Evaluate x and y if `[(x,2),(-3,y)]=2[(x^(2),1),(-(3)/(2),3y-5)]`.

Options

A. `x=0,y=2`
B. `x=1,y=2`
C. `x=-1,1,y=(5)/(3)`
D. `x=0,(1)/(2),y=2`

Explanation

To solve the equation given by the matrix equality \(\begin{pmatrix} x & 2 \\ -3 & y \end{pmatrix} = 2 \begin{pmatrix} x^2 & 1 \\ -\frac{3}{2} & 3y - 5 \end{pmatrix}\), we will follow these steps:

### Step 1: Expand the Right Side
First, we need to multiply the right side of the equation by 2:

\[ 2 \begin{pmatrix} x^2 & 1 \\ -\frac{3}{2} & 3y - 5 \end{pmatrix} = \begin{pmatrix} 2x^2 & 2 \\ -3 & 6y - 10 \end{pmatrix} \]

### Step 2: Set Up the Equation
Now we can set the two matrices equal to each other:

\[ \begin{pmatrix} x & 2 \\ -3 & y \end{pmatrix} = \begin{pmatrix} 2x^2 & 2 \\ -3 & 6y - 10 \end{pmatrix} \]

### Step 3: Equate Corresponding Entries
We can equate the corresponding entries of the matrices:

1. From the first row, first column:
\[ x = 2x^2 \quad (1) \]

2. From the first row, second column:
\[ 2 = 2 \quad (2) \]

3. From the second row, first column:
\[ -3 = -3 \quad (3) \]

4. From the second row, second column:
\[ y = 6y - 10 \quad (4) \]

### Step 4: Solve for \(x\)
From equation (1):
\[ x = 2x^2 \]
Rearranging gives:
\[ 2x^2 - x = 0 \]
Factoring out \(x\):
\[ x(2x - 1) = 0 \]
This gives us two possible solutions:
\[ x = 0 \quad \text{or} \quad 2x - 1 = 0 \Rightarrow x = \frac{1}{2} \]

### Step 5: Solve for \(y\)
From equation (4):
\[ y = 6y - 10 \]
Rearranging gives:
\[ y - 6y = -10 \Rightarrow -5y = -10 \Rightarrow y = \frac{-10}{-5} = 2 \]

### Final Solutions
Thus, we find:
- \(x = 0\) or \(x = \frac{1}{2}\)
- \(y = 2\)

### Summary of Solutions
The solutions are:
- \(x = 0\) or \(x = \frac{1}{2}\)
- \(y = 2\)