Solving Simultaneous Equations Using the Matrix Method: A Step-by-Step Guide
In this article, we will explore how to solve a system of simultaneous linear equations using the matrix method. The equations given are:
14x 18y 6 114y 2Step 1: Represent the Equations in Matrix Form
To solve this system, we first represent it in matrix form. The system can be expressed as: [ Amathbf{x} mathbf{b} ]where
[ A begin{bmatrix} 14 18 0 114 end{bmatrix} ] [ mathbf{x} begin{bmatrix} x y end{bmatrix} ] [ mathbf{b} begin{bmatrix} 6 2 end{bmatrix} ]Step 2: Find the Inverse of Matrix A
To find the values of (x) and (y), we need to find the inverse of matrix (A), denoted as (A^{-1}). For a 2x2 matrix [ A begin{bmatrix} a b c d end{bmatrix} ] the inverse is given by}A^{-1} frac{1}{ad - bc} begin{bmatrix} d -b -c a end{bmatrix} ]
For our matrix
[ A begin{bmatrix} 14 18 10 14 end{bmatrix} ]the elements are:
[ a 14, b 18, c 10, d 14 ]Step 3: Calculate the Determinant
The determinant of matrix (A) is: [ text{det}A ad - bc 14cdot 14 - 18cdot 10 196 - 180 16 ]Step 4: Calculate the Inverse of A
Now, we can find the inverse of (A):A^{-1} frac{1}{16} begin{bmatrix} 14 -18 -10 14 end{bmatrix} begin{bmatrix} frac{14}{16} -frac{18}{16} -frac{10}{16} frac{14}{16} end{bmatrix} begin{bmatrix} frac{7}{8} -frac{9}{8} -frac{5}{8} frac{7}{8} end{bmatrix} ]
Step 5: Multiply (A^{-1}) by (mathbf{b})
We now multiply (A^{-1}) by (mathbf{b}):A^{-1} mathbf{b} begin{bmatrix} frac{7}{8} -frac{9}{8} -frac{5}{8} frac{7}{8} end{bmatrix} begin{bmatrix} 6 2 end{bmatrix} ]
Calculating the components:
[ x frac{7}{8} cdot 6 - frac{9}{8} cdot 2 frac{42}{8} - frac{18}{8} frac{24}{8} 3 ] [ y -frac{5}{8} cdot 6 frac{7}{8} cdot 2 -frac{30}{8} frac{14}{8} -frac{16}{8} -2 ]