\begin{tabular}{l} 2) Find \( b \) * \\ Given that \( A=\left(\begin{array}{cc}2 & -3 \\ 4 & 5\end{array}\right), B=\left(\begin{array}{ll}7 & 6 \\ 8 & 1\end{array}\right), C=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right) \) and \( \frac{1}{2} A-\frac{1}{4} B=C \), \\ calculate the values of \( a, b, c \) and \( d \). \\ Your answer \\ \hline\end{tabular}
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To solve for \( a, b, c, \) and \( d \) in the equation \( \frac{1}{2} A - \frac{1}{4} B = C \), we first need to calculate \( \frac{1}{2} A \) and \( \frac{1}{4} B \). 1. Calculate \( \frac{1}{2} A \): \[ \frac{1}{2} A = \frac{1}{2} \left(\begin{array}{cc} 2 & -3 \\ 4 & 5 \end{array}\right) = \left(\begin{array}{cc} 1 & -\frac{3}{2} \\ 2 & \frac{5}{2} \end{array}\right) \] 2. Calculate \( \frac{1}{4} B \): \[ \frac{1}{4} B = \frac{1}{4} \left(\begin{array}{cc} 7 & 6 \\ 8 & 1 \end{array}\right) = \left(\begin{array}{cc} \frac{7}{4} & \frac{3}{2} \\ 2 & \frac{1}{4} \end{array}\right) \] 3. Now we can subtract \( \frac{1}{4} B \) from \( \frac{1}{2} A \): \[ \frac{1}{2} A - \frac{1}{4} B = \left(\begin{array}{cc} 1 & -\frac{3}{2} \\ 2 & \frac{5}{2} \end{array}\right) - \left(\begin{array}{cc} \frac{7}{4} & \frac{3}{2} \\ 2 & \frac{1}{4} \end{array}\right) \] 4. Perform the subtraction: For the first row, first column: \[ 1 - \frac{7}{4} = \frac{4}{4} - \frac{7}{4} = -\frac{3}{4} \] For the first row, second column: \[ -\frac{3}{2} - \frac{3}{2} = -3 \] For the second row, first column: \[ 2 - 2 = 0 \] For the second row, second column: \[ \frac{5}{2} - \frac{1}{4} = \frac{10}{4} - \frac{1}{4} = \frac{9}{4} \] 5. Thus, we have: \[ C = \left(\begin{array}{cc} -\frac{3}{4} & -3 \\ 0 & \frac{9}{4} \end{array}\right) \] From here, we can identify the values of \( a, b, c, d \): \[ a = -\frac{3}{4}, \quad b = -3, \quad c = 0, \quad d = \frac{9}{4} \] To answer the question explicitly, the values are: \[ a = -\frac{3}{4}, \quad b = -3, \quad c = 0, \quad d = \frac{9}{4} \]