Advertisements
Advertisements
Question
Given: x + 2y = 1
3x + y = 4
Solution
\[D = \begin{vmatrix}1 & 2 \\ 3 & 1\end{vmatrix} = - 5\]
\[ D_1 = \begin{vmatrix}1 & 2 \\ 4 & 1\end{vmatrix} = - 7\]
\[ D_2 = \begin{vmatrix}1 & 1 \\ 3 & 4\end{vmatrix} = 1\]
\[Now, \]
\[x = \frac{D_1}{D} = \frac{7}{5}\]
\[y = \frac{D_2}{D} = - \frac{1}{5}\]
\[ \therefore x = \frac{7}{5}\text{ and }y = - \frac{1}{5}\]
APPEARS IN
RELATED QUESTIONS
Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to ______.
Examine the consistency of the system of equations.
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
Solve system of linear equations, using matrix method.
5x + 2y = 4
7x + 3y = 5
Solve system of linear equations, using matrix method.
4x – 3y = 3
3x – 5y = 7
Solve system of linear equations, using matrix method.
5x + 2y = 3
3x + 2y = 5
Solve the system of linear equations using the matrix method.
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
Find the value of x, if
\[\begin{vmatrix}3 & x \\ x & 1\end{vmatrix} = \begin{vmatrix}3 & 2 \\ 4 & 1\end{vmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38\end{vmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}1 & - 3 & 2 \\ 4 & - 1 & 2 \\ 3 & 5 & 2\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}1 & a & a^2 - bc \\ 1 & b & b^2 - ac \\ 1 & c & c^2 - ab\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3\end{vmatrix}\]
Prove the following identities:
\[\begin{vmatrix}x + \lambda & 2x & 2x \\ 2x & x + \lambda & 2x \\ 2x & 2x & x + \lambda\end{vmatrix} = \left( 5x + \lambda \right) \left( \lambda - x \right)^2\]
Solve the following determinant equation:
If \[\begin{vmatrix}a & b - y & c - z \\ a - x & b & c - z \\ a - x & b - y & c\end{vmatrix} =\] 0, then using properties of determinants, find the value of \[\frac{a}{x} + \frac{b}{y} + \frac{c}{z}\] , where \[x, y, z \neq\] 0
Using determinants show that the following points are collinear:
(1, −1), (2, 1) and (4, 5)
2y − 3z = 0
x + 3y = − 4
3x + 4y = 3
x + 2y = 5
3x + 6y = 15
Solve each of the following system of homogeneous linear equations.
2x + 3y + 4z = 0
x + y + z = 0
2x − y + 3z = 0
If a, b, c are non-zero real numbers and if the system of equations
(a − 1) x = y + z
(b − 1) y = z + x
(c − 1) z = x + y
has a non-trivial solution, then prove that ab + bc + ca = abc.
The value of the determinant
If \[∆_1 = \begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{vmatrix}, ∆_2 = \begin{vmatrix}1 & bc & a \\ 1 & ca & b \\ 1 & ab & c\end{vmatrix},\text{ then }\]}
Solve the following system of equations by matrix method:
Show that the following systems of linear equations is consistent and also find their solutions:
2x + 3y = 5
6x + 9y = 15
Show that the following systems of linear equations is consistent and also find their solutions:
x + y + z = 6
x + 2y + 3z = 14
x + 4y + 7z = 30
Show that each one of the following systems of linear equation is inconsistent:
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prices at the rate of Rs. x, y and z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total price money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with a total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of x, y and z. What values are described in this equations?
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.
x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0
The number of solutions of the system of equations:
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
If A = `[[1,1,1],[0,1,3],[1,-2,1]]` , find A-1Hence, solve the system of equations:
x +y + z = 6
y + 3z = 11
and x -2y +z = 0
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 ______
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)) =` ____________.
Let A = `[(1,sin α,1),(-sin α,1,sin α),(-1,-sin α,1)]`, where 0 ≤ α ≤ 2π, then:
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`
Let the system of linear equations x + y + az = 2; 3x + y + z = 4; x + 2z = 1 have a unique solution (x*, y*, z*). If (α, x*), (y*, α) and (x*, –y*) are collinear points, then the sum of absolute values of all possible values of α is ______.