Find the inverse of the matrix `[{:(-3,2),(5,-3):}]` Hence, find the matrix `P` satisfying the matrix equation :
`P[{:(-3,2),(5,-3):}]=[{:(1,2),(2,-1):}]`

Answer

To solve the problem, we need to find the inverse of the matrix \( A = \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix} \) and then use this inverse to find the matrix \( P \) that satisfies the equation \( P A = B \), where \( B = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} \).

### Step 1: Find the determinant of matrix \( A \)

The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is calculated as:

\[ \text{det}(A) = ad - bc \]

For our matrix \( A \):

- \( a = -3 \)
- \( b = 2 \)
- \( c = 5 \)
- \( d = -3 \)

Calculating the determinant:

\[ \text{det}(A) = (-3)(-3) - (2)(5) = 9 - 10 = -1 \]

### Step 2: Find the adjoint of matrix \( A \)

The adjoint of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:

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

For our matrix \( A \):

\[ \text{adj}(A) = \begin{pmatrix} -3 & -2 \\ -5 & -3 \end{pmatrix} \]

### Step 3: Find the inverse of matrix \( A \)

The inverse of a matrix \( A \) is given by:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]

Substituting the values we found:

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

### Step 4: Find the matrix \( P \)

We know from the equation \( P A = B \) that:

\[ P = B A^{-1} \]

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

\[ B = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}, \quad A^{-1} = \begin{pmatrix} 3 & 2 \\ 5 & 3 \end{pmatrix} \]

Now we will calculate \( P \):

\[ P = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} 3 & 2 \\ 5 & 3 \end{pmatrix} \]

Calculating the elements of \( P \):

- First row, first column: \( 1 \cdot 3 + 2 \cdot 5 = 3 + 10 = 13 \)
- First row, second column: \( 1 \cdot 2 + 2 \cdot 3 = 2 + 6 = 8 \)
- Second row, first column: \( 2 \cdot 3 + (-1) \cdot 5 = 6 - 5 = 1 \)
- Second row, second column: \( 2 \cdot 2 + (-1) \cdot 3 = 4 - 3 = 1 \)

Thus, we have:

\[ P = \begin{pmatrix} 13 & 8 \\ 1 & 1 \end{pmatrix} \]

### Final Answer

The inverse of the matrix \( A \) is:

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

And the matrix \( P \) is:

\[ P = \begin{pmatrix} 13 & 8 \\ 1 & 1 \end{pmatrix} \]