English

Evaluate: abcaabbcabcccab|a-b-c2a2a2bb-c-a2b2c2cc-a-b| - Mathematics

Advertisements
Advertisements

Question

Evaluate: `|("a" - "b" - "c", 2"a", 2"a"),(2"b", "b" - "c" - "a", 2"b"),(2"c", 2"c", "c" - "a" - "b")|`

Sum
Advertisements

Solution

We have, `|("a" - "b" - "c", 2"a", 2"a"),(2"b", "b" - "c" - "a", 2"b"),(2"c", 2"c", "c" - "a" - "b")|`

[Applying R1 → R1 + R2 + R3]

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

[Taking (a + b + c) common from the first row]

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

[Applying C1 → C1 – Cand C2 → C2 – C3]

= `("a" + "b" + "c")|(0, 0, 1),(0, -("a" + "b" + "c"), 2"b"),("a" + "b" + "c", "a" + "b" + "c", "c" - "a" - "b")|`

Expanding along R1,

= (a + b + c) [1 × 0 + (a + b + c)2]

= (a + b + c)3 

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Determinants - Exercise [Page 77]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 4 Determinants
Exercise | Q 6 | Page 77

RELATED QUESTIONS

Using the properties of determinants, prove the following:

`|[1,x,x+1],[2x,x(x-1),x(x+1)],[3x(1-x),x(x-1)(x-2),x(x+1)(x-1)]|=6x^2(1-x^2)`


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:

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


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`


By using properties of determinants, show that:

`|(1,x,x^2),(x^2,1,x),(x,x^2,1)| = (1-x^3)^2`


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


Using properties of determinants, prove the following :

\[\begin{vmatrix}1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1\end{vmatrix} = \left( 1 - a^3 \right)^2\].

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


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. 


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


Using properties of determinant show that

`|(1, log_x y, log_x z),(log_y x, 1, log_y z),(log_z x, log_z y, 1)|` = 0


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.


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:

Which of the following is correct


Evaluate: `|("a" + x, y, z),(x, "a" + y, z),(x, y, "a" + z)|`


Evaluate: `|(x + 4, x, x),(x, x + 4, x),(x, x, x + 4)|`


Prove that: `|(y + z, z, y),(z, z + x, x),(y, x, x + y)|` = 4xyz


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


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.


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


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.


If a, b, c are the roots of the equation x3 - 3x2 + 3x + 7 = 0, then the value of `abs((2 "bc - a"^2, "c"^2, "b"^2),("c"^2, 2 "ac - b"^2, "a"^2),("b"^2, "a"^2, 2 "ab - c"^2))` is ____________.


If `abs ((2"x",5),(8, "x")) = abs ((6,-2),(7,3)),`  then the value of x is ____________.


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


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:


The value of the determinant `|(1, cos(β - α), cos(γ - α)),(cos(α - β), 1, cos(γ - β)),(cos(α - γ), cos(β - γ), 1)|` is equal to ______.


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`


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 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`


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


By using properties of determinants, prove that 

`|(x+y, y+z, z+x),(z, x, y),(1, 1, 1)|` = 0 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×