English

Using properties of determinant show that |a+babaa+ccbcb+c| = 4abc - Mathematics and Statistics

Advertisements
Advertisements

Question

Using properties of determinant show that

`|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc

Sum
Advertisements

Solution

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

By C1 + (C2 + C3), we get,

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

By taking 2 common from C1, we get,

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

By C1 – (C2 + C3), we get,

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

By C2 + C1 and C3 + C1, we get

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

By taking c, a, b common from C1, C2, C3 respectively, we get,

L.H.S. = `2("c"  "a"  "b")|(0, 1, 1),(-1, 1, 0),(-1, 0, 1)|`

= 2abc [0 (1 – 0) – 1 ( – 1 + 0) + 1 (0 + 1)]

= 2abc (0 + 1 + 1)

= 4abc

= RHS.

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Determinants and Matrices - Exercise 4.2 [Page 68]

APPEARS IN

RELATED QUESTIONS

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, show that ΔABC is isosceles if:`|[1,1,1],[1+cosA,1+cosB,1+cosC],[cos^2A+cosA,cos^B+cosB,cos^2C+cosC]|=0​`


 

If ` f(x)=|[a,-1,0],[ax,a,-1],[ax^2,ax,a]| ` , using properties of determinants find the value of f(2x) − f(x).

 

Using properties of determinants, prove that

`|[x+y,x,x],[5x+4y,4x,2x],[10x+8y,8x,3x]|=x^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]|`

 

Using the property of determinants and without expanding, prove that:

`|(2,7,65),(3,8,75),(5,9,86)| = 0`


Using the property of determinants and without expanding, prove that:

`|(1, bc, a(b+c)),(1, ca, b(c+a)),(1, ab, c(a+b))| = 0`


By using properties of determinants, show that:

`|(1,a,a^2),(1,b,b^2),(1,c,c^2)| = (a - b)(b-c)(c-a)`


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

`|[b+c , a ,a  ] ,[ b , a+c, b ] ,[c , c, a+b ]|` = 4abc 


Using properties of determinants, prove that:

`|(a,b,b+c),(c,a,c+a),(b,c,a+b)|` = (a+b+c)(a-c)2 


 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`


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`.


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.


Find the value (s) of x, if `|(1, 4, 20),(1, -2, -5),(1, 2x, 5x^2)|` = 0


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


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


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")|`


Without expanding the determinants, show that `|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|` = 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


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 system 3x – y + 4z = 3, x + 2y – 3z = –2 and 6x + 5y + λz = –3 has at least one Solution when


Select the correct option from the given alternatives:

If `|(6"i", -3"i", 1),(4, 3"i", -1),(20, 3, "i")|` = x + iy then


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


Prove that: `|("a"^2 + 2"a", 2"a" + 1, 1),(2"a" + 1, "a" + 2, 1),(3, 3, 1)| = ("a" - 1)^3`


Find the value of θ satisfying `[(1, 1, sin3theta),(-4, 3, cos2theta),(7, -7, -2)]` = 0


If cos2θ = 0, then `|(0, costheta, sin theta),(cos theta, sin theta,0),(sin theta, 0, cos theta)|^2` = ______.


If the determinant `|(x + "a", "p" + "u", "l" + "f"),("y" + "b", "q" + "v", "m" + "g"),("z" + "c", "r" + "w", "n" + "h")|` splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.


`abs(("x", -7),("x", 5"x" + 1))`


If the ratio of the H.M. and GM. between two numbers a and bis 4 : 5, then a: b is


If f(α) = `[(cosα, -sinα, 0),(sinα, cosα, 0),(0, 0, 1)]`, prove that f(α) . f(– β) = f(α – β).


If `|(α, 3, 4),(1, 2, 1),(1, 4, 1)|` = 0, then the value of α is ______.


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 determinants, find the value of  `|(10, 57, 107), (12, 64, 124), (15, 78, 153)|`


Without expanding the 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)|`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×