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

To find the matrix \( X \) such that \( 2X + 3A - 4B = I \), we first need to compute \( 3A \) and \( -4B \). Given: \[ A = \begin{bmatrix} 3 & -1 \\ 2 & -4 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} \] ### Step 1: Calculate \( 3A \) \[ 3A = 3 \begin{bmatrix} 3 & -1 \\ 2 & -4 \end{bmatrix} = \begin{bmatrix} 9 & -3 \\ 6 & -12 \end{bmatrix} \] ### Step 2: Calculate \( -4B \) \[ -4B = -4 \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} = \begin{bmatrix} -4 & -8 \\ 12 & 0 \end{bmatrix} \] ### Step 3: Calculate \( 3A - 4B \) Now we add \( 3A \) and \( -4B \): \[ 3A - 4B = \begin{bmatrix} 9 & -3 \\ 6 & -12 \end{bmatrix} + \begin{bmatrix} -4 & -8 \\ 12 & 0 \end{bmatrix} = \begin{bmatrix} 9 - 4 & -3 - 8 \\ 6 + 12 & -12 + 0 \end{bmatrix} = \begin{bmatrix} 5 & -11 \\ 18 & -12 \end{bmatrix} \] ### Step 4: Set up the equation Now we substitute back into the equation: \[ 2X + 3A - 4B = I \implies 2X + \begin{bmatrix} 5 & -11 \\ 18 & -12 \end{bmatrix} = I \] ### Step 5: Calculate the identity matrix \( I \) The identity matrix \( I \) for a \( 2 \times 2 \) matrix is: \[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] ### Step 6: Rearranging the equation Rearranging gives: \[ 2X = I - \begin{bmatrix} 5 & -11 \\ 18 & -12 \end{bmatrix} = \begin{bmatrix} 1 - 5 & 0 + 11 \\ 0 - 18 & 1 + 12 \end{bmatrix} = \begin{bmatrix} -4 & 11 \\ -18 & 13 \end{bmatrix} \] ### Step 7: Solve for \( X \) Now, divide by 2: \[ X = \frac{1}{2} \begin{bmatrix} -4 & 11 \\ -18 & 13 \end{bmatrix} = \begin{bmatrix} -2 & \frac{11}{2} \\ -9 & \frac{13}{2} \end{bmatrix} \] Thus, the matrix \( X \) is: \[ X = \begin{bmatrix} -2 & \frac{11}{2} \\ -9 & \frac{13}{2} \end{bmatrix} \]