Question

Solve the equation by inversion method, x + y=4, 2x - y = 5

Answer 1

x + y=4,

2x - y = 5

Write in matrix:

$\left(\begin{array}{ccc}1 \ \ \ \ 1 \\ 2 \ \ -1\end{array}\right) \left(\begin{array}{ccc}x \\ y \end{array}\right) = \left(\begin{array}{ccc}4 \\ 5 \end{array}\right)$

Find the determinant:

$\left|\begin{array}{ccc}1 \ \ \ \ 1 \\ 2 \ \ -1\end{array}\right| = (1)(-1) - (2)(1) = -3$

Find the inverse matrix:

$\text {Inverse of } \left(\begin{array}{ccc}1 \ \ \ \ 1 \\ 2 \ \ -1\end{array}\right) = (-3) \left(\begin{array}{ccc}-1 \ \ -1 \\ -2 \ \ \ \ 1\end{array}\right) =\left(\begin{array}{ccc}3 \ \ \ \ 3 \\ 6 \ \ -3\end{array}\right)$

Multiply the inverse matrix to both sides:

$\left(\begin{array}{ccc}1 \ \ \ \ 1 \\ 2 \ \ -1\end{array}\right)\left(\begin{array}{ccc}3 \ \ \ \ 3 \\ 6 \ \ -3\end{array}\right) \left(\begin{array}{ccc}x \\ y \end{array}\right) = \left(\begin{array}{ccc}4 \\ 5 \end{array}\right)\left(\begin{array}{ccc}3 \ \ \ \ 3 \\ 6 \ \ -3\end{array}\right)$

Left Hand Side (LHS):

$\left(\begin{array}{ccc}1 \ \ \ \ 1 \\ 2 \ \ -1\end{array}\right)\left(\begin{array}{ccc}3 \ \ \ \ 3 \\ 6 \ \ -3\end{array}\right) \left(\begin{array}{ccc}x \\ y \end{array}\right)$

$= \left(\begin{array}{ccc}(1)(3) + (1)(6)\ \ \ \ \ (1)(3) + (1)(-3) \\ 2(3) + (-1)(6) \ \ -2(3) + (-1)(-3) \end{array}\right)\left(\begin{array}{ccc}x \\ y \end{array}\right)$

$= \left(\begin{array}{ccc}9\ \ \ \ \ 0 \\ 0 \ \ -3 \end{array}\right)\left(\begin{array}{ccc}x \\ y \end{array}\right)$

$= \left(\begin{array}{ccc}9x \\ -3y \end{array}\right)$

Right Hand Side (RHS):

$\left(\begin{array}{ccc}4 \\ 5 \end{array}\right)\left(\begin{array}{ccc}3 \ \ \ \ 3 \\ 6 \ \ -3\end{array}\right)$

$= \left(\begin{array}{ccc}4(3) + 4(6) \\ 5(3) + 5(-3) \end{array}\right)$

$= \left(\begin{array}{ccc}36\\ 0 \end{array}\right)$

Solve x:

LHS = RHS

$\left(\begin{array}{ccc}9x \\ -3y \end{array}\right) = \left(\begin{array}{ccc}36\\ 0 \end{array}\right)$

$9x = 36$

$x = 4$

$-3y = 0$

$y = 0$

Answer: x = 4, y = 0