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If  \begin{bmatrix} x-4 & 2x &2x \\ 2x &x-4 &2x \\ 2x&2x &x-4 \end{bmatrix} = \left ( A+ Bx \right )\left ( x-A \right )^{2},   

then the ordered pair \left (A,B \right ) is equal to :

  • Option 1)

    \left ( 4,5 \right )

  • Option 2)

    \left ( -4,-5 \right )

  • Option 3)

    \left ( -4,3 \right )

  • Option 4)

    \left ( -4,5 \right )

 

Answers (2)

best_answer

we can put values of x in both the sides \begin{bmatrix} -4 &0 &0 \\ 0 & -4& 0\\ 0& 0 & -4 \end{bmatrix}= A(-A^{2})

(-4)^{3}= A\Rightarrow A= -4

or x=4

\begin{bmatrix} 0 & 8&8 \\ 8& 0&8 \\ 8&8 & 0 \end{bmatrix}= (A+4B)(4-A)^{2}

-8(-8^{2})+8(8^{2})= \left ( 4B-4 \right )(8^{2})= 16\times 8^{2}= (4B-4\times 8^{2})

B=5

(4,5)

 

Property of determinant -

If a determinant D variables for x=a , then \left (x-a \right ) is a factor of D , in other words if two rows ( or two columns ) becomes identical for x=a , Then \left (x-a \right ) is a factor of D

-

 

 


Option 1)

\left ( 4,5 \right )

Option 2)

\left ( -4,-5 \right )

Option 3)

\left ( -4,3 \right )

Option 4)

\left ( -4,5 \right )

Posted by

gaurav

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