Find all values of x such that x4 0 0 0 4 5 x28x15 0 7 3x 1

Find all values of x such that [x-4 0 0 0 -4 5 x2-8x+15 0 7 3x - 12 0 0 9 -6 3 3x] = 0

Solution

Expanding the determinant across the first row since three enteries are zero

(x-4) * (Matrix from 2,2 to 4,4)

(x-4) * [5(0*3x-0*3) + (x^2-8x+15)(0 + (12-3x)3x)]

(x-4)*(x^2-8x+15)(12-3x)(3x)

3x(x-4)(12-3x)(x^2-3x-5x+15)

=> 3x(x-4)(12-3x)(x-3)(x-5)

Hence the matrix will be equal to zero when the given expression is equal to 0

Hence the possible values of x are 0,3,4,5

Hene for these four values the 4X4 matrix determinant will be equal to zero