English

Without expanding the determinants, show that |b+cbcb2c2c+acac2a2a+ baba2b2| = 0 - Mathematics and Statistics

Advertisements
Advertisements

Question

Without expanding the determinants, show that `|("b" + "c", "bc", "b"^2"c"^2),("c" + "a", "ca", "c"^2"a"^2),("a" +  "b", "ab", "a"^2"b"^2)|` = 0

Sum
Advertisements

Solution

L.H.S. = `|("b" + "c", "bc", "b"^2"c"^2),("c" + "a", "ca", "c"^2"a"^2),("a" +  "b", "ab", "a"^2"b"^2)|` 

Taking bc, ca, ab common from R1, R2, R3 respectively, we get

L.H.S. = `("bc")("ca")("ab")|(("b" - "c")/"bc" , 1, "bc"),(("c" + "a")/"ca", 1, "ca"),(("a" + "b")/"ab", 1, "ab")|`

Taking abc common from C3, we get

L.H.S. = `("a"^2"b"^2"c"^2)("abc") |(1/"c" + 1/"b", 1, 1/"a"),(1/"a" + 1/"c", 1, 1/"b"),(1/"b" + 1/"a", 1, 1/"c")|`

Applying C1 → C1 + C3, we get

L.H.S. = `"a"^3"b"^3"c"^3|(1/"a" + 1/"b" + 1/"c", 1, 1/"a"),(1/"a" + 1/"b"+ 1/"c", 1, 1/"b"),(1/"a" + 1/"b" + 1/"c", 1, 1/"c")|`

Taking `(1/"a" + 1/"b" + 1/"c")` common from C1, we get

L.H.S. = `"a"^3"b"^3"c"^3 (1/"a" + 1/"b" + 1/"c")|(1, 1, 1/"a"),(1, 1, 1/"b"),(1, 1, 1/"c")|`

= `"a"^3"b"^3"c"^3 (1/"a" + 1/"b" + 1/"c")(0)` ...[C1 and C2 are identical]

= 0
= R.H.S.

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Determinants - MISCELLANEOUS EXERCISE - 6 [Page 95]

APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) [English] Standard 11 Maharashtra State Board
Chapter 6 Determinants
MISCELLANEOUS EXERCISE - 6 | Q 4) i) | Page 95

RELATED QUESTIONS

By using properties of determinants, show that:

`|(0,a, -b),(-a,0, -c),(b, c,0)| = 0`


By using properties of determinants, show that:

`|(1,1,1),(a,b,c),(a^3, b^3,c^3)|` = (a-b)(b-c)(c-a)(a+b+c)


Solve for x : `|("a"+"x","a"-"x","a"-"x"),("a"-"x","a"+"x","a"-"x"),("a"-"x","a"-"x","a"+"x")| = 0`, using properties of determinants. 


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 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.


Without expanding the determinants, show that `|(l, "m", "n"),("e", "d", "f"),("u", "v", "w")| = |("n", "f", "w"),(l, "e", "u"),("m", "d", "v")|`


Without expanding evaluate the following determinant:

`|(2, 7, 65),(3, 8, 75),(5, 9, 86)|`


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)|`


Answer the following question:

Without expanding determinant show that

`|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|` = 0


Prove that: `|(y^2z^2, yz, y + z),(z^2x^2, zx, z + x),(x^2y^2, xy, x + y)|` = 0


If A + B + C = 0, then prove that `|(1, cos"c", cos"B"),(cos"C", 1, cos"A"),(cos"B", cos"A", 1)|` = 0


The maximum value of Δ = `|(1, 1, 1),(1, 1 + sin theta, 1),(1 + cos theta, 1, 1)|` is ______. (θ is real number)


If x = – 9 is a root of `|(x, 3, 7),(2, x, 2),(7, 6, x)|` = 0, then other two roots are ______.


The value of the determinant `abs ((alpha, beta, gamma),(alpha^2, beta^2, gamma^2),(beta + gamma, gamma + alpha, alpha + beta)) =` ____________.


In a third order matrix B, bij denotes the element in the ith row and jth column. If

bij = 0 for i = j

= 1 for > j

= – 1 for i < j

Then the matrix is


By using properties of determinant prove that `|(x+y, y+z,z+x),(z,x,y),(1,1,1)|=0`


Without expanding determinant find the value of `|(10,57,107),(12,64,124),(15,78,153)|`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×