What is the product in solution of equation  (1 2 3 4)(-2 2-x 1 7)(3 6 x+4 12)(- x-14 2 3)?

1 Answer | Add Yours

Top Answer

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to solve the matrix equation. Notice that the matrix has 4 rows and columns.

`((1, 2, 3, 4),(-2, 2-x, 1, 7),(3, 6, x+4, 12),(- x,-14, 2, 3))`

`` Evaluate the determinant of this matrix using the first line:

`Delta = (-1)^(1+1)*1*[[2-x,1,7],[6,x+4,12],[-14,2,3]] + (-1)^(1+2)*2* [[-2,1,7],[3,x+4,12],[-x,2,3]]+ (-1)^(1+3)*3*[[-2,2-x,7],[3,6,12],[x,-14,3]] + (-1)^(1+4)*4*[[-2,2-x,1],[3,6,x+4],[-x,-14,2]]`

`Delta = delta_1 - 2delta_2 + 3delta_3 - 4delta_4`

`` Calculating each minor yields:

`[[2-x,1,7],[6,x+4,12],[-14,2,3]]` = `3(2-x)(x+4) + 84- 168 + 98(x+4) + 24(x-2) - 18`

`` `delta_1 = (x+4)(6-3x+98) - 102 =gt delta_1 = (x+4)(104-3x) - 102`

`` `delta_1 = 104x - 3x^2 + 116 - 12x - 102`

`` `delta_1 = - 3x^2 + 92x + 14`

`delta_2 = [[-2,1,7],[3,x+4,12],[-x,2,3]]`

`delta_2 = -6(x+4) + 42 - 12x + 7x^2 + 28x + 48 - 9`

`delta_2 = -6x - 24+ 42 - 12x + 7x^2 + 28x + 48 - 9`

`delta_2 = 7x^2 + 10x + 57`

`` `delta_3 = [[-2,2-x,7],[3,6,12],[x,-14,3]]`

`delta_3 = -36 - 294 + 24x - 12x^2 - 42x - 336 - 18 - 9x`

`delta_3 =- 12x^2 - 27x - 684`

`` `delta_4 = [[-2,2-x,1],[3,6,x+4],[-x,-14,2]]`

`delta_4 = -24 - 42+ x(x-2)(x+4)+ 6x - 28(x+4) - 12 + 6x`

`delta_4 = -78 + x^3 + 2x^2 - 8x + 12x - 28x - 112`

`delta_4 = x^3 + 2x^2 - 24x - 190`

`Delta = - 3x^2 + 92x + 14 - 2*(7x^2 + 10x + 57) + 3*(- 12x^2 - 27x - 684) - 4*(x^3 + 2x^2 - 24x - 190)`

Opening the brackets yields:

`Delta = - 3x^2 + 92x + 14 - 14x^2 - 20x - 114 - 36x^2 - 81x - 2052 - 4x^3 - 8x^2 + 96x + 760`

`Delta = - 4x^3 - 61x^2 + 87x - 1392`

The product of the roots of the equation

`- 4x^3 - 61x^2 + 87x - 1392 = 0`  is `x_1*x_2*x_3 = -(1392)/(-4)`

`x_1*x_2*x_3 = -348`

The product of the three roots of the equation is `x_1*x_2*x_3 = -348` .

We’ve answered 318,957 questions. We can answer yours, too.

Ask a question