If $A = \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}$ then using $A^{-1}$, solve the following system of equations: $x - 2y = -1, 2x + y = 2$. |
$x = 3, y = 4$ $x = \frac{3}{5}, y = \frac{4}{5}$ $x = 1, y = 1$ $x = -1, y = 2$ |
$x = \frac{3}{5}, y = \frac{4}{5}$ |
The correct answer is Option (2) → $x = \frac{3}{5}, y = \frac{4}{5}$ ## $|A| = 5$ $\text{adj. } A = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}$ $A^{-1} = \frac{\text{adj. } A}{|A|} = \frac{1}{5} \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}$ Given system of equations is $AX = B$, where $A = \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix}, B = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$ $\begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$ $\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 2 \end{bmatrix}$ $\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{5} \begin{bmatrix} -1+4 \\ 2+2 \end{bmatrix} = \begin{bmatrix} 3/5 \\ 4/5 \end{bmatrix}$ $x = \frac{3}{5}$ and $y = \frac{4}{5}$ |