Advertisements
Advertisements
प्रश्न
Evaluate: `|(0, xy^2, xz^2),(x^2y, 0, yz^2),(x^2z, zy^2, 0)|`
उत्तर
We have, `|(0, xy^2, xz^2),(x^2y, 0, yz^2),(x^2z, zy^2, 0)|`
[Taking x2, y2 and z2 common from C1, C2 and C3, respectively]
= `x^2y^2z^2|(0, x, x),(y, 0, y),(z, z, 0)|`
[Applying C1 → C2 – C3]
= `x^2y^2z^2|(0, 0, x),(y, -y, y),(z, z, 0)|`
= `x^2y^2z^2 (x(yz + yz))`
= `x^2y^2z^2 * (2xyz)`
= 2x3y3z3
APPEARS IN
संबंधित प्रश्न
Using properties of determinants prove the following: `|[1,x,x^2],[x^2,1,x],[x,x^2,1]|=(1-x^3)^2`
Using properties of determinants, prove that
`|((x+y)^2,zx,zy),(zx,(z+y)^2,xy),(zy,xy,(z+x)^2)|=2xyz(x+y+z)^3`
Using properties of determinants, prove that
`|[b+c,c+a,a+b],[q+r,r+p,p+q],[y+z,z+x,x+y]|=2|[a,b,c],[p,q,r],[x,y,z]|`
By using properties of determinants, show that:
`|(y+k,y, y),(y, y+k, y),(y, y, y+k)| = k^2(3y + k)`
Using properties of determinants, prove that:
`|(3a, -a+b, -a+c),(-b+a, 3b, -b+c),(-c+a, -c+b, 3c)|`= 3(a + b + c) (ab + bc + ca)
Using properties of determinants, prove that:
`|(1+a^2-b^2, 2ab, -2b),(2ab, 1-a^2+b^2, 2a),(2b, -2a, 1-a^2-b^2)| = (1 + a^2 + b^2)^3`
Using properties of determinants, prove that
`|[b+c , a ,a ] ,[ b , a+c, b ] ,[c , c, a+b ]|` = 4abc
Using properties of determinants, find the value of x for which
`|(4-"x",4+"x",4+"x"),(4+"x",4-"x",4+"x"),(4+"x",4+"x",4-"x")|= 0`
Without expanding determinants, find the value of `|(2014, 2017, 1),(2020, 2023, 1),(2023, 2026, 1)|`
Find the value (s) of x, if `|(1, 4, 20),(1, -2, -5),(1, 2x, 5x^2)|` = 0
Without expanding determinants show that
`|(1, 3, 6),(6, 1, 4),(3, 7, 12)| + 4|(2, 3, 3),(2, 1, 2),(1, 7, 6)| = 10|(1, 2, 1),(3, 1, 7),(3, 2, 6)|`
Select the correct option from the given alternatives:
The determinant D = `|("a", "b", "a" + "b"),("b", "c", "b" + "c"),("a" + "b", "b" + "c", 0)|` = 0 if
Select the correct option from the given alternatives:
Which of the following is correct
Answer the following question:
Evaluate `|(101, 102, 103),(106, 107, 108),(1, 2, 3)|` by using properties
Answer the following question:
By using properties of determinant prove that `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|` = 0
Answer the following question:
Without expanding determinant show that
`|("b" + "c", "bc", "b"^2"c"^2),("c" + "a", "ca", "c"^2"a"^2),("a" + "b", "ab", "a"^2"b"^2)|` = 0
The value of `|(1, 1, 1),(""^"n""C"_1, ""^("n" + 2)"C"_1, ""^("n" + 4)"C"_1),(""^"n""C"_2, ""^("n" + 2)"C"_2, ""^("n" + 4)"C"_2)|` is 8.
Evaluate: `|("a" - "b" - "c", 2"a", 2"a"),(2"b", "b" - "c" - "a", 2"b"),(2"c", 2"c", "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 values of x.
`|(x + 1, x + 2, x + "a"),(x + 2, x + 3, x + "b"),(x + 3, x + 4, x + "c")|` = 0, where a, b, c are in A.P.
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.
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.
A system of linear equations represented in matrix form Ax = 0, A is n × n matrix, has a non-zero solution if the determinant of A (i.e., det(A)) is
If `|(α, 3, 4),(1, 2, 1),(1, 4, 1)|` = 0, then the value of α is ______.
Evaluate the following determinant without expanding:
`|(5, 5, 5),(a, b, c),(b + c, c + a, a + b)|`
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
Without expanding the determinant, find the value of `|(10,57,107),(12,64,124),(15,78,153)|`
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)|`?
