Advertisements
Advertisements
प्रश्न
Solve the following system of equations by matrix method:
x − y + z = 2
2x − y = 0
2y − z = 1
Advertisements
उत्तर
Here,
\[A = \begin{bmatrix}1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 0 & 2 & - 1\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 0 & 2 & - 1\end{vmatrix}\]
\[ = 1\left( 1 - 0 \right) + 1\left( - 2 - 0 \right) + 1(4 - 0)\]
\[ = 1 - 2 + 4\]
\[ = 3\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & 0 \\ 2 & - 1\end{vmatrix} = 1, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 0 \\ 0 & - 1\end{vmatrix} = 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 1 \\ 0 & 2\end{vmatrix} = 4\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & 1 \\ 2 & - 1\end{vmatrix} = 1, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 0 & - 1\end{vmatrix} = - 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 1 \\ 0 & 2\end{vmatrix} = - 2\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & 1 \\ - 1 & 0\end{vmatrix} = 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 2 & 0\end{vmatrix} = 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 1 \\ 2 & - 1\end{vmatrix} = 1\]
\[adj A = \begin{bmatrix}1 & 2 & 4 \\ 1 & - 1 & - 2 \\ 1 & 2 & 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 2 \\ 4 & - 2 & 1\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{1}\begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 2 \\ 4 & - 2 & 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{3}\begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 2 \\ 4 & - 2 & 1\end{bmatrix}\begin{bmatrix}2 \\ 0 \\ 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{3}\begin{bmatrix}2 + 1 \\ 4 + 2 \\ 8 + 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{1}\begin{bmatrix}3 \\ 6 \\ 9\end{bmatrix}\]
\[ \Rightarrow x = \frac{3}{3}, y = \frac{6}{3}\text{ and }z = \frac{9}{3}\]
\[ \therefore x = 1, y = 2\text{ and }z = 3\]
APPEARS IN
संबंधित प्रश्न
Examine the consistency of the system of equations.
2x − y = 5
x + y = 4
Solve the system of linear equations using the matrix method.
2x + y + z = 1
x – 2y – z = `3/2`
3y – 5z = 9
Show that
\[\begin{vmatrix}\sin 10^\circ & - \cos 10^\circ \\ \sin 80^\circ & \cos 80^\circ\end{vmatrix} = 1\]
Evaluate
\[\begin{vmatrix}2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1\end{vmatrix}\] by two methods.
If \[A = \begin{bmatrix}2 & 5 \\ 2 & 1\end{bmatrix} \text{ and } B = \begin{bmatrix}4 & - 3 \\ 2 & 5\end{bmatrix}\] , verify that |AB| = |A| |B|.
Find the value of x, if
\[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & 5 \\ 8 & 3\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]
\[\begin{vmatrix}b + c & a & a \\ b & c + a & b \\ c & c & a + b\end{vmatrix} = 4abc\]
Prove the following identity:
\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix} = a^2 \left( a + x + y + z \right)\]
Show that x = 2 is a root of the equation
Solve the following determinant equation:
Solve the following determinant equation:
Using determinants show that the following points are collinear:
(3, −2), (8, 8) and (5, 2)
Prove that :
\[\begin{vmatrix}\left( b + c \right)^2 & a^2 & bc \\ \left( c + a \right)^2 & b^2 & ca \\ \left( a + b \right)^2 & c^2 & ab\end{vmatrix} = \left( a - b \right) \left( b - c \right) \left( c - a \right) \left( a + b + c \right) \left( a^2 + b^2 + c^2 \right)\]
3x − y + 2z = 3
2x + y + 3z = 5
x − 2y − z = 1
Solve each of the following system of homogeneous linear equations.
2x + 3y + 4z = 0
x + y + z = 0
2x − y + 3z = 0
Solve each of the following system of homogeneous linear equations.
3x + y + z = 0
x − 4y + 3z = 0
2x + 5y − 2z = 0
Find the real values of λ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions
\[2 \lambda x - 2y + 3z = 0\]
\[ x + \lambda y + 2z = 0\]
\[ 2x + \lambda z = 0\]
If \[A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 0 \\ - 1 & 0\end{bmatrix}\] , find |AB|.
Write the value of \[\begin{vmatrix}a + ib & c + id \\ - c + id & a - ib\end{vmatrix} .\]
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 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 + 7y + 2 = 0
4x + 6y + 3 = 0
Solve the following system of equations by matrix method:
3x + y = 19
3x − y = 23
Solve the following system of equations by matrix method:
\[\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10\]
\[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10\]
\[\frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13\]
If A = `[(1, 2, 0), (-2, -1, -2), (0, -1, 1)]`, find A−1. Using A−1, solve the system of linear equations x − 2y = 10, 2x − y − z = 8, −2y + z = 7.
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 prices of three commodities P, Q and R are Rs x, y and z per unit respectively. A purchases 4 units of R and sells 3 units of P and 5 units of Q. B purchases 3 units of Q and sells 2 units of P and 1 unit of R. Cpurchases 1 unit of P and sells 4 units of Q and 6 units of R. In the process A, B and C earn Rs 6000, Rs 5000 and Rs 13000 respectively. If selling the units is positive earning and buying the units is negative earnings, find the price per unit of three commodities by using matrix method.
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.
2x − y + z = 0
3x + 2y − z = 0
x + 4y + 3z = 0
3x − y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0
3x + y − 2z = 0
x + y + z = 0
x − 2y + z = 0
If \[\begin{bmatrix}1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}\], find x, y and z.
Transform `[(1, 2, 4),(3, -1, 5),(2, 4, 6)]` into an upper triangular matrix by using suitable row transformations
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 a, b, c are non-zeros, then the system of equations (α + a)x + αy + αz = 0, αx + (α + b)y + αz = 0, αx+ αy + (α + c)z = 0 has a non-trivial solution if
Let A = `[(i, -i),(-i, i)], i = sqrt(-1)`. Then, the system of linear equations `A^8[(x),(y)] = [(8),(64)]` has ______.
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
______.
