
5x + 3y + 3z = 10
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|.
Find the value of the determinant
\[\begin{bmatrix}4200 & 4201 \\ 4205 & 4203\end{bmatrix}\]
If |A| = 2, where A is 2 × 2 matrix, find |adj A|.
If \[A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\] , then for any natural number, find the value of Det(An).
If a > 0 and discriminant of ax2 + 2bx + c is negative, then
\[∆ = \begin{vmatrix}a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & 0\end{vmatrix} is\]
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
Let \[A = \begin{bmatrix}1 & \sin \theta & 1 \\ - \sin \theta & 1 & \sin \theta \\ - 1 & - \sin \theta & 1\end{bmatrix},\text{ where 0 }\leq \theta \leq 2\pi . \text{ Then,}\]
If x, y, z are different from zero and \[\begin{vmatrix}1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z\end{vmatrix} = 0\] , then the value of x−1 + y−1 + z−1 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:
8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5
Show that the following systems of linear equations is consistent and also find their solutions:
2x + 3y = 5
6x + 9y = 15
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.
A shopkeeper has 3 varieties of pens 'A', 'B' and 'C'. Meenu purchased 1 pen of each variety for a total of Rs 21. Jeevan purchased 4 pens of 'A' variety 3 pens of 'B' variety and 2 pens of 'C' variety for Rs 60. While Shikha purchased 6 pens of 'A' variety, 2 pens of 'B' variety and 3 pens of 'C' variety for Rs 70. Using matrix method, find cost of each variety of pen.
2x − y + z = 0
3x + 2y − z = 0
x + 4y + 3z = 0
The system of equation x + y + z = 2, 3x − y + 2z = 6 and 3x + y + z = −18 has
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 + z = 7.
Find the inverse of the following matrix, using elementary transformations:
`A= [[2 , 3 , 1 ],[2 , 4 , 1],[3 , 7 ,2]]`
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?
Transform `[(1, 2, 4),(3, -1, 5),(2, 4, 6)]` into an upper triangular matrix by using suitable row transformations
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`
If the system of equations x + λy + 2 = 0, λx + y – 2 = 0, λx + λy + 3 = 0 is consistent, then
If the following equations
x + y – 3 = 0
(1 + λ)x + (2 + λ)y – 8 = 0
x – (1 + λ)y + (2 + λ) = 0
are consistent then the value of λ can be ______.
