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Matrices and Determinants question

2018 · Shift 0 · Q31
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Matrices and Determinants question

2018 · Shift 0 · Q31

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If ∣x−42x2x2xx−42x2x2xx−4∣=(A+Bx)(x−A)2\left| {\begin{matrix} {x - 4} & {2x} & {2x} \\ {2x} & {x - 4} & {2x} \\ {2x} & {2x} & {x - 4} \\ \end{matrix} } \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2}​x−42x2x​2xx−42x​2x2xx−4​​=(A+Bx)(x−A)2 then the ordered pair (A, B) is equal to :
  1. A
    (4, 5)
  2. B
    (-4, -5)
  3. C
    (-4, 3)
  4. D
    (-4, 5)
View written solutionFree

Correct answer: D

  1. Identify the matrix form

The determinant is

∣x−42x2x2xx−42x2x2xx−4∣.\begin{vmatrix} x-4 & 2x & 2x\\ 2x & x-4 & 2x\\ 2x & 2x & x-4 \end{vmatrix}.​x−42x2x​2xx−42x​2x2xx−4​​.

This is a 3×33\times 33×3 matrix with all diagonal entries equal to a=x−4a=x-4a=x−4 and all off-diagonal entries equal to b=2xb=2xb=2x.

For a matrix of the form

(abbbabbba),\begin{pmatrix} a & b & b\\ b & a & b\\ b & b & a \end{pmatrix},​abb​bab​bba​​,

its determinant is

(a−b)2(a+2b).(a-b)^2(a+2b).(a−b)2(a+2b).
  1. Apply the formula

Here,

a=x−4,b=2x.a=x-4, \qquad b=2x.a=x−4,b=2x.

So,

a−b=(x−4)−2x=−x−4=−(x+4),a-b=(x-4)-2x=-x-4=-(x+4),a−b=(x−4)−2x=−x−4=−(x+4),

and hence

(a−b)2=(x+4)2.(a-b)^2=(x+4)^2.(a−b)2=(x+4)2.

Also,

a+2b=(x−4)+4x=5x−4.a+2b=(x-4)+4x=5x-4.a+2b=(x−4)+4x=5x−4.

Therefore,

det⁡=(x+4)2(5x−4).\det = (x+4)^2(5x-4).det=(x+4)2(5x−4).
  1. Compare with the given factorized form

Given,

det⁡=(A+Bx)(x−A)2.\det = (A+Bx)(x-A)^2.det=(A+Bx)(x−A)2.

We have found

det⁡=(5x−4)(x+4)2.\det = (5x-4)(x+4)^2.det=(5x−4)(x+4)2.

Now match factors:

  • Since (x−A)2=(x+4)2(x-A)^2=(x+4)^2(x−A)2=(x+4)2, we get A=−4.A=-4.A=−4.
  • Then A+Bx=−4+Bx=5x−4,A+Bx=-4+Bx=5x-4,A+Bx=−4+Bx=5x−4, so B=5.B=5.B=5.
  1. Final answer

Thus,

(A,B)=(−4,5).(A,B)=(-4,5).(A,B)=(−4,5).

So the correct option is D.

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