Advertisements
Advertisements
प्रश्न
Using properties of determinants, prove the following:
`|(a, b,c),(a-b, b-c, c-a),(b+c, c+a, a+b)| = a^3 + b^3 + c^3 - 3abc`.
Advertisements
उत्तर
Δ = `|(a, b,c),(a-b, b-c, c-a),(b+c, c+a, a+b)|`
= `|(a,b,c),(a-b, b-c, c-a),(a+b+c, c+a+b, a+b+c)| ...[ "Applying" R_3 -> R_3 + R_1]`
= `(a +b+c) |(a,b,c),(a-b, b-c, c-a),(1,1,1)| ...["Taking" (a +b +c) "common"]`
= `(a +b+c) |(a, b ,c),(-b , -c, -a),(1, 1, 1)| ...["Applying" R_2 -> R_2 - R_1]`
= `(a +b+c) |(a-c, b-c, c),(-b +a , -c+a, -a),(0, 0, 1)| ...[C_1 -> C_1 - C_3 "and" C_2 ->C_2 - C_3]`
= `(a +b+c) [{(a-c) (a-c) -(b-c)(a-b)}]`
= `(a +b+c) ...[{(a - c)^2 - (ab - ac - b^2 + bc)]`
= `(a +b+c) [(a -c ^2- (ab - ac - b^2 + bc)]`
= `(a +b+c) (a^2 + b^2 + c^2 - ab - bc - ca)`
= `a^3 + b^3 + c^3 - 3abc`
APPEARS IN
संबंधित प्रश्न
Using the property of determinants and without expanding, prove that:
`|(b+c, q+r, y+z),(c+a, r+p, z +x),(a+b, p+q, x + y )| = 2|(a,p,x),(b,q,y),(c, r,z)|`
By using properties of determinants, show that:
`|(y+k,y, y),(y, y+k, y),(y, y, y+k)| = k^2(3y + k)`
By using properties of determinants, show that:
`|(1+a^2-b^2, 2ab, -2b),(2ab, 1-a^+b^2, 2a),(2b, -2a, 1-a^2-b^2)| = (1+a^2+b^2)`
Evaluate `|(x, y, x+y),(y, x+y, x),(x+y, x, y)|`
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
`|(a^2 + 2a,2a + 1,1),(2a+1,a+2, 1),(3, 3, 1)| = (a - 1)^3`
Using properties of determinants, prove that `|(1,1,1+3x),(1+3y, 1,1),(1,1+3z,1)| = 9(3xyz + xy + yz+ zx)`
Prove the following using properties of determinants :
\[\begin{vmatrix}a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b\end{vmatrix} = 2\left( a + b + c \right) {}^3\]
Evaluate the following determinants:
`|(x - 1, x, x - 2),(0, x - 2, x - 3),(0, 0, x - 3)| = 0`
Using properties of determinants, show that `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc.
Find the value (s) of x, if `|(1, 2x, 4x),(1, 4, 16),(1, 1, 1)|` = 0
Using properties of determinant show that
`|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc
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.
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:
Let D = `|(sintheta*cosphi, sintheta*sinphi, costheta),(costheta*cosphi, costheta*sinphi, -sintheta),(-sintheta*sinphi, sintheta*cosphi, 0)|` then
Select the correct option from the given alternatives:
`|("b" + "c", "c" + "a", "a" + "b"),("q" + "r", "r" + "p", "p" + "q"),(y + z, z + x, x + y)|` =
Evaluate: `|("a" + x, y, z),(x, "a" + y, z),(x, y, "a" + z)|`
If A + B + C = 0, then prove that `|(1, cos"c", cos"B"),(cos"C", 1, cos"A"),(cos"B", cos"A", 1)|` = 0
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
The A.M., H.M. and G.M. between two numbers are `144/15`, 15 and 12, but not necessarily in this order then, H.M., G.M. and A.M. respectively are
A number consists of two digits and the digit in the ten's place exceeds that in the unit's place by 5. If 5 times the sum of the digits be subtracted from the number, the digits of the number are reversed. Then the sum of digits of the number is:
If f(α) = `[(cosα, -sinα, 0),(sinα, cosα, 0),(0, 0, 1)]`, prove that f(α) . f(– β) = f(α – β).
Without expanding evaluate the following determinant:
`|(1, a, b + c), (1, b, c + a), (1, c, a + b)|`
By using properties of determinant prove that `|(x+y, y+z,z+x),(z,x,y),(1,1,1)|=0`
The value of the determinant of a matrix A of order 3 is 3. If C is the matrix of cofactors of the matrix A, then what is the value of the determinant of C2?
