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# how to solve the given determinant

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

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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