\( A=\left[\begin{array}{ll}1 & 2 \\ 3 & 44\end{array}\right], \quad B=\left[\begin{array}{ll}5 & 6 \\ 7 & 8\end{array}\right] \quad \) Find \( 2 A-3 X=B \)

To solve the equation \( 2A - 3X = B \), we first need to calculate \( 2A \). Starting with the matrix \( A \): \[ A = \begin{bmatrix} 1 & 2 \\ 3 & 44 \end{bmatrix} \] We multiply every element by 2: \[ 2A = 2 \times \begin{bmatrix} 1 & 2 \\ 3 & 44 \end{bmatrix} = \begin{bmatrix} 2 \times 1 & 2 \times 2 \\ 2 \times 3 & 2 \times 44 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 6 & 88 \end{bmatrix} \] Next, we substitute \( 2A \) into the equation: \[ 2A - 3X = B \implies \begin{bmatrix} 2 & 4 \\ 6 & 88 \end{bmatrix} - 3X = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} \] Now, we isolate \( -3X \): \[ -3X = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} - \begin{bmatrix} 2 & 4 \\ 6 & 88 \end{bmatrix} \] Calculating the right side: \[ \begin{bmatrix} 5 - 2 & 6 - 4 \\ 7 - 6 & 8 - 88 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & -80 \end{bmatrix} \] So we have: \[ -3X = \begin{bmatrix} 3 & 2 \\ 1 & -80 \end{bmatrix} \] Multiply both sides by \(-\frac{1}{3}\) to isolate \( X \): \[ X = -\frac{1}{3} \begin{bmatrix} 3 & 2 \\ 1 & -80 \end{bmatrix} = \begin{bmatrix} -1 & -\frac{2}{3} \\ -\frac{1}{3} & \frac{80}{3} \end{bmatrix} \] Thus, the solution is: \[ X = \begin{bmatrix} -1 & -\frac{2}{3} \\ -\frac{1}{3} & \frac{80}{3} \end{bmatrix} \]