Advertisements
Advertisements
प्रश्न
\[If ∆ = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix}, ∆_1 = \begin{vmatrix}1 & 1 & 1 \\ yz & zx & xy \\ x & y & z\end{vmatrix},\text{ then prove that }∆ + ∆_1 = 0 .\]
Advertisements
उत्तर
\[∆ + ∆_1 = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix} + \begin{vmatrix}1 & 1 & 1 \\ yz & zx & xy \\ x & y & z\end{vmatrix}\]
\[ = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix} + \begin{vmatrix}1 & yz & x \\ 1 & zx & y \\ 1 & xy & z\end{vmatrix} \left[\text{ Interchanging rows and coloumns in }∆_1 \right]\]
\[ = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix} - \begin{vmatrix}1 & x & yz \\ 1 & y & zx \\ 1 & z & xy\end{vmatrix} \left[\text{ Applying }C_2 \leftrightarrow C_3\text{ in }∆_1 \right]\]
\[ = \begin{vmatrix}1 & x & x^2 \\ 0 & y - x & y^2 - x^2 \\ 0 & z - x & z^2 - x^2\end{vmatrix} - \begin{vmatrix}1 & x & yz \\ 0 & y - x & zx - yz \\ 0 & z - x & xy - yz\end{vmatrix} \left[ \text{ Applying }R_2 \to R_2 - R_1\text{ and }R_3 \to R_3 - R_1 \right]\]
\[ = \left( y - x \right)\left( z - x \right)\begin{vmatrix}1 & x & x^2 \\ 0 & 1 & y + x \\ 0 & 1 & z + x\end{vmatrix} - \left( y - x \right)\left( z - x \right)\begin{vmatrix}1 & x & yz \\ 0 & 1 & - z \\ 0 & 1 & - y\end{vmatrix} \left[\text{ Taking }\left( y - x \right) \text{ common from }R_2\text{ and }\left( z - x \right)\text{ common from }R_3 \right]\]
\[ = \left( y - x \right)\left( z - x \right)\left( z + x - y - x \right) - \left( y - x \right)\left( z - x \right)\left( - y + z \right) \left[\text{ Expanding along first column }\right]\]
\[ = \left( y - x \right)\left( z - x \right)\left( z - y \right)\left( 1 - 1 \right)\]
\[ = 0\]
\[ \therefore ∆ + ∆_1 = 0 .\]
APPEARS IN
संबंधित प्रश्न
Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
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
Evaluate the following determinant:
\[\begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}\]
Evaluate
\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}^2 .\]
Find the value of x, if
\[\begin{vmatrix}2 & 4 \\ 5 & 1\end{vmatrix} = \begin{vmatrix}2x & 4 \\ 6 & x\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}a & b + c & a^2 \\ b & c + a & b^2 \\ c & a + b & c^2\end{vmatrix}\]
Prove that:
`[(a, b, c),(a - b, b - c, c - a),(b + c, c + a, a + b)] = a^3 + b^3 + c^3 -3abc`
\[\begin{vmatrix}b^2 + c^2 & ab & ac \\ ba & c^2 + a^2 & bc \\ ca & cb & a^2 + b^2\end{vmatrix} = 4 a^2 b^2 c^2\]
Without expanding, prove that
\[\begin{vmatrix}a & b & c \\ x & y & z \\ p & q & r\end{vmatrix} = \begin{vmatrix}x & y & z \\ p & q & r \\ a & b & c\end{vmatrix} = \begin{vmatrix}y & b & q \\ x & a & p \\ z & c & r\end{vmatrix}\]
Find the area of the triangle with vertice at the point:
(2, 7), (1, 1) and (10, 8)
Using determinants show that the following points are collinear:
(5, 5), (−5, 1) and (10, 7)
Using determinants prove that the points (a, b), (a', b') and (a − a', b − b') are collinear if ab' = a'b.
Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?
Prove that :
Prove that :
Prove that
Prove that
2x − y = − 2
3x + 4y = 3
x+ y = 5
y + z = 3
x + z = 4
x + 2y = 5
3x + 6y = 15
A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission
| Month | Sale of units | Total commission drawn (in Rs) |
||
| A | B | C | ||
| Jan | 90 | 100 | 20 | 800 |
| Feb | 130 | 50 | 40 | 900 |
| March | 60 | 100 | 30 | 850 |
Find out the rates of commission on items A, B and C by using determinant method.
Solve each of the following system of homogeneous linear equations.
x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0
If \[A = \begin{bmatrix}0 & i \\ i & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\] , find the value of |A| + |B|.
If I3 denotes identity matrix of order 3 × 3, write the value of its determinant.
Find the maximum value of \[\begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos \theta\end{vmatrix}\]
If a, b, c are distinct, then the value of x satisfying \[\begin{vmatrix}0 & x^2 - a & x^3 - b \\ x^2 + a & 0 & x^2 + c \\ x^4 + b & x - c & 0\end{vmatrix} = 0\text{ is }\]
If ω is a non-real cube root of unity and n is not a multiple of 3, then \[∆ = \begin{vmatrix}1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^n \\ \omega^n & \omega^{2n} & 1\end{vmatrix}\]
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:
x + y + z = 6
x + 2z = 7
3x + y + z = 12
Show that the following systems of linear equations is consistent and also find their solutions:
5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
x + y + z = 0
x − y − 5z = 0
x + 2y + 4z = 0
x + y = 1
x + z = − 6
x − y − 2z = 3
If A = `[(1,-1,0),(2,3,4),(0,1,2)]` and B = `[(2,2,-4),(-4,2,-4),(2,-1,5)]`, then:
If the system of equations x + λy + 2 = 0, λx + y – 2 = 0, λx + λy + 3 = 0 is consistent, then
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:
