If \(A = \begin{bmatrix}3 & 2\\ 4 & 3\end{bmatrix}\), then \(A^{-1} = \begin{bmatrix}3 & -2\\ -4 & 3\end{bmatrix}\). Using this, the solution of the system \(3x + 2y = 7\), \(4x + 3y = 10\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(x = 1,\ y = 2\)
The system is AX = B with \(X = \begin{bmatrix}x\\ y\end{bmatrix}\) and \(B = \begin{bmatrix}7\\ 10\end{bmatrix}\). Since A is invertible, \(X = A^{-1}B\), with the inverse on the left of B.
\(\displaystyle X = \begin{bmatrix}3 & -2\\ -4 & 3\end{bmatrix}\begin{bmatrix}7\\ 10\end{bmatrix}\)
\(\displaystyle X = \begin{bmatrix}21 - 20\\ -28 + 30\end{bmatrix} = \begin{bmatrix}1\\ 2\end{bmatrix}\)
Check: \(3(1) + 2(2) = 7\) and \(4(1) + 3(2) = 10\).