Consider the following two systems.
(a) \begin{array}{ccc} 4 x – 2 y &=& -3 \\ x+ 5 y &=& 1 \end{array}
(b)
\begin{array}{ccc} 4 x – 2 y &=& 2 \\ x+ 5 y &=& 3 \end{array}
(i) Find the inverse of the (common) coefficient matrix of the two systems.
(ii) Find the solutions to the two systems by using the inverse, i.e. by evaluating A^{-1} B where B represents the right hand side (i.e.
B = \begin{array}{c} -3 \\ 1 \end{array} for system (ii) Find the solutions to the two systems by using the inverse, i.e. by evaluating A^{-1} B where B represents the right hand side (i.e. B = \begin{array}{c} -3 \\ 1 \end{array} for system (a) and B = \begin{array}{c} 2 \\ 3 \end{array} for system (b)).
and B = \begin{array}{c} 2 \\ 3 \end{array} for system (b)).
Can anybody help me solve this question? Thanks.
Best Answer
The coefficient matrix is $$\begin{bmatrix}4&-2\\1&5\end{bmatrix}$$ To find the inverse, use the formula $$A^{-1}={1\over {ad-bc}}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$ Then multiply this inverse matrix by the vectors to find the answers. (As in, solve $Ax=b$ by writing $x=A^{-1}b$.)