14Number 14ExplanationGiven the matrix P=(3−x−563)\mathbf{P} = \begin{pmatrix} 3 - x & -5 \\ 6 & 3 \end{pmatrix}P=(3−x6−53). If the determinant of P\mathbf{P}P is 333333, then the value of xxx is ....111222333444555ExplanationGiven the matrix P=(3−x−563)\mathbf{P} = \begin{pmatrix} 3 - x & -5 \\ 6 & 3 \end{pmatrix}P=(3−x6−53). The formula for the determinant of a 2×22 \times 22×2 matrix: det(P)=∣abcd∣=ad−bc\det(\mathbf{P}) = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bcdet(P)=acbd=ad−bc For matrix P\mathbf{P}P: det(P)=(3−x)(3)−(−5)(6)\det(\mathbf{P}) = (3 - x)(3) - (-5)(6)det(P)=(3−x)(3)−(−5)(6)=3(3−x)−(−30)= 3(3 - x) - (-30)=3(3−x)−(−30)=9−3x+30= 9 - 3x + 30=9−3x+30=39−3x= 39 - 3x=39−3x Given det(P)=33\det(\mathbf{P}) = 33det(P)=33, then: 39−3x=3339 - 3x = 3339−3x=33−3x=33−39-3x = 33 - 39−3x=33−39−3x=−6-3x = -6−3x=−6x=−6−3x = \frac{-6}{-3}x=−3−6x=2x = 2x=2 Thus, the value of xxx is 222.
