Advertisements
Advertisements
Question
Solution
Given:α, β, γ are in A.P.
Now,
`Delta=1/2|(x-3,x-4,x-alpha),(2x-4,2x-6,2x-2beta),(x-1,x-2,x-gamma)|` `["Applying" R_2->2R_2]`
`Delta=1/2|(x-3,x-4,x-alpha),(0,0,-2beta+alpha+gamma),(x-1,x-2,x-gamma)|` `[because 2beta=alpha+gamma]` `["Applying" R_2->R_2-(R_1+R_3)]`
`Delta=1/2|(x-3,x-4,x-alpha),(0,0,0),(x-1,x-2,x-gamma)|`
Δ = 0
APPEARS IN
RELATED QUESTIONS
Examine the consistency of the system of equations.
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
Solve the system of linear equations using the matrix method.
x − y + z = 4
2x + y − 3z = 0
x + y + z = 2
Evaluate the following determinant:
\[\begin{vmatrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{vmatrix}\]
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:
\[\begin{vmatrix}8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}a & b + c & a^2 \\ b & c + a & b^2 \\ c & a + b & c^2\end{vmatrix}\]
\[\begin{vmatrix}- a \left( b^2 + c^2 - a^2 \right) & 2 b^3 & 2 c^3 \\ 2 a^3 & - b \left( c^2 + a^2 - b^2 \right) & 2 c^3 \\ 2 a^3 & 2 b^3 & - c \left( a^2 + b^2 - c^2 \right)\end{vmatrix} = abc \left( a^2 + b^2 + c^2 \right)^3\]
Without expanding, prove that
\[\begin{vmatrix}a & b & c \\ x & y & z \\ p & q & r\end{vmatrix} = \begin{vmatrix}x & y & z \\ p & q & r \\ a & b & c\end{vmatrix} = \begin{vmatrix}y & b & q \\ x & a & p \\ z & c & r\end{vmatrix}\]
Solve the following determinant equation:
Solve the following determinant equation:
Using determinants show that the following points are collinear:
(2, 3), (−1, −2) and (5, 8)
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.
If the points (3, −2), (x, 2), (8, 8) are collinear, find x using determinant.
Prove that :
2y − 3z = 0
x + 3y = − 4
3x + 4y = 3
3x − y + 2z = 3
2x + y + 3z = 5
x − 2y − z = 1
For what value of x, the following matrix is singular?
Find the value of the determinant
\[\begin{bmatrix}4200 & 4201 \\ 4205 & 4203\end{bmatrix}\]
If \[\begin{vmatrix}2x + 5 & 3 \\ 5x + 2 & 9\end{vmatrix} = 0\]
If x ∈ N and \[\begin{vmatrix}x + 3 & - 2 \\ - 3x & 2x\end{vmatrix}\] = 8, then find the value of x.
Let \[\begin{vmatrix}x & 2 & x \\ x^2 & x & 6 \\ x & x & 6\end{vmatrix} = a x^4 + b x^3 + c x^2 + dx + e\]
Then, the value of \[5a + 4b + 3c + 2d + e\] is equal to
Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant
The other factor in the value of the determinant is
If ω is a non-real cube root of unity and n is not a multiple of 3, then \[∆ = \begin{vmatrix}1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^n \\ \omega^n & \omega^{2n} & 1\end{vmatrix}\]
The value of the determinant
If \[\begin{vmatrix}a & p & x \\ b & q & y \\ c & r & z\end{vmatrix} = 16\] , then the value of \[\begin{vmatrix}p + x & a + x & a + p \\ q + y & b + y & b + q \\ r + z & c + z & c + r\end{vmatrix}\] is
Solve the following system of equations by matrix method:
3x + 7y = 4
x + 2y = −1
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:
8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5
Show that the following systems of linear equations is consistent and also find their solutions:
2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3
Show that each one of the following systems of linear equation is inconsistent:
2x + 3y = 5
6x + 9y = 10
Write the value of `|(a-b, b- c, c-a),(b-c, c-a, a-b),(c-a, a-b, b-c)|`
Prove that (A–1)′ = (A′)–1, where A is an invertible matrix.
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`
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.
