Given the matrix \(\small B = \begin{pmatrix} 2-x & 1 \\ 4 & 2-x \end{pmatrix}\) , find all values of x that are solutions of det(B) = 0.
det(B) = ad - bc
Therefore:
det(B) = (2−x)(2−x)−(1)(4)
det(B) = (2−x)(2−x)−(1)(4)
det(B) = (2-x)^2 - 4
det(B)= 4 − 4x + x^2 −4
det(B) = x^2 - 4x
x(x-4) = 0
x = 0 or x = 4
Therefore:
det(B) = 0 when x = 0 or x = 4. This makes the matrix invertible
(singular).