Find the value of x (determinants)

Question

|3x-5 1 1 |

|2 3x-5 3 | = 0

|2 3 3x-5. |

Answer 1

Since we are given the determinant as 0. We can use Cramer's rule to determine the determinant of this matrix...

(3x-5)³ + (1)(3)(2) + (1)(2)(3) - (1)(2)(3x-5) - (1)(2)(3x-5) - (3)(3)(3x-5) = 0

(3x-5)³ + 12 - 13(3x-5) = 0

(3x-5)(3x-5)(3x-5) - 13(3x-5) + 13 = 0

Let w = 3x-5

w³ - 13w + 13 = 0

We now use Cardano's equation which allow us to solve cubic equations of the form x³ + px +q = 0

p = -13, q = 13 for our usage case...

w = (-q/2 + sqrt(q²/4 + p³/27))^(1/3) + (-q/2 - sqrt(q²/4 - p³/27))^(1/3)

w = (-13/2 + sqrt(169/4 - 2197/27))^(1/3) + (-13/2 - sqrt(169/4 + 2197/27))^(1/3)

The other two roots are related to this one by multiplication of (-1 + i√3)/2 and (-1 - i√3)/2

Once you have determined w, you can then determine x using the bold faced representation above.

sqrt(169/4 + 2197/27) ≈ 11.12

sqrt(169/4 - 2197/27) ≈ 6.29*i