What's the value of a?
1 answer:
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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.<span>
Calculations
</span>⇒ <span>Rewrite the linear equations above as a matrix
</span>
![\left[\begin{array}{ccc}3&5&42\\6&5&54\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%265%2642%5C%5C6%265%2654%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
⇒ Apply to Row₂ : Row₂ - 2 <span>Row₁
</span>
![\left[\begin{array}{ccc}3&5&42\\0&-5&-30\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%265%2642%5C%5C0%26-5%26-30%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
⇒ <span>Simplify rows
</span>
![\left[\begin{array}{ccc}3&5&42\\0&1&6\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%265%2642%5C%5C0%261%266%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
Note: The matrix is now in echelon form.
<span>The steps below are for back substitution.
</span>
⇒ Apply to Row₁<span> : Row</span>₁<span> - </span>5 Row₂
![\left[\begin{array}{ccc}3&0&12\\0&1&6\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%260%2612%5C%5C0%261%266%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
⇒ <span>Simplify rows
</span>
![\left[\begin{array}{ccc}1&0&4\\0&1&6\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%264%5C%5C0%261%266%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
⇒ <span>Therefore,
</span>
<span>
</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 <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<span> : Row</span>₁<span> - </span>5 Row₂
⇒ <span>Simplify rows
</span>
⇒ <span>Therefore,
</span>
<span>
</span>