1) Find \( a \) * 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

To find the values of \( a, b, c, \) and \( d \), we start by calculating \( \frac{1}{2} A \) and \( \frac{1}{4} B \). First, let's compute \( \frac{1}{2} A \): \[ \frac{1}{2} A = \frac{1}{2} \left(\begin{array}{cc} 2 & -3 \\ 4 & 5 \end{array}\right) = \left(\begin{array}{cc} 1 & -\frac{3}{2} \\ 2 & \frac{5}{2} \end{array}\right) \] Next, we compute \( \frac{1}{4} B \): \[ \frac{1}{4} B = \frac{1}{4} \left(\begin{array}{cc} 7 & 6 \\ 8 & 1 \end{array}\right) = \left(\begin{array}{cc} \frac{7}{4} & \frac{3}{2} \\ 2 & \frac{1}{4} \end{array}\right) \] Now, we can find \( C \) by substituting back into the given equation: \[ C = \frac{1}{2} A - \frac{1}{4} B = \left(\begin{array}{cc} 1 & -\frac{3}{2} \\ 2 & \frac{5}{2} \end{array}\right) - \left(\begin{array}{cc} \frac{7}{4} & \frac{3}{2} \\ 2 & \frac{1}{4} \end{array}\right) \] Let's perform the matrix subtraction: \[ C = \left(\begin{array}{cc} 1 - \frac{7}{4} & -\frac{3}{2} - \frac{3}{2} \\ 2 - 2 & \frac{5}{2} - \frac{1}{4} \end{array}\right) \] Calculating each component: 1. For the first element: \[ 1 - \frac{7}{4} = \frac{4}{4} - \frac{7}{4} = -\frac{3}{4} \] 2. For the second element in the first row: \[ -\frac{3}{2} - \frac{3}{2} = -3 \] 3. For the first element in the second row (simplified): \[ 2 - 2 = 0 \] 4. For the second element in the second row: \[ \frac{5}{2} - \frac{1}{4} = \frac{10}{4} - \frac{1}{4} = \frac{9}{4} \] Thus, we obtain: \[ C = \left(\begin{array}{cc} -\frac{3}{4} & -3 \\ 0 & \frac{9}{4} \end{array}\right) \] From this, we identify: \[ a = -\frac{3}{4}, \quad b = -3, \quad c = 0, \quad d = \frac{9}{4} \] So the final values are: - \( a = -\frac{3}{4} \) - \( b = -3 \) - \( c = 0 \) - \( d = \frac{9}{4} \)