Maths · Evaluation of determinants
then
\( f\left|\begin{array}{ccc}\boldsymbol{x}+\mathbf{1} & \mathbf{3} & \mathbf{5} \\ \mathbf{2} & \boldsymbol{x}+\mathbf{2} & \mathbf{5} \\ \mathbf{2} & \mathbf{3} & \boldsymbol{x}+\mathbf{4}\end{array}\right|=\mathbf{0}, \) then \( \boldsymbol{x}=? \)
- A. 1,9
- B. -1,9
- C. -1,-9
- D. 1,-9
Step-by-step solution
Compute the determinant: (x+1)[(x+2)(x+4)-15] - 3[2(x+4)-10] + 5[6-2(x+2)] = (x+1)(x^2+6x-7) - 3(2x-2) + 5(2-2x) = x^3+7x^2-17x+9 = 0. Factor: (x-1)^2(x+9)=0 => x=1 or x=-9.
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