मराठी

The Determinant ∣ ∣ ∣ ∣ ∣ B 2 − a B B − C B C − a C a B − a 2 a − B B 2 − a B B C − C a C − a A B − a 2 ∣ ∣ ∣ ∣ ∣ (A) a B C (B) ( C − a ) ( a − B ) (D) None of These - Mathematics

Advertisements
Advertisements

प्रश्न

The determinant  \[\begin{vmatrix}b^2 - ab & b - c & bc - ac \\ ab - a^2 & a - b & b^2 - ab \\ bc - ca & c - a & ab - a^2\end{vmatrix}\]


 

पर्याय

  • \[abc\left( b - c \right)\left( c - a \right)\left( a - b \right)\]

  • \[\left( b - c \right)\left( c - a \right)\left( a - b \right)\]

  • \[\left( a + b + c \right)\left( b - c \right)\left( c - a \right)\left( a - b \right)\]

  • none of these

MCQ
Advertisements

उत्तर

\[\begin{vmatrix}b^2 - ab & b - c & bc - ac \\ ab - a^2 & a - b & b^2 - ab \\ bc - ca & c - a & ab - a^2\end{vmatrix}\]
\[ = \begin{vmatrix}b\left( b - a \right) & b - c & c\left( b - a \right) \\ a\left( b - a \right) & a - b & b\left( b - a \right) \\ c\left( b - a \right) & c - a & a\left( b - a \right)\end{vmatrix}\]
\[ = \left( b - a \right)^2 \begin{vmatrix}b & b - c & c \\ a & a - b & b \\ c & c - a & a\end{vmatrix} \left[\text{ Taking }\left( b - a \right)\text{ common from }C_1\text{ and }C_3 \right]\]
\[ = \left( b - a \right)^2 \begin{vmatrix}0 & b - c & c \\ 0 & a - b & b \\ 0 & c - a & a\end{vmatrix} \left[\text{ Applying }C_1 \to C_1 - C_2 - C_3 \right]\]
\[ = 0\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Determinants - Exercise 6.7 [पृष्ठ ९५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 6 Determinants
Exercise 6.7 | Q 26 | पृष्ठ ९५

संबंधित प्रश्‍न

Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to ______.


Solve the system of linear equations using the matrix method.

2x + 3y + 3z = 5

x − 2y + z = −4

3x − y − 2z = 3


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:

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


Evaluate the following:

\[\begin{vmatrix}0 & x y^2 & x z^2 \\ x^2 y & 0 & y z^2 \\ x^2 z & z y^2 & 0\end{vmatrix}\]


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


\[\begin{vmatrix}0 & b^2 a & c^2 a \\ a^2 b & 0 & c^2 b \\ a^2 c & b^2 c & 0\end{vmatrix} = 2 a^3 b^3 c^3\]


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


Prove the following identities:

\[\begin{vmatrix}y + z & z & y \\ z & z + x & x \\ y & x & x + y\end{vmatrix} = 4xyz\]


​Solve the following determinant equation:

\[\begin{vmatrix}3x - 8 & 3 & 3 \\ 3 & 3x - 8 & 3 \\ 3 & 3 & 3x - 8\end{vmatrix} = 0\]

 


​Solve the following determinant equation:

\[\begin{vmatrix}x + 1 & 3 & 5 \\ 2 & x + 2 & 5 \\ 2 & 3 & x + 4\end{vmatrix} = 0\]

 


​Solve the following determinant equation:

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

 


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.

 


If a, b, c are real numbers such that
\[\begin{vmatrix}b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a\end{vmatrix} = 0\] , then show that either
\[a + b + c = 0 \text{ or, } a = b = c\]


Using determinants show that the following points are collinear:

(3, −2), (8, 8) and (5, 2)


Using determinants show that the following points are collinear:

(2, 3), (−1, −2) and (5, 8)


Prove that :

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

 


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 :

\[\begin{vmatrix}\left( a + 1 \right) \left( a + 2 \right) & a + 2 & 1 \\ \left( a + 2 \right) \left( a + 3 \right) & a + 3 & 1 \\ \left( a + 3 \right) \left( a + 4 \right) & a + 4 & 1\end{vmatrix} = - 2\]

 


Prove that :

\[\begin{vmatrix}x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4\end{vmatrix} = 16 \left( 3x + 4 \right)\]

3x − y + 2z = 3
2x + y + 3z = 5
x − 2y − z = 1


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


If \[A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}\text{ and B} = \begin{bmatrix}1 & - 4 \\ 3 & - 2\end{bmatrix},\text{ find }|AB|\]


If \[A = \begin{bmatrix}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{bmatrix}\]. Write the cofactor of the element a32.


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

 


Using the factor theorem it is found that a + bb + c and c + a are three factors of the determinant 

\[\begin{vmatrix}- 2a & a + b & a + c \\ b + a & - 2b & b + c \\ c + a & c + b & - 2c\end{vmatrix}\]
The other factor in the value of the determinant is


The value of the determinant \[\begin{vmatrix}x & x + y & x + 2y \\ x + 2y & x & x + y \\ x + y & x + 2y & x\end{vmatrix}\] is 



Solve the following system of equations by matrix method:
3x + 4y − 5 = 0
x − y + 3 = 0


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


3x − y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0


The system of equation x + y + z = 2, 3x − y + 2z = 6 and 3x + y + z = −18 has


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


Show that  \[\begin{vmatrix}y + z & x & y \\ z + x & z & x \\ x + y & y & z\end{vmatrix} = \left( x + y + z \right) \left( x - z \right)^2\]

 

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


On her birthday Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got ₹ 10 more. However, if there were 16 children more, everyone would have got ₹ 10 less. Using the matrix method, find the number of children and the amount distributed by Seema. What values are reflected by Seema’s decision?


If `|(2x, 5),(8, x)| = |(6, 5),(8, 3)|`, then find x


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


`abs ((2"xy", "x"^2, "y"^2),("x"^2, "y"^2, 2"xy"),("y"^2, 2"xy", "x"^2)) =` ____________.


What is the nature of the given system of equations

`{:(x + 2y = 2),(2x + 3y = 3):}`


Let P = `[(-30, 20, 56),(90, 140, 112),(120, 60, 14)]` and A = `[(2, 7, ω^2),(-1, -ω, 1),(0, -ω, -ω + 1)]` where ω = `(-1 + isqrt(3))/2`, and I3 be the identity matrix of order 3. If the determinant of the matrix (P–1AP – I3)2 is αω2, then the value of α is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×