Advertisements
Advertisements
Question
Show that x = 2 is a root of the equation
Advertisements
Solution
\[\text{ Let }∆ = \begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 3 & 2x & x + 2\end{vmatrix}\]
\[ = \begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 3 - x & 2x + 6 & x + 3\end{vmatrix} \left[\text{ Applying }R_3\text{ to }R_3 - R_1 \right]\]
\[ = \left( x + 3 \right)\begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 1 & 2 & 1\end{vmatrix} \]
\[ = \left( x + 3 \right)\begin{vmatrix}x - 2 & 3x - 6 & - x + 2 \\ 2 & - 3x & x - 3 \\ - 1 & 2 & 1\end{vmatrix} \left[\text{ Applying } R_1 \text{ to }R_1 - R_2 \right]\]
\[ = \left( x + 3 \right)\left( x - 2 \right)\begin{vmatrix}1 & 3 & - 1 \\ 2 & - 3x & x - 3 \\ - 1 & 2 & 1\end{vmatrix} \]
\[ = \left( x + 3 \right)\left( x - 2 \right)\begin{vmatrix}1 & 3 & 0 \\ 2 & - 3x & x - 1 \\ - 1 & 2 & 0\end{vmatrix} \left[\text{ Applying }C_3 \text{ to }C_3 + C_1 \right]\]
\[ = \left( x + 3 \right)\left( x - 2 \right)\left( x - 1 \right)\begin{vmatrix}1 & 3 & 0 \\ 2 & - 3x & 1 \\ - 1 & 2 & 0\end{vmatrix} \]
\[ = \left( x + 3 \right)\left( x - 2 \right)\left( x - 1 \right)\left\{ - 1\begin{vmatrix}1 & 3 \\ - 1 & 2\end{vmatrix} \right\} \left[ \text{ Expanding along }C_3 \right]\]
\[ = - 5\left( x + 3 \right)\left( x - 2 \right)\left( x - 1 \right)\]
\[x = 2, - 3, 1\]
APPEARS IN
RELATED QUESTIONS
Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
Solve the system of linear equations using the matrix method.
5x + 2y = 3
3x + 2y = 5
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|.
Evaluate the following determinant:
\[\begin{vmatrix}1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38\end{vmatrix}\]
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:
\[\begin{vmatrix}49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix}\]
Prove the following identities:
\[\begin{vmatrix}x + \lambda & 2x & 2x \\ 2x & x + \lambda & 2x \\ 2x & 2x & x + \lambda\end{vmatrix} = \left( 5x + \lambda \right) \left( \lambda - x \right)^2\]
\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & a \\ 1 & 1 & 1 + a\end{vmatrix} = a^3 + 3 a^2\]
Using determinants show that the following points are collinear:
(5, 5), (−5, 1) and (10, 7)
Using determinants show that the following points are collinear:
(3, −2), (8, 8) and (5, 2)
If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.
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 :
2x − y = 17
3x + 5y = 6
x + 2y = 5
3x + 6y = 15
Find the value of the determinant
\[\begin{bmatrix}4200 & 4201 \\ 4205 & 4203\end{bmatrix}\]
The value of \[\begin{vmatrix}1 & 1 & 1 \\ {}^n C_1 & {}^{n + 2} C_1 & {}^{n + 4} C_1 \\ {}^n C_2 & {}^{n + 2} C_2 & {}^{n + 4} C_2\end{vmatrix}\] is
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:
Solve the following system of equations by matrix method:
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
Show that the following systems of linear equations is consistent and also find their solutions:
2x + 3y = 5
6x + 9y = 15
Show that each one of the following systems of linear equation is inconsistent:
2x + 5y = 7
6x + 15y = 13
Given \[A = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}, B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}\] , find BA and use this to solve the system of equations y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17
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.
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.
2x − y + z = 0
3x + 2y − z = 0
x + 4y + 3z = 0
x + y − 6z = 0
x − y + 2z = 0
−3x + y + 2z = 0
The number of solutions of the system of equations:
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
The system of linear equations:
x + y + z = 2
2x + y − z = 3
3x + 2y + kz = 4 has a unique solution if
Transform `[(1, 2, 4),(3, -1, 5),(2, 4, 6)]` into an upper triangular matrix by using suitable row transformations
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
If `|(2x, 5),(8, x)| = |(6, -2),(7, 3)|`, then value of x is ______.
If A = `[(1,-1,0),(2,3,4),(0,1,2)]` and B = `[(2,2,-4),(-4,2,-4),(2,-1,5)]`, then:
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
The value (s) of m does the system of equations 3x + my = m and 2x – 5y = 20 has a solution satisfying the conditions x > 0, y > 0.
