Advertisements
Advertisements
प्रश्न
Solve the following system of equations by matrix method:
\[\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10\]
\[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10\]
\[\frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13\]
Advertisements
उत्तर
\[\text{ Let }\frac{1}{x}\text{ be }a,\frac{1}{y}\text{ be }b \text{ and}\frac{1}{z}\text{ be }c.\]
Here,
\[A = \begin{bmatrix}2 & - 3 & 3 \\ 1 & 1 & 1 \\ 3 & - 1 & 2\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}2 & - 3 & 3 \\ 1 & 1 & 1 \\ 3 & - 1 & 2\end{vmatrix}\]
\[ = 2\left( 2 + 1 \right) + 3\left( 2 - 3 \right) + 3( - 1 - 3)\]
\[ = 6 - 3 - 12\]
\[ = - 9\]
\[ {\text{ Let }C}_{ij} {\text{be the cofactors of the elements a}}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 1 \\ - 1 & 2\end{vmatrix} = 3, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 1 \\ 3 & 2\end{vmatrix} = 1 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 1 \\ 3 & - 1\end{vmatrix} = - 4\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 3 & 3 \\ - 1 & 2\end{vmatrix} = 3 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 3 \\ 3 & 2\end{vmatrix} = - 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & - 3 \\ 3 & - 1\end{vmatrix} = - 7\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 3 & 3 \\ 1 & 1\end{vmatrix} = - 6 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 3 \\ 1 & 1\end{vmatrix} = 1, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & - 3 \\ 1 & 1\end{vmatrix} = 5\]
\[adj A = \begin{bmatrix}3 & 1 & - 4 \\ 3 & - 5 & - 7 \\ - 6 & 1 & 5\end{bmatrix}^T \]
\[ = \begin{bmatrix}3 & 3 & - 6 \\ 1 & - 5 & 1 \\ - 4 & - 7 & 5\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 9}\begin{bmatrix}3 & 3 & - 6 \\ 1 & - 5 & 1 \\ - 4 & - 7 & 5\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}3 & 3 & - 6 \\ 1 & - 5 & 1 \\ - 4 & - 7 & 5\end{bmatrix}\begin{bmatrix}10 \\ 10 \\ 13\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}30 + 30 - 78 \\ 10 - 50 + 13 \\ - 40 - 70 + 65\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}- 18 \\ - 27 \\ - 45\end{bmatrix}\]
\[ \Rightarrow x = \frac{1}{a} = \frac{- 9}{- 18}, y = \frac{1}{b} = \frac{- 9}{- 27}\text{ and }z = \frac{1}{c} = \frac{- 9}{- 45}\]
\[ \therefore x = \frac{1}{a} = \frac{1}{2}, y = \frac{1}{b} = \frac{1}{3}\text{ and }z = \frac{1}{c} = \frac{1}{5}\]
APPEARS IN
संबंधित प्रश्न
If A = `[(2,-3,5),(3,2,-4),(1,1,-2)]` find A−1. Using A−1 solve the system of equations:
2x – 3y + 5z = 11
3x + 2y – 4z = –5
x + y – 2z = –3
Evaluate the following determinant:
\[\begin{vmatrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{vmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}a + ib & c + id \\ - c + id & a - ib\end{vmatrix}\]
Find the value of x, if
\[\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y\end{vmatrix}\]
\[If ∆ = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix}, ∆_1 = \begin{vmatrix}1 & 1 & 1 \\ yz & zx & xy \\ x & y & z\end{vmatrix},\text{ then prove that }∆ + ∆_1 = 0 .\]
Solve the following determinant equation:
Using determinants show that the following points are collinear:
(5, 5), (−5, 1) and (10, 7)
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.
Prove that :
Prove that :
Prove that :
2x − y = − 2
3x + 4y = 3
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 \[A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}\text{ and B} = \begin{bmatrix}1 & - 4 \\ 3 & - 2\end{bmatrix},\text{ find }|AB|\]
Find the value of x from the following : \[\begin{vmatrix}x & 4 \\ 2 & 2x\end{vmatrix} = 0\]
Write the value of the determinant \[\begin{vmatrix}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x\end{vmatrix}\]
The value of the determinant
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 + 7z = 14
2x − y + 3z = 4
x + 2y − 3z = 0
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.
3x − y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0
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)|`
System of equations x + y = 2, 2x + 2y = 3 has ______
Transform `[(1, 2, 4),(3, -1, 5),(2, 4, 6)]` into an upper triangular matrix by using suitable row transformations
Solve the following system of equations by using inversion method
x + y = 1, y + z = `5/3`, z + x = `4/3`
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)) =` ____________.
Solve the following system of equations x − y + z = 4, x − 2y + 2z = 9 and 2x + y + 3z = 1.
The existence of unique solution of the system of linear equations x + y + z = a, 5x – y + bz = 10, 2x + 3y – z = 6 depends on
If the system of equations x + λy + 2 = 0, λx + y – 2 = 0, λx + λy + 3 = 0 is consistent, then
If `|(x + a, beta, y),(a, x + beta, y),(a, beta, x + y)|` = 0, then 'x' is equal to
If the system of linear equations x + 2ay + az = 0; x + 3by + bz = 0; x + 4cy + cz = 0 has a non-zero solution, then a, b, c ______.
Using the matrix method, solve the following system of linear equations:
`2/x + 3/y + 10/z` = 4, `4/x - 6/y + 5/z` = 1, `6/x + 9/y - 20/z` = 2.
