If X and Y are the matrices of order `2xx2` each and `2x-3y=[{:(-7,0),(7,-13):}] and 3x+2y =[{:(9,13),(4,13):}]` then what is Y equal to

Options

A. A. `[{:(1,3),(-2,1):}]`
B. B. `[{:(1,3),(2,1):}]`
C. C. `[{:(3,2),(-1,5):}]`
D. D. `[{:(3,2),(1,-5):}]`

Answer

To find the matrix \( Y \) given the equations \( 2X - 3Y = \begin{pmatrix} -7 & 0 \\ 7 & -13 \end{pmatrix} \) and \( 3X + 2Y = \begin{pmatrix} 9 & 13 \\ 4 & 13 \end{pmatrix} \), we can follow these steps:

### Step 1: Express \( X \) in terms of \( Y \) from the first equation.
Starting with the equation:
\[ 2X - 3Y = \begin{pmatrix} -7 & 0 \\ 7 & -13 \end{pmatrix} \]
We can rearrange this to express \( X \):
\[ 2X = \begin{pmatrix} -7 & 0 \\ 7 & -13 \end{pmatrix} + 3Y \]
\[ X = \frac{1}{2} \left( \begin{pmatrix} -7 & 0 \\ 7 & -13 \end{pmatrix} + 3Y \right) \]

### Step 2: Substitute \( X \) into the second equation.
Now we substitute \( X \) into the second equation:
\[ 3X + 2Y = \begin{pmatrix} 9 & 13 \\ 4 & 13 \end{pmatrix} \]
Substituting for \( X \):
\[ 3 \left( \frac{1}{2} \left( \begin{pmatrix} -7 & 0 \\ 7 & -13 \end{pmatrix} + 3Y \right) \right) + 2Y = \begin{pmatrix} 9 & 13 \\ 4 & 13 \end{pmatrix} \]
This simplifies to:
\[ \frac{3}{2} \left( \begin{pmatrix} -7 & 0 \\ 7 & -13 \end{pmatrix} + 3Y \right) + 2Y = \begin{pmatrix} 9 & 13 \\ 4 & 13 \end{pmatrix} \]

### Step 3: Distribute and combine like terms.
Distributing \( \frac{3}{2} \):
\[ \begin{pmatrix} -\frac{21}{2} & 0 \\ \frac{21}{2} & -\frac{39}{2} \end{pmatrix} + \frac{9}{2} Y + 2Y = \begin{pmatrix} 9 & 13 \\ 4 & 13 \end{pmatrix} \]
Now, convert \( 2Y \) to have a common denominator:
\[ 2Y = \frac{4}{2} Y \]
So, we have:
\[ \begin{pmatrix} -\frac{21}{2} & 0 \\ \frac{21}{2} & -\frac{39}{2} \end{pmatrix} + \left( \frac{9}{2} + \frac{4}{2} \right) Y = \begin{pmatrix} 9 & 13 \\ 4 & 13 \end{pmatrix} \]
This simplifies to:
\[ \begin{pmatrix} -\frac{21}{2} & 0 \\ \frac{21}{2} & -\frac{39}{2} \end{pmatrix} + \frac{13}{2} Y = \begin{pmatrix} 9 & 13 \\ 4 & 13 \end{pmatrix} \]

### Step 4: Isolate \( Y \).
Now, isolate \( Y \):
\[ \frac{13}{2} Y = \begin{pmatrix} 9 & 13 \\ 4 & 13 \end{pmatrix} - \begin{pmatrix} -\frac{21}{2} & 0 \\ \frac{21}{2} & -\frac{39}{2} \end{pmatrix} \]
Calculating the right-hand side:
\[ = \begin{pmatrix} 9 + \frac{21}{2} & 13 - 0 \\ 4 - \frac{21}{2} & 13 + \frac{39}{2} \end{pmatrix} \]
Calculating each element:
- For the first element: \( 9 + \frac{21}{2} = \frac{18}{2} + \frac{21}{2} = \frac{39}{2} \)
- For the second element: \( 13 = \frac{26}{2} \)
- For the third element: \( 4 - \frac{21}{2} = \frac{8}{2} - \frac{21}{2} = -\frac{13}{2} \)
- For the fourth element: \( 13 + \frac{39}{2} = \frac{26}{2} + \frac{39}{2} = \frac{65}{2} \)

Thus, we have:
\[ \frac{13}{2} Y = \begin{pmatrix} \frac{39}{2} & \frac{26}{2} \\ -\frac{13}{2} & \frac{65}{2} \end{pmatrix} \]

### Step 5: Solve for \( Y \).
Multiply both sides by \( \frac{2}{13} \):
\[ Y = \frac{2}{13} \begin{pmatrix} \frac{39}{2} & \frac{26}{2} \\ -\frac{13}{2} & \frac{65}{2} \end{pmatrix} \]
Calculating each element:
- For the first element: \( \frac{2}{13} \cdot \frac{39}{2} = \frac{39}{13} = 3 \)
- For the second element: \( \frac{2}{13} \cdot \frac{26}{2} = \frac{26}{13} = 2 \)
- For the third element: \( \frac{2}{13} \cdot -\frac{13}{2} = -1 \)
- For the fourth element: \( \frac{2}{13} \cdot \frac{65}{2} = 5 \)

Thus, we find:
\[ Y = \begin{pmatrix} 3 & 2 \\ -1 & 5 \end{pmatrix} \]

### Final Answer:
The matrix \( Y \) is:
\[ Y = \begin{pmatrix} 3 & 2 \\ -1 & 5 \end{pmatrix} \]