Find the matrix X so that `X[(1,2),(5,3)]=[(5,10),(2,0)]`

Answer

To find the matrix \( X \) such that

\[ X \begin{pmatrix} 1 & 2 \\ 5 & 3 \end{pmatrix} = \begin{pmatrix} 5 & 10 \\ 2 & 0 \end{pmatrix}, \]

we can follow these steps:

### Step 1: Define the matrices
Let
\[ A = \begin{pmatrix} 1 & 2 \\ 5 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 5 & 10 \\ 2 & 0 \end{pmatrix}. \]

### Step 2: Find the inverse of matrix \( A \)
To find \( X \), we need to compute \( A^{-1} \). The formula for the inverse of a 2x2 matrix

\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

is given by

\[ \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. \]

For our matrix \( A \):
- \( a = 1, b = 2, c = 5, d = 3 \)

First, we calculate the determinant \( \text{det}(A) \):

\[ \text{det}(A) = ad - bc = (1)(3) - (2)(5) = 3 - 10 = -7. \]

Now, we can find the adjoint of \( A \):

\[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -5 & 1 \end{pmatrix}. \]

Thus, the inverse of \( A \) is:

\[ A^{-1} = \frac{1}{-7} \begin{pmatrix} 3 & -2 \\ -5 & 1 \end{pmatrix} = \begin{pmatrix} -\frac{3}{7} & \frac{2}{7} \\ \frac{5}{7} & -\frac{1}{7} \end{pmatrix}. \]

### Step 3: Calculate \( X \)
Now, we can find \( X \) using the formula:

\[ X = B A^{-1}. \]

Substituting the values of \( B \) and \( A^{-1} \):

\[ X = \begin{pmatrix} 5 & 10 \\ 2 & 0 \end{pmatrix} \begin{pmatrix} -\frac{3}{7} & \frac{2}{7} \\ \frac{5}{7} & -\frac{1}{7} \end{pmatrix}. \]

### Step 4: Perform the matrix multiplication
Now, we perform the multiplication:

1. For the first row, first column:

\[ 5 \cdot -\frac{3}{7} + 10 \cdot \frac{5}{7} = -\frac{15}{7} + \frac{50}{7} = \frac{35}{7} = 5. \]

2. For the first row, second column:

\[ 5 \cdot \frac{2}{7} + 10 \cdot -\frac{1}{7} = \frac{10}{7} - \frac{10}{7} = 0. \]

3. For the second row, first column:

\[ 2 \cdot -\frac{3}{7} + 0 \cdot \frac{5}{7} = -\frac{6}{7} + 0 = -\frac{6}{7}. \]

4. For the second row, second column:

\[ 2 \cdot \frac{2}{7} + 0 \cdot -\frac{1}{7} = \frac{4}{7} + 0 = \frac{4}{7}. \]

Putting it all together, we have:

\[ X = \begin{pmatrix} 5 & 0 \\ -\frac{6}{7} & \frac{4}{7} \end{pmatrix}. \]

### Final Result
Thus, the matrix \( X \) is

\[ X = \begin{pmatrix} 5 & 0 \\ -\frac{6}{7} & \frac{4}{7} \end{pmatrix}. \]

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