English

Evaluate ∣ ∣ ∣ ∣ 2 3 − 5 7 1 − 2 − 3 4 1 ∣ ∣ ∣ ∣ by Two Methods. - Mathematics

Advertisements
Advertisements

Question

Evaluate

\[\begin{vmatrix}2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1\end{vmatrix}\] by two methods.

 
Advertisements

Solution

Let 

\[∆ = \begin{vmatrix}2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1\end{vmatrix}\]
First method
\[∆ = \left( - 1 \right)^{1 + 1} 2\left( 1 + 8 \right) + \left( - 1 \right)^{1 + 2} 3\left( 7 - 6 \right) + \left( - 1 \right)^{1 + 3} \left( - 5 \right)\left( 28 + 3 \right)\]
\[ = 2\left( 1 + 8 \right) - 3\left( 7 - 6 \right) - 5\left( 28 + 3 \right)\]
\[ = 18 - 3 - 155\]
\[ = - 140\]
Second method is the Sarus Method, where we adjoin the first two columns to the right to get
`|[2,3,-5,2,3],[7,1,-2,7,1],[-3,4,1,-3,4]|`

=(2 x 1 x 1 + 3 x -2 x -3 -5 x 7 x 4)-(-5 x 1 x -3 + 2 x -2 x 4 + 3 x 7 x 1)

=(2 + 18 - 140) - (15 - 16 + 21)

= -120 - 20

= -140

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Determinants - Exercise 6.1 [Page 10]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.1 | Q 5 | Page 10

RELATED QUESTIONS

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.

5x + 2y = 3

3x + 2y = 5


Evaluate the following determinant:

\[\begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}\]


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:

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


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

\[\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y\end{vmatrix}\]


Evaluate :

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


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


Prove the following identities:

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


Prove the following identity:

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

 


Prove the following identity:

`|(a^3,2,a),(b^3,2,b),(c^3,2,c)| = 2(a-b) (b-c) (c-a) (a+b+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:

\[\begin{vmatrix}x + a & x & x \\ x & x + a & x \\ x & x & x + a\end{vmatrix} = 0, a \neq 0\]

 


​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:

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

 


Using determinants, find the value of k so that the points (k, 2 − 2 k), (−k + 1, 2k) and (−4 − k, 6 − 2k) may be collinear.


Prove that :

\[\begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix} = \begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{vmatrix}\]

 


Prove that :

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

2x − y = − 2
3x + 4y = 3


3x + ay = 4
2x + ay = 2, a ≠ 0


2x + 3y = 10
x + 6y = 4


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


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 
\[\begin{bmatrix}2 & 3 & 4 \\ 2x & 3x & 4x \\ 5 & 6 & 8\end{bmatrix} .\]

 


State whether the matrix 
\[\begin{bmatrix}2 & 3 \\ 6 & 4\end{bmatrix}\] is singular or non-singular.


If A = [aij] is a 3 × 3 scalar matrix such that a11 = 2, then write the value of |A|.

 

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




There are two values of a which makes the determinant  \[∆ = \begin{vmatrix}1 & - 2 & 5 \\ 2 & a & - 1 \\ 0 & 4 & 2a\end{vmatrix}\]  equal to 86. The sum of these two values is

 


Solve the following system of equations by matrix method:
3x + 4y + 2z = 8
2y − 3z = 3
x − 2y + 6z = −2


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


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.

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


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 = `[[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


The value of λ, such that the following system of equations has no solution, is

`2x - y - 2z = - 5`

`x - 2y + z = 2`

`x + y + lambdaz = 3`


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:


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×