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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 

[tex] \left[\begin{array}{ccc}7&-4&13\\2&-5&-4\\\end{array}\right] [/tex] 

⇒  Apply to Row₂ : Row₂ - 2/7 Row₁ 

[tex] \left[\begin{array}{ccc}7&-4&13\\0&-27/7&-54/7\\\end{array}\right] [/tex] 

⇒ Simplify rows 

[tex] \left[\begin{array}{ccc}7&-4&13\\0&1&2\\\end{array}\right] [/tex] 

Note: The matrix is now in echelon form.
The steps below are for back substitution. 

⇒ Apply to Row₁ : Row₁ + 4 Row₂ 

[tex] \left[\begin{array}{ccc}7&0&21\\0&1&2\\\end{array}\right] [/tex]

⇒ Simplify rows 

[tex] \left[\begin{array}{ccc}1&0&3\\0&1&2\\\end{array}\right] [/tex] 

⇒ Therefore, 

[tex]x=3 y=2[/tex]




dalendrk
[tex]\left\{\begin{array}{ccc}7x-4y=13\\2x-5y=-4\end{array}\right\\\begin{array}{ccc}(i)\\(ii)\end{array}\left[\begin{array}{ccc}7&-4\\2&-5\end{array}\right\left|\begin{array}{ccc}13\\-4\end{array}\right]\xrightarrow{(i):7}\left[\begin{array}{ccc}1&-\frac{4}{7}\\2&-5\end{array}\right\left|\begin{array}{ccc}\frac{13}{7}\\-4\end{array}\right]\xrightarrow{(ii)-2(i)}[/tex]
[tex]\left[\begin{array}{ccc}1&-\frac{4}{7}\\0&-\frac{27}{7}\end{array}\right\left|\begin{array}{ccc}\frac{13}{7}\\-\frac{54}{7}\end{array}\right]\xrightarrow{(ii)\cdot(-\frac{7}{27})}\\\\\left[\begin{array}{ccc}1&-\frac{4}{7}\\0&1\end{array}\right\left|\begin{array}{ccc}\frac{13}{7}\\2\end{array}\right]\xrightarrow{(i)+(ii)\cdot\frac{4}{7}}[/tex]
[tex]\left[\begin{array}{ccc}1&0\\0&1\end{array}\right\left|\begin{array}{ccc}3\\2\end{array}\right]\\\\Answer:\boxed{ \left \{ {{x=3} \atop {y=2}} \right. }\to\fbox{B.}[/tex]

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