2. If \( A=\left[\begin{array}{cc}3 & -1 \\ 2 & -4\end{array}\right], B=\left[\begin{array}{cc}1 & 2 \\ -3 & 0\end{array}\right] \) then find the matrix ' \( X \) ' such that \( 2 X+3 A-4 B=I \) where, \( I \) is identity matrix. 3. If \( A=\left[\begin{array}{ccc}1 & 3 & 2 \\ -1 & 2 & 0 \\ 4 & 0 & 3\end{array}\right], B=\left[\begin{array}{ccc}1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 0 & 3\end{array}\right] \& C=\left[\begin{array}{ccc}2 & 1 & 2 \\ 2 & 2 & 1 \\ 1 & 2 & 2\end{array}\right] \) then find the matrix \( D \) such that \( 2 A-3 B-D=C \).

To solve the problems, we will work through each one step by step. ### Problem 2 We need to find the matrix \( X \) such that: \[ 2X + 3A - 4B = I \] where \( A = \begin{bmatrix} 3 & -1 \\ 2 & -4 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} \), and \( I \) is the identity matrix \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \). First, we calculate \( 3A \) and \( -4B \): 1. Calculate \( 3A \): \[ 3A = 3 \begin{bmatrix} 3 & -1 \\ 2 & -4 \end{bmatrix} = \begin{bmatrix} 9 & -3 \\ 6 & -12 \end{bmatrix} \] 2. Calculate \( -4B \): \[ -4B = -4 \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} = \begin{bmatrix} -4 & -8 \\ 12 & 0 \end{bmatrix} \] Now, we add \( 3A \) and \( -4B \): \[ 3A - 4B = \begin{bmatrix} 9 & -3 \\ 6 & -12 \end{bmatrix} + \begin{bmatrix} -4 & -8 \\ 12 & 0 \end{bmatrix} = \begin{bmatrix} 9 - 4 & -3 - 8 \\ 6 + 12 & -12 + 0 \end{bmatrix} = \begin{bmatrix} 5 & -11 \\ 18 & -12 \end{bmatrix} \] Now we substitute this back into the equation: \[ 2X + \begin{bmatrix} 5 & -11 \\ 18 & -12 \end{bmatrix} = I \] Subtract \( \begin{bmatrix} 5 & -11 \\ 18 & -12 \end{bmatrix} \) from both sides: \[ 2X = I - \begin{bmatrix} 5 & -11 \\ 18 & -12 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 5 & -11 \\ 18 & -12 \end{bmatrix} = \begin{bmatrix} 1 - 5 & 0 + 11 \\ 0 - 18 & 1 + 12 \end{bmatrix} = \begin{bmatrix} -4 & 11 \\ -18 & 13 \end{bmatrix} \] Now, divide by 2 to find \( X \): \[ X = \frac{1}{2} \begin{bmatrix} -4 & 11 \\ -18 & 13 \end{bmatrix} = \begin{bmatrix} -2 & \frac{11}{2} \\ -9 & \frac{13}{2} \end{bmatrix} \] Thus, the matrix \( X \) is: \[ X = \begin{bmatrix} -2 & \frac{11}{2} \\ -9 & \frac{13}{2} \end{bmatrix} \] ### Problem 3 We need to find the matrix \( D \) such that: \[ 2A - 3B - D = C \] where \( A = \begin{bmatrix} 1 & 3 & 2 \\ -1 & 2 & 0 \\ 4 & 0 & 3 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 0 & 3 \end{bmatrix} \), and \( C = \begin{bmatrix} 2 & 1 & 2 \\ 2 & 2 & 1 \\ 1 & 2 & 2 \end{bmatrix} \). First, we calculate \( 2A \) and \( -3B \): 1. Calculate \( 2A \): \[ 2A = 2 \begin{bmatrix} 1 & 3 & 2 \\ -1 & 2 & 0 \\ 4 & 0 & 3