Advertisements
Advertisements
प्रश्न
Solve the following system of equations by matrix method:
x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9
Advertisements
उत्तर
Here,
\[A = \begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 1 \\ 2 & 1 & - 3\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}1 & 1 & 1 \\ 2 & - 1 & 1 \\ 2 & 1 & - 3\end{vmatrix}\]
\[ = 1\left( 3 - 1 \right) - 1\left( - 6 - 2 \right) + 1(2 + 2)\]
\[ = 2 + 8 + 4\]
\[ = 14\]
\[ {\text{ Let }C}_{ij} {\text{be the cofactors of the elements a}}_{ij}\text{ in }A\left[ a_{ij} \right]. \text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & 1 \\ 1 & - 3\end{vmatrix} = 2 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ 2 & - 3\end{vmatrix} = 8, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 1 \\ 2 & 1\end{vmatrix} = 4\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & - 3\end{vmatrix} = 4, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 2 & - 3\end{vmatrix} = - 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = 1\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ - 1 & 1\end{vmatrix} = 2 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = 1, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 2 & - 1\end{vmatrix} = - 3\]
\[adj A = \begin{bmatrix}2 & 8 & 4 \\ 4 & - 5 & 1 \\ 2 & 1 & - 3\end{bmatrix}^T \]
\[ = \begin{bmatrix}2 & 4 & 2 \\ 8 & - 5 & 1 \\ 4 & 1 & - 3\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{14}\begin{bmatrix}2 & 4 & 2 \\ 8 & - 5 & 1 \\ 4 & 1 & - 3\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{14}\begin{bmatrix}2 & 4 & 2 \\ 8 & - 5 & 1 \\ 4 & 1 & - 3\end{bmatrix}\begin{bmatrix}3 \\ - 1 \\ - 9\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{14}\begin{bmatrix}6 - 4 - 18 \\ 24 + 5 - 9 \\ 12 - 1 + 27\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{14}\begin{bmatrix}- 16 \\ 20 \\ 38\end{bmatrix}\]
\[ \Rightarrow x = \frac{- 16}{14}, y = \frac{20}{14}\text{ and }z = \frac{38}{14}\]
\[ \therefore x = \frac{- 8}{7}, y = \frac{10}{7}\text{ and }z = \frac{19}{7}\]
APPEARS IN
संबंधित प्रश्न
Examine the consistency of the system of equations.
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
Solve the system of the following equations:
`2/x+3/y+10/z = 4`
`4/x-6/y + 5/z = 1`
`6/x + 9/y - 20/x = 2`
\[∆ = \begin{vmatrix}\cos \alpha \cos \beta & \cos \alpha \sin \beta & - \sin \alpha \\ - \sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{vmatrix}\]
Find the value of x, if
\[\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}a + b & 2a + b & 3a + b \\ 2a + b & 3a + b & 4a + b \\ 4a + b & 5a + b & 6a + b\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}a & b & c \\ a + 2x & b + 2y & c + 2z \\ x & y & z\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin^2 23^\circ & \sin^2 67^\circ & \cos180^\circ \\ - \sin^2 67^\circ & - \sin^2 23^\circ & \cos^2 180^\circ \\ \cos180^\circ & \sin^2 23^\circ & \sin^2 67^\circ\end{vmatrix}\]
Evaluate the following:
\[\begin{vmatrix}0 & x y^2 & x z^2 \\ x^2 y & 0 & y z^2 \\ x^2 z & z y^2 & 0\end{vmatrix}\]
If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.
Prove that :
Prove that :
3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20.
2x + y − 2z = 4
x − 2y + z = − 2
5x − 5y + z = − 2
Solve each of the following system of homogeneous linear equations.
2x + 3y + 4z = 0
x + y + z = 0
2x − y + 3z = 0
If \[A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}\text{ and B} = \begin{bmatrix}1 & - 4 \\ 3 & - 2\end{bmatrix},\text{ find }|AB|\]
Write the value of the determinant \[\begin{vmatrix}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x\end{vmatrix}\]
If \[\begin{vmatrix}3x & 7 \\ - 2 & 4\end{vmatrix} = \begin{vmatrix}8 & 7 \\ 6 & 4\end{vmatrix}\] , find the value of x.
Find the maximum value of \[\begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos \theta\end{vmatrix}\]
If ω is a non-real cube root of unity and n is not a multiple of 3, then \[∆ = \begin{vmatrix}1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^n \\ \omega^n & \omega^{2n} & 1\end{vmatrix}\]
The number of distinct real roots of \[\begin{vmatrix}cosec x & \sec x & \sec x \\ \sec x & cosec x & \sec x \\ \sec x & \sec x & cosec x\end{vmatrix} = 0\] lies in the interval
\[- \frac{\pi}{4} \leq x \leq \frac{\pi}{4}\]
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:
3x + 4y − 5 = 0
x − y + 3 = 0
A company produces three products every day. Their production on a certain day is 45 tons. It is found that the production of third product exceeds the production of first product by 8 tons while the total production of first and third product is twice the production of second product. Determine the production level of each product using matrix method.
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 A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A 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 ₹1,600. School B wants to spend ₹2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is ₹900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.
A total amount of ₹7000 is deposited in three different saving bank accounts with annual interest rates 5%, 8% and \[8\frac{1}{2}\] % respectively. The total annual interest from these three accounts is ₹550. Equal amounts have been deposited in the 5% and 8% saving accounts. Find the amount deposited in each of the three accounts, with the help of matrices.
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.
The number of solutions of the system of equations:
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
Find the inverse of the following matrix, using elementary transformations:
`A= [[2 , 3 , 1 ],[2 , 4 , 1],[3 , 7 ,2]]`
System of equations x + y = 2, 2x + 2y = 3 has ______
Three chairs and two tables cost ₹ 1850. Five chairs and three tables cost ₹2850. Find the cost of four chairs and one table by using matrices
Solve the following system of equations by using inversion method
x + y = 1, y + z = `5/3`, z + x = `4/3`
If `|(2x, 5),(8, x)| = |(6, 5),(8, 3)|`, then find x
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 ____________.
`abs ((2"xy", "x"^2, "y"^2),("x"^2, "y"^2, 2"xy"),("y"^2, 2"xy", "x"^2)) =` ____________.
The number of values of k for which the linear equations 4x + ky + 2z = 0, kx + 4y + z = 0 and 2x + 2y + z = 0 possess a non-zero solution is
If `|(x + a, beta, y),(a, x + beta, y),(a, beta, x + y)|` = 0, then 'x' is equal to
The number of real value of 'x satisfying `|(x, 3x + 2, 2x - 1),(2x - 1, 4x, 3x + 1),(7x - 2, 17x + 6, 12x - 1)|` = 0 is
