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.<span>
Calculations </span>
⇒ <span>Rewrite the linear equations above as a matrix
</span>![\left[\begin{array}{ccc}7&-4&13\\2&-5&-4\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-4%2613%5C%5C2%26-5%26-4%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
⇒ Apply to Row₂ : Row₂ - 2/7 <span>Row₁
</span>
![\left[\begin{array}{ccc}7&-4&13\\0&-27/7&-54/7\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-4%2613%5C%5C0%26-27%2F7%26-54%2F7%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
⇒ <span>Simplify rows
</span>
![\left[\begin{array}{ccc}7&-4&13\\0&1&2\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-4%2613%5C%5C0%261%262%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%20)
Note: The matrix is now in echelon form.
<span>The steps below are for back substitution.
</span>
⇒ Apply to Row₁ : Row₁ + 4 Row₂
![\left[\begin{array}{ccc}7&0&21\\0&1&2\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%260%2621%5C%5C0%261%262%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
⇒ <span>Simplify rows
</span>
![\left[\begin{array}{ccc}1&0&3\\0&1&2\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%263%5C%5C0%261%262%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
⇒ <span>Therefore,
</span>
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.<span>
Calculations </span>
⇒ <span>Rewrite the linear equations above as a matrix
</span>
⇒ Apply to Row₂ : Row₂ - 2/7 <span>Row₁
</span>
⇒ <span>Simplify rows
</span>
Note: The matrix is now in echelon form.
<span>The steps below are for back substitution.
</span>
⇒ Apply to Row₁ : Row₁ + 4 Row₂
⇒ <span>Simplify rows
</span>
⇒ <span>Therefore,
</span>