Advertisements
Advertisements
प्रश्न
Write the value of the determinant
Advertisements
उत्तर
\[\text{ Let }∆ = \begin{vmatrix}a & 1 & b + c \\ b & 1 & c + a \\ c & 1 & a + b\end{vmatrix} \]
\[ = \begin{vmatrix} a + b + c & 1 & b + c\\a + b + c & 1 & c + a\\a + b + c & 1 & a + b \end{vmatrix} \left[\text{ Applying }C_1 \to C_1 + C_3 \right]\]
\[ = a + b + c \begin{vmatrix} 1 & 1& b + c\\1 & 1 & c + a\\1 & 1 & a + b \end{vmatrix}\]
\[ = \left( a + b + c \right) \times 0\]
\[ = 0\]
APPEARS IN
संबंधित प्रश्न
Examine the consistency of the system of equations.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Examine the consistency of the system of equations.
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
If A \[\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}\] , then show that |3 A| = 27 |A|.
Find the value of x, if
\[\begin{vmatrix}3x & 7 \\ 2 & 4\end{vmatrix} = 10\] , find the value of x.
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y\end{vmatrix}\]
\[\begin{vmatrix}b + c & a & a \\ b & c + a & b \\ c & c & a + b\end{vmatrix} = 4abc\]
Solve the following determinant equation:
Find the area of the triangle with vertice at the point:
(3, 8), (−4, 2) and (5, −1)
Using determinants, find the area of the triangle with vertices (−3, 5), (3, −6), (7, 2).
If the points (3, −2), (x, 2), (8, 8) are collinear, find x using determinant.
Find values of k, if area of triangle is 4 square units whose vertices are
(k, 0), (4, 0), (0, 2)
Find values of k, if area of triangle is 4 square units whose vertices are
(−2, 0), (0, 4), (0, k)
2x − y = 1
7x − 2y = −7
Prove that :
Prove that :
x+ y = 5
y + z = 3
x + z = 4
2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11
3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20.
If \[\begin{vmatrix}2x & x + 3 \\ 2\left( x + 1 \right) & x + 1\end{vmatrix} = \begin{vmatrix}1 & 5 \\ 3 & 3\end{vmatrix}\], then write the value of x.
Solve the following system of equations by matrix method:
5x + 2y = 3
3x + 2y = 5
Solve the following system of equations by matrix method:
x − y + z = 2
2x − y = 0
2y − z = 1
Solve the following system of equations by matrix method:
Show that the following systems of linear equations is consistent and also find their solutions:
2x + 3y = 5
6x + 9y = 15
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.
x + y − 6z = 0
x − y + 2z = 0
−3x + y + 2z = 0
The existence of the unique solution of the system of equations:
x + y + z = λ
5x − y + µz = 10
2x + 3y − z = 6
depends on
The system of equations:
x + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ
has a unique solution, if
(a) λ = 5, µ = 13
(b) λ ≠ 5
(c) λ = 5, µ ≠ 13
(d) µ ≠ 13
Show that \[\begin{vmatrix}y + z & x & y \\ z + x & z & x \\ x + y & y & z\end{vmatrix} = \left( x + y + z \right) \left( x - z \right)^2\]
Transform `[(1, 2, 4),(3, -1, 5),(2, 4, 6)]` into an upper triangular matrix by using suitable row transformations
Solve the following by inversion method 2x + y = 5, 3x + 5y = −3
`abs ((1, "a"^2 + "bc", "a"^3),(1, "b"^2 + "ca", "b"^3),(1, "c"^2 + "ab", "c"^3))`
Solve the following system of equations x − y + z = 4, x − 2y + 2z = 9 and 2x + y + 3z = 1.
`abs ((("b" + "c"^2), "a"^2, "bc"),(("c" + "a"^2), "b"^2, "ca"),(("a" + "b"^2), "c"^2, "ab")) =` ____________.
If A = `[(1,-1,0),(2,3,4),(0,1,2)]` and B = `[(2,2,-4),(-4,2,-4),(2,-1,5)]`, then:
