If `A = [[1,2],[3,-5]]` and `B=[[1,0],[0,2]]` and X is a matrix such that `A=BX`, then X=

Explanation

To solve for the matrix \( X \) given that \( A = BX \), where \( A = \begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \), we will follow these steps:

### Step 1: Write the equation
We start with the equation:
\[ A = BX \]

### Step 2: Multiply both sides by \( B^{-1} \)
To isolate \( X \), we multiply both sides of the equation by the inverse of \( B \):
\[ B^{-1}A = B^{-1}BX \]
Since \( B^{-1}B \) is the identity matrix \( I \), this simplifies to:
\[ B^{-1}A = X \]

### Step 3: Find \( B^{-1} \)
To find \( B^{-1} \), we first calculate the determinant of \( B \):
\[ \text{det}(B) = (1)(2) - (0)(0) = 2 \]
Next, we find the adjoint of \( B \). For a \( 2 \times 2 \) matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the adjoint is given by:
\[ \text{adj}(B) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]
Thus, for \( B = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \):
\[ \text{adj}(B) = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} \]
Now, we can find \( B^{-1} \) using the formula:
\[ B^{-1} = \frac{1}{\text{det}(B)} \text{adj}(B) = \frac{1}{2} \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \]

### Step 4: Substitute \( B^{-1} \) and \( A \) into the equation for \( X \)
Now we substitute \( B^{-1} \) and \( A \) into the equation \( X = B^{-1}A \):
\[ X = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix} \]

### Step 5: Perform the matrix multiplication
Now we perform the multiplication:
\[ X = \begin{bmatrix} (1)(1) + (0)(3) & (1)(2) + (0)(-5) \\ (0)(1) + \left(\frac{1}{2}\right)(3) & (0)(2) + \left(\frac{1}{2}\right)(-5) \end{bmatrix} \]
Calculating each element:
- First row, first column: \( 1 \)
- First row, second column: \( 2 \)
- Second row, first column: \( \frac{3}{2} \)
- Second row, second column: \( -\frac{5}{2} \)

Thus, we have:
\[ X = \begin{bmatrix} 1 & 2 \\ \frac{3}{2} & -\frac{5}{2} \end{bmatrix} \]

### Final Answer
Therefore, the matrix \( X \) is:
\[ X = \begin{bmatrix} 1 & 2 \\ \frac{3}{2} & -\frac{5}{2} \end{bmatrix} \]