English

Find the Value of the Determinant ⎡ ⎢ ⎣ 101 102 103 104 105 106 107 108 109 ⎤ ⎥ ⎦

Advertisements
Advertisements

Question

Find the value of the determinant 
\[\begin{bmatrix}101 & 102 & 103 \\ 104 & 105 & 106 \\ 107 & 108 & 109\end{bmatrix}\]

 

Advertisements

Solution

\[\text{ Let }\Delta = \begin{vmatrix} 101 & 102 & 103\\104 & 105 & 106 \\107 & 108 & 109 \end{vmatrix}\] 
\[ \Delta = \begin{vmatrix} 101 & 1 & 2\\104 & 1 & 2\\107 & 1 & 2 \end{vmatrix} \left[\text{ Applying }C_2 \to C_2 - C_1\text{ and }C_3 \to C_3 - C_1 \right]\] 
\[ = 2\begin{vmatrix} 101 & 1 & 1\\104 & 1 & 1\\107 & 1 & 1 \end{vmatrix}\] 
\[ = 0 \] 
Since two columns are identitical, the value of the determinant is zero . 

\[ \Rightarrow \Delta = \begin{vmatrix} 101 & 102 & 103\\104 & 105 & 106 \\107 & 108 & 109 \end{vmatrix} = 0\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Determinants - Exercise 6.6 [Page 90]

APPEARS IN

RD Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 5 Determinants
Exercise 6.6 | Q 6 | Page 90

RELATED QUESTIONS

If A = `[(2,-3,5),(3,2,-4),(1,1,-2)]` find A−1. Using A−1 solve the system of equations:

2x – 3y + 5z = 11

3x + 2y – 4z = –5

x + y – 2z = –3


Find the value of x, if

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


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:

\[\begin{vmatrix}1^2 & 2^2 & 3^2 & 4^2 \\ 2^2 & 3^2 & 4^2 & 5^2 \\ 3^2 & 4^2 & 5^2 & 6^2 \\ 4^2 & 5^2 & 6^2 & 7^2\end{vmatrix}\]


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:

\[\begin{vmatrix}1 & x & x^2 \\ 1 & a & a^2 \\ 1 & b & b^2\end{vmatrix} = 0, a \neq b\]

 


​Solve the following determinant equation:
\[\begin{vmatrix}15 - 2x & 11 - 3x & 7 - x \\ 11 & 17 & 14 \\ 10 & 16 & 13\end{vmatrix} = 0\]

Using determinants, find the area of the triangle with vertices (−3, 5), (3, −6), (7, 2).


If the points (x, −2), (5, 2), (8, 8) are collinear, find x using determinants.


Prove that

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

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


3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20.


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.


If \[A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 0 \\ - 1 & 0\end{bmatrix}\] , find |AB|.

 

Evaluate \[\begin{vmatrix}4785 & 4787 \\ 4789 & 4791\end{vmatrix}\]


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


If |A| = 2, where A is 2 × 2 matrix, find |adj A|.


If \[A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\] , then for any natural number, find the value of Det(An).


If a > 0 and discriminant of ax2 + 2bx + c is negative, then
\[∆ = \begin{vmatrix}a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & 0\end{vmatrix} is\]




\[\begin{vmatrix}\log_3 512 & \log_4 3 \\ \log_3 8 & \log_4 9\end{vmatrix} \times \begin{vmatrix}\log_2 3 & \log_8 3 \\ \log_3 4 & \log_3 4\end{vmatrix}\]


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 xyare different from zero and \[\begin{vmatrix}1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z\end{vmatrix} = 0\] , then the value of x−1 + y−1 + z−1 is





Solve the following system of equations by matrix method:
5x + 7y + 2 = 0
4x + 6y + 3 = 0


Solve the following system of equations by matrix method:
3x + y = 7
5x + 3y = 12


Solve the following system of equations by matrix method:
 8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5


Show that each one of the following systems of linear equation is inconsistent:
4x − 5y − 2z = 2
5x − 4y + 2z = −2
2x + 2y + 8z = −1


If \[A = \begin{bmatrix}2 & 3 & 1 \\ 1 & 2 & 2 \\ 3 & 1 & - 1\end{bmatrix}\] , find A–1 and hence solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2y – z = 8.


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.


The sum of three numbers is 2. If twice the second number is added to the sum of first and third, the sum is 1. By adding second and third number to five times the first number, we get 6. Find the three numbers by using matrices.


x + y − 6z = 0
x − y + 2z = 0
−3x + y + 2z = 0


Solve the following for x and y: \[\begin{bmatrix}3 & - 4 \\ 9 & 2\end{bmatrix}\binom{x}{y} = \binom{10}{ 2}\]


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

 

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?


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


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 system of simultaneous linear equations kx + 2y – z = 1,  (k – 1)y – 2z = 2 and (k + 2)z = 3 have a unique solution if k equals:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×