मराठी

If D K = ∣ ∣ ∣ ∣ ∣ 1 N N 2 K N 2 + N + 2 N 2 + N 2 K − 1 N 2 N 2 + N + 2 ∣ ∣ ∣ ∣ ∣ a N D N ∑ K = 1 D K = 48 , Then N Equals (A) 4 (B) 6 (C) 8 (D) None of These - Mathematics

Advertisements
Advertisements

प्रश्न

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

 

पर्याय

  • 4

  • 6

  • 8

  •  none of these

MCQ
Advertisements

उत्तर

(a) 4
\[D_k = \begin{vmatrix} 1 & n & n\\ 2k & n^2 + n + 2 & n^2 + n\\2k - 1 & n^2 & n^2 + n + 2 \end{vmatrix}\]
\[ = \begin{vmatrix} 1 & n & n\\ 1 & n + 2 & - 2\\2k - 1 & n^2 & n^2 + n + 2 \end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_3 \right]\]
\[ = \begin{vmatrix} 1 & n & n\\ 0 & 2 & - 2 - n\\2k - 1 & n^2 & n^2 + n + 2 \end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\]
\[ \sum\nolimits_{k = 1}^n D_k = \begin{vmatrix} 1 & n & n\\ 0 & 2 & - 2 - n\\ 1 & n^2 & n^2 + n + 2 \end{vmatrix} + \begin{vmatrix} 1 & n & n\\ 0 & 2 & - 2 - n\\ 3 & n^2 & n^2 + n + 2 \end{vmatrix} + . . . + \begin{vmatrix} 1 & n & n\\ 0 & 2 & - 2 - n\\ n & n^2 & n^2 + n + 2 \end{vmatrix}\]
\[ \sum\nolimits_{k = 1}^n D_k = 1\left( 2\left( n^2 + n + 2 \right) + \left( 2 + n \right) n^2 \right) + 1\left( n\left( - 2 - n \right) - 2n \right) + 1\left( 2\left( n^2 + n + 2 \right) + \left( 2 + n \right) n^2 \right) + 2\left( n\left( - 2 - n \right) - 2n \right) + . . . + 1\left( 2\left( n^2 + n + 2 \right) + \left( 2 + n \right) n^2 \right) + n\left( n\left( - 2 - n \right) - 2n \right)\]
\[ \sum\nolimits_{k = 1}^n D_k = n\left( 2\left( n^2 + n + 2 \right) + \left( 2 + n \right) n^2 \right) + \left( n\left( - 2 - n \right) - 2n \right)\left( 1 + 3 + 5 + 7 + . . . + n \right)\]
\[ \sum\nolimits_{k = 1}^n D_k = n\left( 2\left( n^2 + n + 2 \right) + \left( 2 + n \right) n^2 \right) + \left( n\left( - 2 - n \right) - 2n \right)\left( n^2 \right)\]
\[ \sum\nolimits_{k = 1}^n D_k = 2 n^2 + 4n\]
\[ \Rightarrow 2 n^2 + 4n = 48\]
\[ \Rightarrow \left( n - 6 \right)\left( n - 4 \right) = 0\]
\[ \Rightarrow n = 4\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Determinants - Exercise 6.7 [पृष्ठ ९३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 6 Determinants
Exercise 6.7 | Q 8 | पृष्ठ ९३

संबंधित प्रश्‍न

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 = 4

7x + 3y = 5


Solve the system of linear equations using the matrix method.

4x – 3y = 3

3x – 5y = 7


Solve the system of linear equations using the matrix method.

x − y + 2z = 7

3x + 4y − 5z = −5

2x − y + 3z = 12


Find the value of x, if

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


For what value of x the matrix A is singular? 

\[A = \begin{bmatrix}x - 1 & 1 & 1 \\ 1 & x - 1 & 1 \\ 1 & 1 & x - 1\end{bmatrix}\]


Evaluate :

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


Evaluate the following:

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


Prove that

\[\begin{vmatrix}\frac{a^2 + b^2}{c} & c & c \\ a & \frac{b^2 + c^2}{a} & a \\ b & b & \frac{c^2 + a^2}{b}\end{vmatrix} = 4abc\]


Prove the following identities:

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


\[If \begin{vmatrix}p & b & c \\ a & q & c \\ a & b & r\end{vmatrix} = 0,\text{ find the value of }\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}, p \neq a, q \neq b, r \neq c .\]

 


Show that x = 2 is a root of the equation

\[\begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 3 & 2x & x + 2\end{vmatrix} = 0\]  and solve it completely.
 

