मराठी

​Solve the Following Determinant Equation: ∣ ∣ ∣ ∣ ∣ 1 X X 3 1 B B 3 1 C C 3 ∣ ∣ ∣ ∣ ∣ = 0 , B ≠ C - Mathematics

Advertisements
Advertisements

प्रश्न

​Solve the following determinant equation:

\[\begin{vmatrix}1 & x & x^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{vmatrix} = 0, b \neq c\]

 

Advertisements

उत्तर

\[\text{ Let }∆ = \begin{vmatrix}1 & x & x^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{vmatrix}\] 

\[ = \begin{vmatrix}1 & x & x^3 \\ 0 & b - x & b^3 - x^3 \\ 1 & c & c^3\end{vmatrix} \left[\text{ Applying }R_2 \text{ to }R_2 - R_1 \right]\] 

\[ = \begin{vmatrix}1 & x & x^3 \\ 0 & b - x & b^3 - x^3 \\ 0 & c - x & c^3 - x^3\end{vmatrix} \left[\text{ Applying }R_3 \text{ to }R_3 - R_1 \right]\] 

\[ = \begin{vmatrix}1 & x & x^3 \\ 0 & x - b & x^3 - b^3 \\ 0 & x - c & x^3 - c^3\end{vmatrix}\] 

\[ = \left( x - b \right)\left( x - c \right)\begin{vmatrix}1 & x & x^2 \\ 0 & 1 & x^2 + xb + b^2 \\ 0 & 1 & x^2 + xc + c^2\end{vmatrix} \] 

\[ ∆ = \left( x - b \right)\left( x - c \right)\left( x\left( c - b \right) - b^2 + c^2 \right) = 0\] 

\[x = b, c, - \left( b + c \right)\] 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Determinants - Exercise 6.2 [पृष्ठ ६१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 6 Determinants
Exercise 6.2 | Q 52.6 | पृष्ठ ६१

संबंधित प्रश्‍न

Solve the system of linear equations using the matrix method.

x − y + z = 4

2x + y − 3z = 0

x + y + z = 2


\[∆ = \begin{vmatrix}\cos \alpha \cos \beta & \cos \alpha \sin \beta & - \sin \alpha \\ - \sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{vmatrix}\]


If A \[\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}\] , then show that |3 A| = 27 |A|.

 

Find the value of x, if

\[\begin{vmatrix}x + 1 & x - 1 \\ x - 3 & x + 2\end{vmatrix} = \begin{vmatrix}4 & - 1 \\ 1 & 3\end{vmatrix}\]


Evaluate the following determinant:

\[\begin{vmatrix}67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26\end{vmatrix}\]


Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2\end{vmatrix}\]


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:

\[\begin{vmatrix}0 & x & y \\ - x & 0 & z \\ - y & - z & 0\end{vmatrix}\]


Evaluate :

\[\begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix}\]


Evaluate the following:

\[\begin{vmatrix}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{vmatrix}\]


Prove that
\[\begin{vmatrix}- bc & b^2 + bc & c^2 + bc \\ a^2 + ac & - ac & c^2 + ac \\ a^2 + ab & b^2 + ab & - ab\end{vmatrix} = \left( ab + bc + ca \right)^3\]


If \[a, b\] and c  are all non-zero and 

\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c\end{vmatrix} =\] 0, then prove that 
\[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} +\]1
= 0

 


If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.


Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?


Prove that :

\[\begin{vmatrix}a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b\end{vmatrix} = 2 \left( a + b + c \right)^3\]

 


3x + y = 19
3x − y = 23


2x − y = − 2
3x + 4y = 3


6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8


x − y + 3z = 6
x + 3y − 3z = − 4
5x + 3y + 3z = 10


Solve each of the following system of homogeneous linear equations.
x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 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.


Write the value of the determinant 
\[\begin{bmatrix}2 & 3 & 4 \\ 2x & 3x & 4x \\ 5 & 6 & 8\end{bmatrix} .\]

 


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|.


Find the value of the determinant \[\begin{vmatrix}243 & 156 & 300 \\ 81 & 52 & 100 \\ - 3 & 0 & 4\end{vmatrix} .\]


Find the value of x from the following : \[\begin{vmatrix}x & 4 \\ 2 & 2x\end{vmatrix} = 0\]


Write the value of the determinant \[\begin{vmatrix}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x\end{vmatrix}\]


If \[D_k = \begin{vmatrix}1 & n & n \\ 2k & n^2 + n + 2 & n^2 + n \\ 2k - 1 & n^2 & n^2 + n + 2\end{vmatrix} and \sum^n_{k = 1} D_k = 48\], then n equals

 


The value of \[\begin{vmatrix}5^2 & 5^3 & 5^4 \\ 5^3 & 5^4 & 5^5 \\ 5^4 & 5^5 & 5^6\end{vmatrix}\]

 


Solve the following system of equations by matrix method:
3x + 7y = 4
x + 2y = −1


Solve the following system of equations by matrix method:
2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6


Solve the following system of equations by matrix method:

\[\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0\]

 


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:
2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3


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.


2x − y + 2z = 0
5x + 3y − z = 0
x + 5y − 5z = 0


If \[\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}2 \\ - 1 \\ 3\end{bmatrix}\], find x, y, z.

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 + = 7.


If A = `[(2, 0),(0, 1)]` and B = `[(1),(2)]`, then find the matrix X such that A−1X = B.


Let A = `[(1,sin α,1),(-sin α,1,sin α),(-1,-sin α,1)]`, where 0 ≤ α ≤ 2π, then:


The number of real value of 'x satisfying `|(x, 3x + 2, 2x - 1),(2x - 1, 4x, 3x + 1),(7x - 2, 17x + 6, 12x - 1)|` = 0 is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×