Given matrices A = \(\begin{bmatrix} 2 & 3 \\ 4 & 7 \end{bmatrix}\) and B = \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\), solve for X in the system of equations AX = B using the inverse matrix method.
Answer
Thus, X = \(\begin{bmatrix} 7 & -3 \\ -4 & 2 \end{bmatrix}\) \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\) = \(\begin{bmatrix} 7*1 - 3*2 \\ -4*1 + 2*2 \end{bmatrix}\) = \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\).
Step 1 ：1. First, we need to find the inverse of matrix A, denoted as A^{-1}. The formula for a 2x2 matrix is \(\frac{1}{ad - bc}\) \(\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\).
Step 2 ：So, the inverse of A, A^{-1}, is \(\frac{1}{2*7 - 3*4}\) \(\begin{bmatrix} 7 & -3 \\ -4 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 7 & -3 \\ -4 & 2 \end{bmatrix}\).
Step 3 ：2. We can solve for X by multiplying A^{-1} by B. Therefore, X = A^{-1}B.
Step 4 ：Thus, X = \(\begin{bmatrix} 7 & -3 \\ -4 & 2 \end{bmatrix}\) \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\) = \(\begin{bmatrix} 7*1 - 3*2 \\ -4*1 + 2*2 \end{bmatrix}\) = \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\).
