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how to solve the given determinant

Asked by mammyji 21st May 2010, 11:04 PM
Answered by Expert
Answer:

Dear student,

 

First of all we find the determinant of given matrix

 

7

6

x

2

x

2

x

3

7

 

 

Expanding by first row, we get the determinant as:

 

   7(7x - 3×2) – 6(2×7 – 2x) + x(2×3 + x×x)

= 7(7x - 6) – 6(14 – 2x) + x(6 – x2)

= 49x – 42 – 84 + 12x + 6x – x3

Rearranging the terms we get

– x3 + (49 + 12 + 6)x – 42 – 84

= – x3 + 67x – 126

 

Now, this determinant is given to be zero

So,

 

–x3 + 67x – 126 = 0

 

Multiplying both sides of above equation by –1, we get

           

x3 – 67x + 126 = 0 ………………………..(i)

 

This equation is a cubic, so it must have three roots.

 

 

Now, it is also given that one root of the given determinant is x = –9. So, by factor theorem  

(x + 9) must be a factor of the polynomial on left hand side of equation (i).

 

Let us divide x3 – 67x + 126 with (x + 9) to find the other two factors

 

 

 

Divisor = x + 9 and Dividend = x3 – 67x + 126

                     

So, we get x3 – 67x + 126 = (x + 9)(x2 – 9x + 14)

                                          = (x + 9)(x2 – 7x – 2x + 14)

                                          = (x + 9)[x(x – 7) – 2(x – 7)]

                 x3 – 67x + 126  = (x + 9)(x – 7)(x – 2) ……………..(ii)

Substituting this value in (i), we get

       

         (x + 9)(x – 7)(x – 2) = 0

So the other two roots are given by x – 7 = 0 and x – 2 = 0  i.e.     x = 7 and x = 2

                                         

Regards,
Team
TopperLearning

Answered by Expert 22nd May 2010, 12:21 PM
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