Given that \(p = \begin{bmatrix} x&4\\3&7\end{bmatrix} Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}\) and the determinant of \(Q\) is three more than that of \(P\) , find the values of \(x\).
\(p = \begin{bmatrix} x&4\\3&7\end{bmatrix}, Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}\)
\(p = \begin{bmatrix} x&4\\3&7\end{bmatrix}=7x-12, Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}=2x^2-3\)
|Q| = |P| + 3 (Given)
\(= 2\times2 - 3 = 7x - 12 + 3\)
\(= 2\times2 - 3 - 7x + 12 - 3 = 0\)
\(= 2\times2 - 7x + 6 = 0\)
\(= 2\times2 - 4x - 3x + 6 = 0\)
\(= 2x(x - 2) - 3(x - 2) = 0\)
\(= (x - 2)(2x - 3) = 0\)
\(\therefore x=2,\frac{3}{2}\)
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