हिंदी

Using properties of determinants, show that |a+babaa+ccbcb+c| = 4abc.

Advertisements
Advertisements

प्रश्न

Using properties of determinants, show that `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc.

योग
Advertisements

उत्तर

L.H.S. = `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|`

Applying C1 → C1 – (C2 + C3), we get

L.H.S. = `|(0, "a", "b"),(-2"c", "a" + "c", "c"),(-2"c", "c", "b" + "c")|`

Taking (– 2) common from C1, we get

L.H.S. = ` - 2|(0, "a", "b"),("c", "a" + "c", "c"),("c", "c", "b" + "c")|`

Applying C2 → C2 – C1 and C3 → C3 – C1, we get

L.H.S. = `-2|(0, "a", "b"),("c", "a", 0),("c", 0, "b")|`

= - 2[0(ab – 0) – a(bc – 0) + b(0 – ac)]
= – 2(0 – abc – abc)
= – 2(– 2abc)
= 4abc
= R.H.S.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Determinants - EXERCISE 6.2 [पृष्ठ ८९]

APPEARS IN

बालभारती Mathematics and Statistics (Commerce) Part 1 [English] Standard 11 Maharashtra State Board
अध्याय 6 Determinants
EXERCISE 6.2 | Q 2) | पृष्ठ ८९

संबंधित प्रश्न

By using properties of determinants, show that:

`|(x,x^2,yz),(y,y^2,zx),(z,z^2,xy)| = (x-y)(y-z)(z-x)(xy+yz+zx)`


By using properties of determinants, show that:

`|(x+y+2z, x, y),(z, y+z+2z,y),(z,x,z+x+2y)| = 2(x+y+z)^3`


Evaluate `|(x, y, x+y),(y, x+y, x),(x+y, x, y)|`


Using properties of determinants, prove that:

`|(1, 1+p, 1+p+q),(2, 3+2p, 4+3p+2q),(3,6+3p,10+6p+3q)| =  1`                 


Using properties of determinants, prove the following:

\[\begin{vmatrix}x^2 + 1 & xy & xz \\ xy & y^2 + 1 & yz \\ xz & yz & z^2 + 1\end{vmatrix} = 1 + x^2 + y^2 + z^2\] .

Using properties of determinants, prove that \[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix} = a^2 \left( a + x + y + z \right)\] .


Without expanding the determinants, show that `|(x"a", y"b", z"c"),("a"^2, "b"^2, "c"^2),(1, 1, 1)| = |(x, y, z),("a", "b", "c"),("bc", "ca", "ab")|`


Solve the following equation: 

`|(x + 2, x + 6, x - 1),(x + 6, x - 1, x + 2),(x - 1, x + 2, x + 6)|` = 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)|`


Answer the following question:

Evaluate `|(101, 102, 103),(106, 107, 108),(1, 2, 3)|` by using properties


Evaluate: `|(0, xy^2, xz^2),(x^2y, 0, yz^2),(x^2z, zy^2, 0)|`


The value of the determinant `|(x , x + y, x + 2y),(x + 2y, x, x + y),(x + y, x + 2y, x)|` is ______.


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


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


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


Without expanding determinants 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 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")|`


Without expanding evaluate the following determinant.

`|(1, a, b+c),(1, b, c+a),(1, c, a+b)|`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×