Solutions
In Matrix we use initially based on systems of linear equations.The matrix method is similar to the method of Elimination as but is a lot cleaner than the elimination method.Solving systems of equations by Matrix Method involves expressing the system of equations in form of a matrix and then reducing that matrix into what is known as Row Echelon Form.
Calculations
⇒ Rewrite the linear equations above as a matrix
\(\left[\begin{array}{ccc}3&5&42\\6&5&54\\\end{array}\right]\)
⇒ Apply to Row₂ : Row₂ - 2 Row₁
\(\left[\begin{array}{ccc}3&5&42\\0&-5&-30\\\end{array}\right]\)
⇒ Simplify rows
\(\left[\begin{array}{ccc}3&5&42\\0&1&6\\\end{array}\right]\)
Note: The matrix is now in echelon form.
The steps below are for back substitution.
⇒ Apply to Row₁ : Row₁ - 5 Row₂
\(\left[\begin{array}{ccc}3&0&12\\0&1&6\\\end{array}\right]\)
⇒ Simplify rows
\(\left[\begin{array}{ccc}1&0&4\\0&1&6\\\end{array}\right]\)
⇒ Therefore,
\(x=4
y = 6\)
