Advertisements
Advertisements
प्रश्न
Write the value of `|(a-b, b- c, c-a),(b-c, c-a, a-b),(c-a, a-b, b-c)|`
Advertisements
उत्तर
Let Δ = `|(a-b, b- c, c-a),(b-c, c-a, a-b),(c-a, a-b, b-c)|`
By R1 → R1 + R2 + R3
⇒ Δ = `|(0, 0, 0),(b-c, c-a, a-b),(c-a, a-b, b-c)|` = 0 ...(as all elements of R1 are zero).
APPEARS IN
संबंधित प्रश्न
If `|[2x,5],[8,x]|=|[6,-2],[7,3]|`, write the value of x.
Solve the system of linear equations using the matrix method.
x − y + z = 4
2x + y − 3z = 0
x + y + z = 2
Evaluate
\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}^2 .\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}0 & x & y \\ - x & 0 & z \\ - y & - z & 0\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2\end{vmatrix}\]
\[\begin{vmatrix}b^2 + c^2 & ab & ac \\ ba & c^2 + a^2 & bc \\ ca & cb & a^2 + b^2\end{vmatrix} = 4 a^2 b^2 c^2\]
If \[\begin{vmatrix}a & b - y & c - z \\ a - x & b & c - z \\ a - x & b - y & c\end{vmatrix} =\] 0, then using properties of determinants, find the value of \[\frac{a}{x} + \frac{b}{y} + \frac{c}{z}\] , where \[x, y, z \neq\] 0
Find the area of the triangle with vertice at the point:
(−1, −8), (−2, −3) and (3, 2)
Using determinants prove that the points (a, b), (a', b') and (a − a', b − b') are collinear if ab' = a'b.
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)
Prove that
x+ y = 5
y + z = 3
x + z = 4
Solve each of the following system of homogeneous linear equations.
x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0
Find the value of the determinant \[\begin{vmatrix}243 & 156 & 300 \\ 81 & 52 & 100 \\ - 3 & 0 & 4\end{vmatrix} .\]
The value of the determinant
If \[∆_1 = \begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{vmatrix}, ∆_2 = \begin{vmatrix}1 & bc & a \\ 1 & ca & b \\ 1 & ab & c\end{vmatrix},\text{ then }\]}
Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant
The other factor in the value of the determinant is
If a > 0 and discriminant of ax2 + 2bx + c is negative, then
\[∆ = \begin{vmatrix}a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & 0\end{vmatrix} is\]
Solve the following system of equations by matrix method:
8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5
2x − y + z = 0
3x + 2y − z = 0
x + 4y + 3z = 0
Consider the system of equations:
a1x + b1y + c1z = 0
a2x + b2y + c2z = 0
a3x + b3y + c3z = 0,
if \[\begin{vmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix}\]= 0, then the system has
For the system of equations:
x + 2y + 3z = 1
2x + y + 3z = 2
5x + 5y + 9z = 4
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
If \[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\] ,find A–1 and hence solve the system of equations x – 2y = 10, 2x + y + 3z = 8 and –2y + z = 7.
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
Solve the following system of equations by using inversion method
x + y = 1, y + z = `5/3`, z + x = `4/3`
The number of values of k for which the linear equations 4x + ky + 2z = 0, kx + 4y + z = 0 and 2x + 2y + z = 0 possess a non-zero solution is
If a, b, c are non-zeros, then the system of equations (α + a)x + αy + αz = 0, αx + (α + b)y + αz = 0, αx+ αy + (α + c)z = 0 has a non-trivial solution if
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 ______.
