Advertisements
Advertisements
प्रश्न
5x + 7y = − 2
4x + 6y = − 3
Advertisements
उत्तर
\[\text{ Given }: \hspace{0.167em} 5x + 7y = - 2\]
\[ 4x + 6y = - 3\]
Using Cramer's Rule, we get
\[D = \begin{vmatrix} 5 & 7 \\4 & 6 \end{vmatrix} = 30 - 28 = 2\]
\[ D_1 = \begin{vmatrix} - 2 & 7\\ - 3 & 6 \end{vmatrix} = - 12 + 21 = 9\]
\[ D_2 = \begin{vmatrix} 5 & - 2 \\4 & - 3 \end{vmatrix} = - 15 + 8 = - 7\]
Now,
\[x = \frac{D_1}{D} = \frac{9}{2}\]
\[y = \frac{D_2}{D} = \frac{- 7}{2}\]
\[ \therefore x = \frac{9}{2}\text{ and }y = \frac{- 7}{2}\]
APPEARS IN
संबंधित प्रश्न
Solve the system of linear equations using the matrix method.
4x – 3y = 3
3x – 5y = 7
Solve the system of linear equations using the matrix method.
2x + y + z = 1
x – 2y – z = `3/2`
3y – 5z = 9
Evaluate
\[∆ = \begin{vmatrix}0 & \sin \alpha & - \cos \alpha \\ - \sin \alpha & 0 & \sin \beta \\ \cos \alpha & - \sin \beta & 0\end{vmatrix}\]
For what value of x the matrix A is singular?
\[ A = \begin{bmatrix}1 + x & 7 \\ 3 - x & 8\end{bmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}a & b & c \\ a + 2x & b + 2y & c + 2z \\ x & y & z\end{vmatrix}\]
Using properties of determinants prove that
\[\begin{vmatrix}x + 4 & 2x & 2x \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix} = \left( 5x + 4 \right) \left( 4 - x \right)^2\]
Find the area of the triangle with vertice at the point:
(3, 8), (−4, 2) and (5, −1)
Using determinants show that the following points are collinear:
(2, 3), (−1, −2) and (5, 8)
If the points (3, −2), (x, 2), (8, 8) are collinear, find x using determinant.
2x − y = 1
7x − 2y = −7
Prove that :
Prove that :
Prove that :
6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8
For what value of x, the following matrix is singular?
If \[A = \begin{bmatrix}0 & i \\ i & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\] , find the value of |A| + |B|.
Write the value of
Find the value of the determinant \[\begin{vmatrix}243 & 156 & 300 \\ 81 & 52 & 100 \\ - 3 & 0 & 4\end{vmatrix} .\]
The value of the determinant
The maximum value of \[∆ = \begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin\theta & 1 \\ 1 + \cos\theta & 1 & 1\end{vmatrix}\] is (θ is real)
Solve the following system of equations by matrix method:
5x + 2y = 3
3x + 2y = 5
Solve the following system of equations by matrix method:
3x + y = 7
5x + 3y = 12
Solve the following system of equations by matrix method:
x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9
Solve the following system of equations by matrix method:
x − y + z = 2
2x − y = 0
2y − z = 1
Solve the following system of equations by matrix method:
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
The sum of three numbers is 2. If twice the second number is added to the sum of first and third, the sum is 1. By adding second and third number to five times the first number, we get 6. Find the three numbers by using matrices.
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.
The number of solutions of the system of equations
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
is
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.
x + y = 1
x + z = − 6
x − y − 2z = 3
The cost of 4 dozen pencils, 3 dozen pens and 2 dozen erasers is ₹ 60. The cost of 2 dozen pencils, 4 dozen pens and 6 dozen erasers is ₹ 90. Whereas the cost of 6 dozen pencils, 2 dozen pens and 3 dozen erasers is ₹ 70. Find the cost of each item per dozen by using matrices
If the system of equations x + λy + 2 = 0, λx + y – 2 = 0, λx + λy + 3 = 0 is consistent, then
The system of simultaneous linear equations kx + 2y – z = 1, (k – 1)y – 2z = 2 and (k + 2)z = 3 have a unique solution if k equals:
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 ______.
