If The matrix \( x=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right) \) and \( y=\left(\begin{array}{ll}4 & 2 \\ 1 & 3\end{array}\right. \) find \( 4 x-3 y \)
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To find \( 4x - 3y \), we first need to calculate \( 4x \) and \( 3y \). Starting with \( x = \left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right) \), \[ 4x = 4 \cdot \left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right) = \left(\begin{array}{ll}4 \cdot 1 & 4 \cdot 2 \\ 4 \cdot 3 & 4 \cdot 4\end{array}\right) = \left(\begin{array}{ll}4 & 8 \\ 12 & 16\end{array}\right). \] Now for \( y = \left(\begin{array}{ll}4 & 2 \\ 1 & 3\end{array}\right) \), \[ 3y = 3 \cdot \left(\begin{array}{ll}4 & 2 \\ 1 & 3\end{array}\right) = \left(\begin{array}{ll}3 \cdot 4 & 3 \cdot 2 \\ 3 \cdot 1 & 3 \cdot 3\end{array}\right) = \left(\begin{array}{ll}12 & 6 \\ 3 & 9\end{array}\right). \] Finally, we subtract \( 3y \) from \( 4x \): \[ 4x - 3y = \left(\begin{array}{ll}4 & 8 \\ 12 & 16\end{array}\right) - \left(\begin{array}{ll}12 & 6 \\ 3 & 9\end{array}\right) = \left(\begin{array}{ll}4 - 12 & 8 - 6 \\ 12 - 3 & 16 - 9\end{array}\right) = \left(\begin{array}{ll}-8 & 2 \\ 9 & 7\end{array}\right). \] Thus, the final result is: \[ 4x - 3y = \left(\begin{array}{ll}-8 & 2 \\ 9 & 7\end{array}\right). \]