Advertisements
Advertisements
प्रश्न
The determinant `|(sin"A", cos"A", sin"A" + cos"B"),(sin"B", cos"A", sin"B" + cos"B"),(sin"C", cos"A", sin"C" + cos"B")|` is equal to zero.
विकल्प
True
False
उत्तर
This statement is True.
Explanation:
Let Δ = `|(sin"A", cos"A", sin"A" + cos"B"),(sin"B", cos"A", sin"B" + cos"B"),(sin"C", cos"A", sin"C" + cos"B")|`
Splitting up C3
= `|(sin"A", cos"A", cos"B"),(sin"B", cos"A", cos"B"),(sin"C", cos"A", cos"B")| + |(sin"A", cos"A", cos"B"),(sin"B", cos"A", cos"B"),(sin"C", cos"A", cos"B")|`
= `0 + |(sin"A", cos"A", cos"B"),(sin"B", cos"A", cos"B"),(sin"C", cos"A", cos"B")|` ....[∵ C1 and C3 are identical]
= `cos"A" cos"B" |(sin"A", 1, 1),(sin"B", 1, 1),(sin"C", 1, 1)|`
[Taking cos A and cos B common from C2 and C3 respectively]
= cos A cos B (0) ....[∵ C2 and C3 are identical]
= 0
APPEARS IN
संबंधित प्रश्न
Using properties of determinants, prove that `|[2y,y-z-x,2y],[2z,2z,z-x-y],[x-y-z,2x,2x]|=(x+y+z)^3`
Without expanding the determinant, prove that
`|(a, a^2,bc),(b,b^2, ca),(c, c^2,ab)| = |(1, a^2, a^3),(1, b^2, b^3),(1, c^2, c^3)|`
Using properties of determinants, prove that:
`|(alpha, alpha^2,beta+gamma),(beta, beta^2, gamma+alpha),(gamma, gamma^2, alpha+beta)|` = (β – γ) (γ – α) (α – β) (α + β + γ)
Prove the following using properties of determinants :
\[\begin{vmatrix}a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b\end{vmatrix} = 2\left( a + b + c \right) {}^3\]
Using properties of determinants, prove the following:
Using properties of determinants, prove that \[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix} = a^2 \left( a + x + y + z \right)\] .
Using properties of determinants, prove that:
`|[a^2 + 1, ab, ac], [ba, b^2 + 1, bc ], [ca, cb, c^2+1]| = a^2 + b^2 + c^2 + 1`
Without expanding evaluate the following determinant:
`|(1, "a", "b" + "c"),(1, "b", "c" + "a"),(1, "c", "a" + "b")|`
If `|(4 + x, 4 - x, 4 - x),(4 - x,4 + x,4 - x),(4 - x,4 - x, 4 + x)|` = 0, then find the values of x.
If `|("x"^"k", "x"^("k" + 2), "x"^("k" + 3)),("y"^"k", "y"^("k" + 2), "y"^("k" + 3)),("z"^"k", "z"^("k" + 2), "z"^("k" + 3))|` = (x - y) (y - z) (z - x)`(1/"x"+ 1/"y" + 1/"z") ` then
Select the correct option from the given alternatives:
The value of a for which system of equation a3x + (a + 1)3 y + (a + 2)3z = 0 ax + (a +1)y + (a + 2)z = 0 and x + y + z = 0 has non zero Soln. is
Answer the following question:
Evaluate `|(2, 3, 5),(400, 600, 1000),(48, 47, 18)|` by using properties
Answer the following question:
Evaluate `|(101, 102, 103),(106, 107, 108),(1, 2, 3)|` by using properties
If x, y, z ∈ R, then the value of determinant `|((2x^2 + 2^(-x))^2, (2^x - 2^(-x))^2, 1),((3^x + 3^(-x))^2, (3^x -3^(-x))^2, 1),((4^x + 4^(-x))^2, (4^x - 4^(-x))^2, 1)|` is equal to ______.
If the value of a third order determinant is 12, then the value of the determinant formed by replacing each element by its co-factor will be 144.
Let Δ = `|("a", "p", x),("b", "q", y),("c", "r", z)|` = 16, then Δ1 = `|("p" + x, "a" + x, "a" + "p"),("q" + y, "b" + y, "b" + "q"),("r" + z, "c" + z, "c" + "r")|` = 32.
If `abs ((2"x",5),(8, "x")) = abs ((6,-2),(7,3)),` then the value of x is ____________.
Let 'A' be a square matrix of order 3 × 3, then |KA| is equal to:
If A, B and C are the angles of a triangle ABC, then `|(sin2"A", sin"C", sin"B"),(sin"C", sin2"B", sin"A"),(sin"B", sin"A", sin2"C")|` = ______.
The value of the determinant `|(6, 0, -1),(2, 1, 4),(1, 1, 3)|` is ______.
Without expanding determinants find the value of `|(10, 57, 107),(12, 64, 124),(15, 78, 153)|`
Without expanding evaluate the following determinant:
`|(1, a, b + c), (1, b, c + a), (1, c, a + b)|`
Without expanding determinant find the value of `|(10,57,107),(12,64,124),(15,78,153)|`
Without expanding evaluate the following determinant.
`|(1, a, b+c), (1, b, c+a), (1, c, a+b)|`
if `|(a, b, c),(m, n, p),(x, y, z)| = k`, then what is the value of `|(6a, 2b, 2c),(3m, n, p),(3x, y, z)|`?
Without expanding determinant find the value of `|(10, 57, 107),(12, 64, 124),(15, 78, 153)|`
By using properties of determinant prove that `|(x+y,y+z,z+x),(z,x,y),(1,1,1)|` = 0.