हिंदी

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

Advertisements
Advertisements

प्रश्न

Evaluate

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

 

उत्तर

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Determinants - Exercise 6.1 [पृष्ठ १०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 6 Determinants
Exercise 6.1 | Q 5 | पृष्ठ १०

संबंधित प्रश्न

Find the value of a if `[[a-b,2a+c],[2a-b,3c+d]]=[[-1,5],[0,13]]`


Examine the consistency of the system of equations.

x + 2y = 2

2x + 3y = 3


Solve the system of linear equations using the matrix method.

x − y + z = 4

2x + y − 3z = 0

x + y + z = 2


Solve the system of linear equations using the matrix method.

x − y + 2z = 7

3x + 4y − 5z = −5

2x − y + 3z = 12


Evaluate
\[∆ = \begin{vmatrix}0 & \sin \alpha & - \cos \alpha \\ - \sin \alpha & 0 & \sin \beta \\ \cos \alpha & - \sin \beta & 0\end{vmatrix}\]


If \[A = \begin{bmatrix}2 & 5 \\ 2 & 1\end{bmatrix} \text{ and } B = \begin{bmatrix}4 & - 3 \\ 2 & 5\end{bmatrix}\] , verify that |AB| = |A| |B|.

 

Find the value of x, if

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


Evaluate the following determinant:

\[\begin{vmatrix}6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2\end{vmatrix}\]


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

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


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

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


Evaluate :

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


Evaluate :

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


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 the following identities:
\[\begin{vmatrix}x + \lambda & 2x & 2x \\ 2x & x + \lambda & 2x \\ 2x & 2x & x + \lambda\end{vmatrix} = \left( 5x + \lambda \right) \left( \lambda - x \right)^2\]


​Solve the following determinant equation:

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

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, find the equation of the line joining the points

(1, 2) and (3, 6)


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 :

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

2x + 3y = 10
x + 6y = 4


x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1


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


If \[\begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8\] , write the value of x.


The value of the determinant

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

 


Let \[\begin{vmatrix}x^2 + 3x & x - 1 & x + 3 \\ x + 1 & - 2x & x - 4 \\ x - 3 & x + 4 & 3x\end{vmatrix} = a x^4 + b x^3 + c x^2 + dx + e\] 
be an identity in x, where abcde are independent of x. Then the value of e is


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


 


Solve the following system of equations by matrix method:
2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6


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

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.


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.

 

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


If \[A = \begin{bmatrix}2 & 4 \\ 4 & 3\end{bmatrix}, X = \binom{n}{1}, B = \binom{ 8}{11}\]  and AX = B, then find n.

The system of linear equations:
x + y + z = 2
2x + y − z = 3
3x + 2y + kz = 4 has a unique solution if


Solve the following by inversion method 2x + y = 5, 3x + 5y = −3


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


`abs ((1, "a"^2 + "bc", "a"^3),(1, "b"^2 + "ca", "b"^3),(1, "c"^2 + "ab", "c"^3))`


If c < 1 and the system of equations x + y – 1 = 0, 2x – y – c = 0 and – bx+ 3by – c = 0 is consistent, then the possible real values of b are


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×