Advertisements
Advertisements
प्रश्न
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin^2 A & \cot A & 1 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C & \cot C & 1\end{vmatrix}, where A, B, C \text{ are the angles of }∆ ABC .\]
उत्तर
\[\begin{vmatrix}\sin^2 A & \cot A & 1 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C & \cot C & 1\end{vmatrix}\]
\[ = \begin{vmatrix}\sin^2 A - \sin^2 B & \cot A - \cot B & 0 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C - \sin^2 B & \cot C - \cot B & 0\end{vmatrix} \left[ \text{ Applying } R_1 \to R_1 - R_2 \text{ and }R_3 \to R_3 - R_2 \right]\]
\[ = \begin{vmatrix}\sin\left( A + B \right)\sin\left( A - B \right) & \frac{\cos A\sin B - \cos B\sin A}{\sin A\sin B} & 0 \\ \sin^2 B & \cot B & 1 \\ \sin\left( C + B \right)\sin\left( C - B \right) & \frac{\cos C\sin B - \cos B\sin C}{\sin B\sin C} & 0\end{vmatrix}\]
\[ = \begin{vmatrix}\sin\left( \pi - C \right)\sin\left( A - B \right) & \frac{- \sin\left( A - B \right)}{\sin A\sin B} & 0 \\ \sin^2 B & cot B & 1 \\ \sin\left( \pi - A \right)\sin\left( C - B \right) & \frac{- \sin\left( C - B \right)}{\sin B\sin C} & 0\end{vmatrix} \left[ \because A + B + C = \pi \right]\]
\[ = \begin{vmatrix}\sin C\sin\left( A - B \right) & \frac{- \sin\left( A - B \right)}{\sin A\sin B} & 0 \\ \sin^2 B & \frac{\cos B}{\sin B} & 1 \\ \sin A\sin\left( C - B \right) & \frac{- \sin\left( C - B \right)}{\sin B\sin C} & 0\end{vmatrix}\]
\[ = \frac{\sin\left( A - B \right)\sin\left( C - B \right)}{\sin B}\begin{vmatrix}\sin C & \frac{- 1}{\sin A} & 0 \\ \sin^2 B & \cos B & 1 \\ \sin A & \frac{- 1}{\sin C} & 0\end{vmatrix}\]
\[ = \frac{\sin\left( A - B \right)\sin\left( C - B \right)}{\sin B\sin A\sin C}\begin{vmatrix}\sin C\sin A & - 1 & 0 \\ \sin^2 B & \cos B & 1 \\ \sin A\sin C & - 1 & 0\end{vmatrix} \left[ \text{ Applying }R_1 \to \sin A R_1\text{ and }R_3 \to \sin C R_3 \right]\]
\[ = \frac{\sin\left( A - B \right)\sin\left( C - B \right)}{\sin B\sin A\sin C}\begin{vmatrix}0 & 0 & 0 \\ \sin^2 B & \cos B & 1 \\ \sin A\sin C & - 1 & 0\end{vmatrix} \left[ \text{ Applying }R_1 \to R_1 - R_3 \right]\]
\[ = 0\]
APPEARS IN
संबंधित प्रश्न
If `|[2x,5],[8,x]|=|[6,-2],[7,3]|`, write the value of x.
Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to ______.
Evaluate the following determinant:
\[\begin{vmatrix}x & - 7 \\ x & 5x + 1\end{vmatrix}\]
Find the value of x, if
\[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & 5 \\ 8 & 3\end{vmatrix}\]
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:
\[\begin{vmatrix}1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2\end{vmatrix}\]
Evaluate the following:
\[\begin{vmatrix}0 & x y^2 & x z^2 \\ x^2 y & 0 & y z^2 \\ x^2 z & z y^2 & 0\end{vmatrix}\]
\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & a \\ 1 & 1 & 1 + a\end{vmatrix} = a^3 + 3 a^2\]
Solve the following determinant equation:
Find the area of the triangle with vertice at the point:
(3, 8), (−4, 2) and (5, −1)
Find the area of the triangle with vertice at the point:
(2, 7), (1, 1) and (10, 8)
Find the area of the triangle with vertice at the point:
(−1, −8), (−2, −3) and (3, 2)
Using determinants show that the following points are collinear:
(3, −2), (8, 8) and (5, 2)
x − 2y = 4
−3x + 5y = −7
Prove that :
Prove that :
Prove that :
Prove that :
3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20.
If A = [aij] is a 3 × 3 scalar matrix such that a11 = 2, then write the value of |A|.
Write the value of the determinant \[\begin{vmatrix}2 & - 3 & 5 \\ 4 & - 6 & 10 \\ 6 & - 9 & 15\end{vmatrix} .\]
For what value of x is the matrix \[\begin{bmatrix}6 - x & 4 \\ 3 - x & 1\end{bmatrix}\] singular?
The value of the determinant
If ω is a non-real cube root of unity and n is not a multiple of 3, then \[∆ = \begin{vmatrix}1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^n \\ \omega^n & \omega^{2n} & 1\end{vmatrix}\]
If \[\begin{vmatrix}a & p & x \\ b & q & y \\ c & r & z\end{vmatrix} = 16\] , then the value of \[\begin{vmatrix}p + x & a + x & a + p \\ q + y & b + y & b + q \\ r + z & c + z & c + r\end{vmatrix}\] is
Solve the following system of equations by matrix method:
3x + 7y = 4
x + 2y = −1
Solve the following system of equations by matrix method:
8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5
Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award ₹x each, ₹y each and ₹z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹1,600. School B wants to spend ₹2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is ₹900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.
If A = `[(2, 0),(0, 1)]` and B = `[(1),(2)]`, then find the matrix X such that A−1X = B.
Solve the following by inversion method 2x + y = 5, 3x + 5y = −3
If A = `[(1, -1, 2),(3, 0, -2),(1, 0, 3)]`, verify that A(adj A) = (adj A)A
The existence of unique solution of the system of linear equations x + y + z = a, 5x – y + bz = 10, 2x + 3y – z = 6 depends on
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`
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
If the following equations
x + y – 3 = 0
(1 + λ)x + (2 + λ)y – 8 = 0
x – (1 + λ)y + (2 + λ) = 0
are consistent then the value of λ can be ______.
