Advertisements
Advertisements
प्रश्न
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin^2 23^\circ & \sin^2 67^\circ & \cos180^\circ \\ - \sin^2 67^\circ & - \sin^2 23^\circ & \cos^2 180^\circ \\ \cos180^\circ & \sin^2 23^\circ & \sin^2 67^\circ\end{vmatrix}\]
Advertisements
उत्तर
\[\begin{vmatrix}\sin^2 23^\circ & \sin^2 67^\circ & \cos180^\circ \\ - \sin^2 67^\circ & - \sin^2 23^\circ & \cos^2 180^\circ \\ \cos180^\circ & \sin^2 23^\circ & \sin^2 67^\circ\end{vmatrix}\]
\[ = \begin{vmatrix}\sin^2 23^\circ & \sin^2 \left( 90 - 23 \right)^\circ & - 1 \\ - \sin^2 \left( 90 - 23 \right)^\circ & - \sin^2 23^\circ & 1 \\ - 1 & \sin^2 23^\circ & \sin^2 \left( 90 - 23 \right)^\circ\end{vmatrix}\]
\[ = \begin{vmatrix}\sin^2 23^\circ & \cos^2 23^\circ & - 1 \\ - \cos^2 23^\circ & - \sin^2 23^\circ & 1 \\ - 1 & \sin^2 23^\circ & \cos^2 23^\circ\end{vmatrix}\]
\[ = \begin{vmatrix}\sin^2 23^\circ + \cos^2 23^\circ & \cos^2 23^\circ & - 1 \\ - \cos^2 23^\circ - \sin^2 23^\circ & - \sin^2 23^\circ & 1 \\ - 1 + \sin^2 23^\circ & \sin^2 23^\circ & \cos^2 23^\circ\end{vmatrix} \left[ \text{ Applying }C_1 \to C_1 + C_2 \right]\]
\[ = \begin{vmatrix}1 & 1 & - 1 \\ - 1 & - \sin^2 23^\circ & 1 \\ - \cos^2 23^\circ & \sin^2 23^\circ & \cos^2 23^\circ\end{vmatrix}\]
\[ = \left( - 1 \right)\begin{vmatrix}- 1 & 1 & - 1 \\ 1 & - \sin^2 23^\circ & 1 \\ \cos^2 23^\circ & \sin^2 23^\circ & \cos^2 23^\circ\end{vmatrix}\]
\[ = 0\]
APPEARS IN
संबंधित प्रश्न
Solve the system of linear equations using the matrix method.
2x – y = –2
3x + 4y = 3
Evaluate the following determinant:
\[\begin{vmatrix}1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y\end{vmatrix}\]
\[\begin{vmatrix}b^2 + c^2 & ab & ac \\ ba & c^2 + a^2 & bc \\ ca & cb & a^2 + b^2\end{vmatrix} = 4 a^2 b^2 c^2\]
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)`
Without expanding, prove that
\[\begin{vmatrix}a & b & c \\ x & y & z \\ p & q & r\end{vmatrix} = \begin{vmatrix}x & y & z \\ p & q & r \\ a & b & c\end{vmatrix} = \begin{vmatrix}y & b & q \\ x & a & p \\ z & c & r\end{vmatrix}\]
Using determinants show that the following points are collinear:
(5, 5), (−5, 1) and (10, 7)
If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.
Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?
Using determinants, find the equation of the line joining the points
(3, 1) and (9, 3)
x − 2y = 4
−3x + 5y = −7
Prove that :
Given: x + 2y = 1
3x + y = 4
3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11
x + y + z + 1 = 0
ax + by + cz + d = 0
a2x + b2y + x2z + d2 = 0
x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0
Write the value of the determinant
\[\begin{bmatrix}2 & 3 & 4 \\ 2x & 3x & 4x \\ 5 & 6 & 8\end{bmatrix} .\]
Write the value of the determinant
Write the value of \[\begin{vmatrix}a + ib & c + id \\ - c + id & a - ib\end{vmatrix} .\]
Find the value of x from the following : \[\begin{vmatrix}x & 4 \\ 2 & 2x\end{vmatrix} = 0\]
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 a, b, c, d, e are independent of x. Then the value of e is
Let \[A = \begin{bmatrix}1 & \sin \theta & 1 \\ - \sin \theta & 1 & \sin \theta \\ - 1 & - \sin \theta & 1\end{bmatrix},\text{ where 0 }\leq \theta \leq 2\pi . \text{ Then,}\]
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 + 4y + 7z = 14
2x − y + 3z = 4
x + 2y − 3z = 0
Given \[A = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}, B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}\] , find BA and use this to solve the system of equations y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17
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.
The system of linear equations:
x + y + z = 2
2x + y − z = 3
3x + 2y + kz = 4 has a unique solution if
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
Find the inverse of the following matrix, using elementary transformations:
`A= [[2 , 3 , 1 ],[2 , 4 , 1],[3 , 7 ,2]]`
Transform `[(1, 2, 4),(3, -1, 5),(2, 4, 6)]` into an upper triangular matrix by using suitable row transformations
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 ____________.
Solve the following system of equations x − y + z = 4, x − 2y + 2z = 9 and 2x + y + 3z = 1.
If the system of linear equations
2x + y – z = 7
x – 3y + 2z = 1
x + 4y + δz = k, where δ, k ∈ R has infinitely many solutions, then δ + k is equal to ______.
Let A = `[(i, -i),(-i, i)], i = sqrt(-1)`. Then, the system of linear equations `A^8[(x),(y)] = [(8),(64)]` has ______.