 


​Solve the following determinant equation:

\[\begin{vmatrix}3 & - 2 & \sin\left( 3\theta \right) \\ - 7 & 8 & \cos\left( 2\theta \right) \\ - 11 & 14 & 2\end{vmatrix} = 0\]

 


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
\[a + b + c = 0 \text{ or, } a = b = c\]


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

(3, 8), (−4, 2) and (5, −1)


Using determinants show that the following points are collinear:

(2, 3), (−1, −2) and (5, 8)


Prove that :

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

 


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 :

\[\begin{vmatrix}1 & a^2 + bc & a^3 \\ 1 & b^2 + ca & b^3 \\ 1 & c^2 + ab & c^3\end{vmatrix} = - \left( a - b \right) \left( b - c \right) \left( c - a \right) \left( a^2 + b^2 + c^2 \right)\]

 


3x + ay = 4
2x + ay = 2, a ≠ 0


5x − 7y + z = 11
6x − 8y − z = 15
3x + 2y − 6z = 7


x + y + z + 1 = 0
ax + by + cz + d = 0
a2x + b2y + x2z + d2 = 0


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


An automobile company uses three types of steel S1S2 and S3 for producing three types of cars C1C2and C3. Steel requirements (in tons) for each type of cars are given below : 

  Cars
C1
C2 C3
Steel S1 2 3 4
S2 1 1 2
S3 3 2 1

Using Cramer's rule, find the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively.


The value of the determinant

\[\begin{vmatrix}a^2 & a & 1 \\ \cos nx & \cos \left( n + 1 \right) x & \cos \left( n + 2 \right) x \\ \sin nx & \sin \left( n + 1 \right) x & \sin \left( n + 2 \right) x\end{vmatrix}\text{ is independent of}\]

 


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


 


A shopkeeper has 3 varieties of pens 'A', 'B' and 'C'. Meenu purchased 1 pen of each variety for a total of Rs 21. Jeevan purchased 4 pens of 'A' variety 3 pens of 'B' variety and 2 pens of 'C' variety for Rs 60. While Shikha purchased 6 pens of 'A' variety, 2 pens of 'B' variety and 3 pens of 'C' variety for Rs 70. Using matrix method, find cost of each variety of pen.

 

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

 

Solve the following system of equations by using inversion method

x + y = 1, y + z = `5/3`, z + x = `4/3`


The cost of 4 dozen pencils, 3 dozen pens and 2 dozen erasers is ₹ 60. The cost of 2 dozen pencils, 4 dozen pens and 6 dozen erasers is ₹ 90. Whereas the cost of 6 dozen pencils, 2 dozen pens and 3 dozen erasers is ₹ 70. Find the cost of each item per dozen by using matrices


Prove that (A–1)′ = (A′)–1, where A is an invertible matrix.


If ` abs((1 + "a"^2 "x", (1 + "b"^2)"x", (1 + "c"^2)"x"),((1 + "a"^2) "x", 1 + "b"^2 "x", (1 + "c"^2) "x"), ((1 + "a"^2) "x", (1 + "b"^2) "x", 1 + "c"^2 "x"))`, then f(x) is apolynomial of degree ____________.


If the system of equations 2x + 3y + 5 = 0, x + ky + 5 = 0, kx - 12y - 14 = 0 has non-trivial solution, then the value of k is ____________.


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


If A = `[(1,-1,0),(2,3,4),(0,1,2)]` and B = `[(2,2,-4),(-4,2,-4),(2,-1,5)]`, then:


In system of equations, if inverse of matrix of coefficients A is multiplied by right side constant B vector then resultant will be?


If `|(x + 1, x + 2, x + a),(x + 2, x + 3, x + b),(x + 3, x + 4, x + c)|` = 0, then a, b, care in


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 ab + bc + ca equals ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×