Question 2 (1 mark) Solve this matrix cquation for $$ X $$ $$ \left[\begin{array}{cc}2 & 2 \ 2 & -5\end{array} ight] X=\left[\begin{array}{cc}8 & 6 \ -13 & -1\end{array} ight] $$ A. $$ \left[\begin{array}{cc}-1 & -2 \ -3 & -1\end{array} ight] $$ B. $$ \left[\begin{array}{cc}1 & 2 \ 3 & 1\end{array} ight] $$ C. $$ \frac{1}{-14}\left[\begin{array}{cc}-14 & -28 \ -42 & -14\end{array} ight] $$ D. $$ \frac{1}{-14}\left[\begin{array}{cc}-14 & -28 \ -42 & -14\end{array} ight] $$ E. $$ \frac{1}{-14}\left[\begin{array}{cc}-52 & -4 \ 68 & 24\end{array} ight] $$ .

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Question 2 (1 mark) Solve this matrix cquation for $$ X $$ $$ \left[\begin{array}{cc}2 & 2 \ 2 & -5\end{array}ight] X=\left[\begin{array}{cc}8 & 6 \ -13 & -1\end{array}ight] $$ A. $$ \left[\begin{array}{cc}-1 & -2 \ -3 & -1\end{array}ight] $$ B. $$ \left[\begin{array}{cc}1 & 2 \ 3 & 1\end{array}ight] $$ C. $$ \frac{1}{-14}\left[\begin{array}{cc}-14 & -28 \ -42 & -14\end{array}ight] $$ D. $$ \frac{1}{-14}\left[\begin{array}{cc}-14 & -28 \ -42 & -14\end{array}ight] $$ E. $$ \frac{1}{-14}\left[\begin{array}{cc}-52 & -4 \ 68 & 24\end{array}ight] $$ .

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To solve the matrix equation $$ \left[\begin{array}{cc}2 & 2 \ 2 & -5\end{array}\right] X = \left[\begin{array}{cc}8 & 6 \ -13 & -1\end{array}\right] $$ for $$ X $$, we need to find the inverse of the coefficient matrix $$ \left[\begin{array}{cc}2 & 2 \ 2 & -5\end{array}\right] $$ and then multiply it by the constant matrix $$ \left[\begin{array}{cc}8 & 6 \ -13 & -1\end{array}\right] $$.

Let's denote the coefficient matrix as $$ A $$ and the constant matrix as $$ B $$. The equation can be written as $$ AX = B $$.

1. Find the inverse of matrix $$ A $$:
$$ A = \left[\begin{array}{cc}2 & 2 \ 2 & -5\end{array}\right] $$

2. Calculate the inverse of matrix $$ A $$:
$$ A^{-1} = \frac{1}{ad - bc} \left[\begin{array}{cc}d & -b \ -c & a\end{array}\right] $$
where $$ a = 2 $$, $$ b = 2 $$, $$ c = 2 $$, and $$ d = -5 $$.

3. Multiply the inverse of matrix $$ A $$ by matrix $$ B $$ to find matrix $$ X $$:
$$ X = A^{-1}B $$

Let's calculate the inverse of matrix $$ A $$ and then find matrix $$ X $$.
Matrices by following steps:
- step0: Find the matrix inverse:
  $$\left[\begin{array}{rr}{2}&{2}\{2}&{-5}\end{array}\right]$$
- step1: Evaluate the determinant:
  $$-14$$
- step2: Using the formula:
  $$\frac{1}{2\left(-5\right)-2\times 2}\times \left[\begin{array}{rr}{-5}&{-2}\{-2}&{2}\end{array}\right]$$
- step3: Evaluate:
  $$-\frac{1}{14}\times \left[\begin{array}{rr}{-5}&{-2}\{-2}&{2}\end{array}\right]$$
- step4: Multiply the terms:
  $$\left[\begin{array}{rr}{-\frac{1}{14}\left(-5\right)}&{-\frac{1}{14}\left(-2\right)}\{-\frac{1}{14}\left(-2\right)}&{-\frac{1}{14}\times 2}\end{array}\right]$$
- step5: Evaluate:
  $$\left[\begin{array}{rr}{\frac{5}{14}}&{\frac{1}{7}}\{\frac{1}{7}}&{-\frac{1}{7}}\end{array}\right]$$
The inverse of matrix $$ A $$ is:
$$ A^{-1} = \left[\begin{array}{rr}\frac{5}{14} & \frac{1}{7} \ \frac{1}{7} & -\frac{1}{7}\end{array}\right] $$

Now, we will multiply the inverse of matrix $$ A $$ by matrix $$ B $$ to find matrix $$ X $$.
Matrices by following steps:
- step0: Solution:
  $$\left[\begin{array}{rr}{\frac{5}{14}}&{\frac{1}{7}}\{\frac{1}{7}}&{-\frac{1}{7}}\end{array}\right]\times \left[\begin{array}{rr}{8}&{6}\{-13}&{-1}\end{array}\right]$$
- step1: Multiply the terms:
  $$\left[\begin{array}{rr}{\frac{5}{14}\times 8+\frac{1}{7}\left(-13\right)}&{\frac{5}{14}\times 6+\frac{1}{7}\left(-1\right)}\{\frac{1}{7}\times 8-\frac{1}{7}\left(-13\right)}&{\frac{1}{7}\times 6-\frac{1}{7}\left(-1\right)}\end{array}\right]$$
- step2: Evaluate:
  $$\left[\begin{array}{rr}{1}&{2}\{3}&{1}\end{array}\right]$$
The solution to the matrix equation $$ \left[\begin{array}{cc}2 & 2 \ 2 & -5\end{array}\right] X = \left[\begin{array}{cc}8 & 6 \ -13 & -1\end{array}\right] $$ is:
$$ X = \left[\begin{array}{cc}1 & 2 \ 3 & 1\end{array}\right] $$

Therefore, the correct answer is B. $$ \left[\begin{array}{cc}1 & 2 \ 3 & 1\end{array}\right] $$.
The solution for $$ X $$ is $$ \left[\begin{array}{cc}1 & 2 \ 3 & 1\end{array}\right] $$. Therefore, the correct answer is B.

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