English

Evaluate : ∣ ∣ ∣ ∣ X + λ X X X X + λ X X X X + λ ∣ ∣ ∣ ∣ - Mathematics

Advertisements
Advertisements

Question

Evaluate :

\[\begin{vmatrix}x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda\end{vmatrix}\]

Advertisements

Solution

\[∆ = \begin{vmatrix}x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda\end{vmatrix}\]

\[ = \begin{vmatrix}\lambda & 0 & x \\ - \lambda & \lambda & x \\ 0 & - \lambda & x + \lambda\end{vmatrix} \left[\text{ Applying }C_1 \to C_1 - C_2 , C_2 \to C_2 - C_3 \right]\]

\[ = \begin{vmatrix}\lambda & 0 & x \\ - \lambda & 0 & 2x + \lambda \\ 0 & - \lambda & x + \lambda\end{vmatrix} \left[ \text{ Applying }R_1 \text{ to } R_2 + R_3 \right]\]

\[ = \lambda\begin{vmatrix}0 & 2x + \lambda \\ - \lambda & x + \lambda\end{vmatrix} + x\begin{vmatrix}- \lambda & 0 \\ 0 & - \lambda\end{vmatrix}\]

\[ = \lambda[\lambda(2x + \lambda)] + x \lambda^2 \]

\[ = \lambda^2 (2x + \lambda + \lambda^2 x)\]

\[ = 3 \lambda^2 x + \lambda^3 \]

\[ = \lambda^2 (3x + \lambda )\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Determinants - Exercise 6.2 [Page 58]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.2 | Q 5 | Page 58

RELATED QUESTIONS

If `|[2x,5],[8,x]|=|[6,-2],[7,3]|`, write the value of x.


Solve the system of linear equations using the matrix method.

5x + 2y = 3

3x + 2y = 5


Solve the system of linear equations using the matrix method.

2x + y + z = 1

x – 2y – z = `3/2`

3y – 5z = 9


Evaluate

\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}^2 .\]


Find the value of x, if

\[\begin{vmatrix}3x & 7 \\ 2 & 4\end{vmatrix} = 10\] , find the value of x.


Find the value of x, if

\[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & 5 \\ 8 & 3\end{vmatrix}\]


Evaluate :

\[\begin{vmatrix}a & b + c & a^2 \\ b & c + a & b^2 \\ c & a + b & c^2\end{vmatrix}\]


\[\begin{vmatrix}b + c & a & a \\ b & c + a & b \\ c & c & a + b\end{vmatrix} = 4abc\]


\[\begin{vmatrix}b^2 + c^2 & ab & ac \\ ba & c^2 + a^2 & bc \\ ca & cb & a^2 + b^2\end{vmatrix} = 4 a^2 b^2 c^2\]


​Solve the following determinant equation:

\[\begin{vmatrix}1 & 1 & x \\ p + 1 & p + 1 & p + x \\ 3 & x + 1 & x + 2\end{vmatrix} = 0\]

Show that
`|(x-3,x-4,x-alpha),(x-2,x-3,x-beta),(x-1,x-2,x-gamma)|=0`, where α, β, γ are in A.P.

 


Find the area of the triangle with vertice at the point:

 (−1, −8), (−2, −3) and (3, 2)


Using determinants show that the following points are collinear:

(2, 3), (−1, −2) and (5, 8)


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


Prove that :

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

 


Prove that :

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


Prove that :

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

Prove that :

\[\begin{vmatrix}1 & 1 + p & 1 + p + q \\ 2 & 3 + 2p & 4 + 3p + 2q \\ 3 & 6 + 3p & 10 + 6p + 3q\end{vmatrix} = 1\]

 


If w is an imaginary cube root of unity, find the value of \[\begin{vmatrix}1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w\end{vmatrix}\]


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


Find the maximum value of \[\begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos \theta\end{vmatrix}\]


If a, b, c are in A.P., then the determinant
\[\begin{vmatrix}x + 2 & x + 3 & x + 2a \\ x + 3 & x + 4 & x + 2b \\ x + 4 & x + 5 & x + 2c\end{vmatrix}\]


If \[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}\] , then x = 

 


Solve the following system of equations by matrix method:

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

 


If \[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\] , find A−1. Using A−1, solve the system of linear equations  x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7.

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 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.

 

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.


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

The number of solutions of the system of equations
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
is


Let a, b, c be positive real numbers. The following system of equations in x, y and z 

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1, \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1, - \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \text { has }\]
(a) no solution
(b) unique solution
(c) infinitely many solutions
(d) finitely many solutions

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.


Show that if the determinant ∆ = `|(3, -2, sin3theta),(-7, 8, cos2theta),(-11, 14, 2)|` = 0, then sinθ = 0 or `1/2`.


If `|(2x, 5),(8, x)| = |(6, -2),(7, 3)|`, then value of x is ______.


If the system of equations x + ky - z = 0, 3x - ky - z = 0 & x - 3y + z = 0 has non-zero solution, then k is equal to ____________.


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


The number of values of k for which the linear equations 4x + ky + 2z = 0, kx + 4y + z = 0 and 2x + 2y + z = 0 possess a non-zero solution is


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×